324
$\begingroup$

I'm ashamed to admit it, but I don't think I've ever been able to genuinely motivate the definition of a topological space in an undergraduate course. Clearly, the definition distills the essence of many examples, but it's never been obvious to me how it came about, compared, for example, to the rather intuitive definition of a metric space. In some ways, the sparseness of the definition is startling as it tries to capture, apparently successfully, the barest notion of 'space' imaginable.

I can try to make this question more precise if necessary, but I'd prefer to leave it slightly vague, and hope that someone who has discussed this successfully in a first course, perhaps using a better understanding of history, might be able to help me out.

Added 24 March:

I'm grateful to everyone for their thoughtful answers so far. I'll have to think over them a bit before I can get a sense of the 'right' answer for myself. In the meanwhile, I thought I'd emphasize again the obvious fact that the standard concise definition has been tremendously successful. For example, when you classify two-manifolds with it, you get equivalence classes that agree exactly with intuition. Then in as divergent a direction as the study of equations over finite fields, there is the etale topology*, which explains very clearly surprising and intricate patterns in the behaviour of solution sets.

*If someone objects that the etale topology goes beyond the usual definition, I would argue that the logical essence is the same. It is notable that the standard definition admits such a generalization so naturally, whereas some of the others do not. (At least not in any obvious way.)

For those who haven't encountered one before, a Grothendieck topology just replaces subsets of a set $X$ by maps $$Y\rightarrow X.$$ The collection of maps that defines the topology on $X$ is required to satisfy some obvious axioms generalizing the usual ones.

Added 25 March:

I hope people aren't too annoyed if I admit I don't quite see a satisfactory answer yet. But thank you for all your efforts. Even though Sigfpe's answer is undoubtedly interesting, invoking the notion of measurement, even a fuzzy one, just doesn't seem to be the best approach. As Qiaochu has pointed out, a topological space is genuinely supposed to be more general than a metric space. If we leave aside the pedagogical issue for a moment and speak as working mathematicians, a general concept is most naturally justified in terms of its consequences. As pointed out earlier, topologies that have no trace of a metric interpretation have been consequential indeed.

When topologies were naturally generalized by Grothendieck, a good deal of emphasis was put on the notion of an open covering, and not just the open sets themselves. I wonder if this was true for Hausdorff as well. (Thanks for the historical information, Donu!) We can see the reason as we visualize a two-manifold. Any sufficiently fine open covering captures a combinatorial skeleton of the space by way of the intersections. Note that this is not true for a closed covering. In fact, I'm not sure what a sensible condition might be on a closed covering of a reasonable space that would allow us to compute homology with it. (Other than just saying they have to be the simplices of a triangulation. Which also reminds me to point out that homology can be computed for ordinary objects without any notion of topology.)

To summarize, a topology relates to analysis with its emphasis on functions and their continuity, and to metric geometry, with its measurements and distances. However, it also interpolates between these and something like combinatorial geometry, where continuous functions and measurements play very minor roles indeed.

For myself, I'm still confused.

Another afterthought: I see what I was trying to say above is that open sets in topology provide an abstract framework for describing local properties of functions. However, an open cover is also able to encode global properties of spaces. It seems the finite intersection property is important for this, but I'm not able to say for sure. And then, when I try to return to the pedagogical question with all this, I'm totally at a loss. There are very few basic concepts that trouble me as much in the classroom.

$\endgroup$
27
  • 23
    $\begingroup$ Do you find the Kuratowski closure axioms intuitive? If so, then the proof of equivalence between the Kuratowski closure axioms and the standard axioms is not hard. $\endgroup$ Mar 23, 2010 at 23:14
  • 7
    $\begingroup$ Maybe "environments" is what some of us call "neighborhoods" (and others of us call "neighbourhoods"). $\endgroup$ Mar 23, 2010 at 23:36
  • 8
    $\begingroup$ filters and nets are equally powerful logically, nets are useful because you can use your intuition of sequences in metric spaces (more or less), filters are useful because the statements about convergence become much shorter and prettier.. $\endgroup$
    – faridrb
    Mar 24, 2010 at 3:08
  • 11
    $\begingroup$ Regarding the effectiveness of the standard definition of "topology": I feel comfortable with its effectiveness in areas such as functional analysis and differential geometry. But I have never understood why the standard definitions of topology should be useful at all when working with finite fields and other discrete objects. Is there any way to motivate that for the non-expert? $\endgroup$
    – Deane Yang
    Mar 24, 2010 at 13:29
  • 116
    $\begingroup$ It's ironic that this question about open sets is now closed. $\endgroup$
    – KConrad
    Jun 18, 2014 at 21:38

34 Answers 34

290
$\begingroup$

Topology is the art of reasoning about imprecise measurements, in a sense I'll try to make precise.

In a perfect world you could imagine rulers that measure lengths exactly. If you wanted to prove that an object had a length of $l$ you could grab your ruler marked $l$, hold it up next to the object, and demonstrate that they are the same length.

In an imperfect world however you have rulers with tolerance. Associated to any ruler is a set $U$ with the property that if your length $l$ lies in $U$, the ruler can tell you it does. Call such a ruler $R_U$.

Given two rulers $R_U$ and $R_V$ you can easily prove a length lies in $U\cup V$. You just hold both rulers up to the length and the length is in $U\cup V$ if one or the other ruler shows a positive match. You can think of $R_{U\cup V}$ as being a kind of virtual ruler.

Similarly you can easily prove that a point lies in $U\cap V$ using two rulers.

If you have an infinite family of rulers, $R_{U_i}$, then you can also prove that a length lies in $\bigcup_i U_i$. The length must lie in one of the $U_i$ and you simply exhibit the ruler $R_{U_i}$ matching for the appropriate $i$.

But you can't always do the same for $\bigcap_i U_i$. To do so might require an infinitely long proof showing that all of the $R_{U_i}$ match your length.

A topology is a (generalised) set of rulers that fits this description.

Your notion of 'measurement' in whatever problem you have might not match the notion that the above description tries to capture. But to the extent that it does, topology will work as a way to reason about your problem.

$\endgroup$
37
  • 60
    $\begingroup$ Very nice explanation! I don't think I've ever seen anyone directly try to explain why open sets should be closed under arbitrary union but not arbitrary intersection. $\endgroup$ Mar 23, 2010 at 23:12
  • 61
    $\begingroup$ Here's something that confuses me about this answer. I'm supposed to be thinking that your 'tolerances' are open conditions, giving open sets. Let's say instead that tolerances are closed conditions (which is to be honest how I always interpreted them: 1m+-5cm would lead me to believe that 95cm is a valid answer for something that was supposed to be 1m). If tolerances were closed then shouldn't your heuristic show that I can take arbitrary intersections but only finite unions? In short: I don't really understand the justification of the infinite unions v intersections bit. $\endgroup$ Mar 24, 2010 at 13:30
  • 23
    $\begingroup$ @gowers: "We could just as well have closed rulers...". This is the point I'm trying to make. If we had closed rulers then does the answer above then give a reasonable heuristic justification of the false statement that an arbitrary union of closed sets is closed? This is precisely what I'm trying to get to the bottom of. Yes I agree that the answer is lovely. But I am not yet convinced that it genuinely gives a conceptual explanation of the axiom that a union of open sets is open, in the sense that it seems to give equal credence to the false statement that a union of closed sets is closed. $\endgroup$ Mar 25, 2010 at 17:23
  • 29
    $\begingroup$ Just because open rules are "more desirable" has, from my point of view, nothing to do with it. Let me try and ask my question again! We have an answer above, and Qiaochu's comment directly below is "this gives a conceptual explanation of why an arbitrary union of open sets is open". I say "oh no it doesn't" and no-one has yet convinced me otherwise. Just because it works, and is "stable"/"desirable" does not convince me. My point is that if you change all the open sets to closed one the analogy seems just as good, but does not work. I'm just repeating myself now though so I'llnotpostanymore $\endgroup$ Mar 26, 2010 at 7:33
  • 58
    $\begingroup$ Very belated comment: Here's a thought experiment to see why rulers should be open. Let's stipulate that no ruler can be exact. I claim this is incompatible with rulers being closed. E.g. the conjunction of the inexact closed rulers for 7 +/- 1 and 9 +/- 1 is an exact ruler for 8. You can certainly constrain the system of closed rulers to prevent such situations from arising. But it will happen invariably whenever a system of rulers for the real numbers is translation invariant. Open rulers don't have this problem. $\endgroup$ Sep 11, 2012 at 1:03
238
$\begingroup$

The textbook presentation of a topology as a collection of open sets is primarily an artefact of the preference for minimalism in the standard foundations of the basic structures of mathematics. This minimalism is a good thing when it comes to analysing or creating such structures, but gets in the way of motivating the foundational definitions of such structures, and can also cause conceptual difficulties when trying to generalise these structures.

An analogy is with Riemannian geometry. The standard, minimalist definition of a Riemannian manifold is a manifold $M$ together with a symmetric positive definite bilinear form $g$ - the metric tensor. There are of course many other important foundational concepts in Riemannian geometry, such as length, angle, volume, distance, isometries, the Levi-Civita connection, and curvature - but it just so happens that they can all be described in terms of the metric tensor $g$, so we omit the other concepts from the standard minimalist definition, viewing them as derived concepts instead. But from a conceptual point of view, it may be better to think of a Riemannian manifold as being an entire package of a half-dozen closely inter-related geometric structures, with the metric tensor merely being a canonical generating element of the package.

Similarly, a topology is really a package of several different structures: the notion of openness, the notion of closedness, the notion of neighbourhoods, the notion of convergence, the notion of continuity, the notion of a homeomorphism, the notion of a homotopy, and so forth. They are all important, and it is somewhat artificial to try to designate one of them as being more "fundamental" than the other. But the notion of openness happens to generate all the other notions, and has a particularly elegant and simple axiomatisation, so we have elected to make it the basis for the standard minimalist definition of a topology. But it is important to realise that this is by no means the only way to define a topology, and adopting a more package-oriented point of view can be preferable in some cases (for instance, when generalising the notion of a topology to more abstract structures, such as topoi, in which open sets no longer are the most convenient foundation to begin with).

$\endgroup$
3
  • 19
    $\begingroup$ The difference between the minimalist and 'packaged' version is exactly what Bill Farmer and I have called axiomatic theories and high-level theories (respectively). A minimalist 'basis' is useful when trying to connect theories (theory interpretations, aka homomorphisms) because less has to be proven, but that makes a really poor "working environment". Any mathematician using a theory (be it topology or Riemannian geometry) will automatically want to use the high-level version. When trying to mechanize mathematics, such issues are no longer philosophical! $\endgroup$ Jul 2, 2010 at 12:38
  • 9
    $\begingroup$ I really like this answer. Perhaps part of the so-called "mathematical maturity" is being able to recover the full package from the minimalist's definition. I'm not always able to do so. It took me a long time to realize the metric tensor really gives you a "package". I wish someone can tell me what the package the minimal definition of a scheme (as a ringed space $(X,\mathcal{O}_X)$ that is locally isomorphic to the spectrum of a ring) gives... Although I feel like I can handle it freely right now, I certainly don't know how to explain it to someone who is new to algebraic geometry. $\endgroup$
    – temp
    May 8, 2012 at 2:46
  • 5
    $\begingroup$ This great answer reminds me of several "unbiased" or "monadic" definitions of structures such as monoids or monoidal categories. The definition of a monoid "should" really introduce $n$-fold products for $n \in \mathbb{N}$, not just $0$-fold and $2$-fold products. Curiously, the monoid axioms are quite easy to state for $n$-fold products; this is even more true for the coherence axioms in the definition of a monoidal category via $n$-fold tensor products. $\endgroup$ Mar 30, 2015 at 5:22
159
$\begingroup$

It may seem hard to add a new answer to all this, but here's mine. How to motivate the open set garbage of topological spaces:

Answer: Don't.

There are many ideas in mathematics that can be easily derived from some real situation, and I would count approximation (ie limits), metric spaces, and neighbourhoods as among these. I think that it is quite easy to motivate the neighbourhood definition of topological spaces, for example, by considering real world examples of needing approximations that can't be controlled by metrics (for example, if you always need your approximations to be greater than the true value).

But one can take this line too far and try to motivate everything in mathematics from real-world situations and this, I think, misses a great opportunity to teach something that all students of mathematics need to learn: that when something is presented to you in a particular way, you don't have to accept that viewpoint but can choose a different one more suited to what you want to do.

We try to teach them this with bases of vector spaces: don't use the basis given, use one that makes the matrix look nice (diagonal if possible!).

So here, we can present topological spaces as sets with lots of declared neighbouhoods satisfying certain simple, intuitive rules. But they are hard to work with so instead we work with open sets (sets which are neighbourhoods of all their points) because it makes life easier.


I should qualify the above a little. It's written as a counterpoint to all the previous replies which try to justify open sets based on some intuition. I'm not saying that those are wrong - far from it - just that with something like this, one should think carefully about the message one is sending to the students about mathematics. At some point, they have to learn that mathematics strives to be clear and elegant rather than intuitive and vague, and it's a good idea to do this with an example like topological spaces where we are still close to the intuition, rather than something like function spaces where intuition often takes a hike.

$\endgroup$
18
  • 23
    $\begingroup$ +1000000 . $\endgroup$ Mar 25, 2010 at 1:58
  • 18
    $\begingroup$ Yay! I got a downvote! Fantastic. With something like this, I don't care whether or not people agree with me. I care only that they have thought about it. I hope that those who think that this is a bad idea do so because they have a better one (if so, please tell me! I like hearing of how others explain stuff.). With teaching styles, everyone needs to develop one that "fits" them best. But often it's hard to do because we don't have many examples of what is possible. So I throw out things like this to help people start thinking about different ideas. $\endgroup$ Mar 27, 2010 at 12:54
  • 24
    $\begingroup$ I don't want to be sarcastic but surely a high school student can answer you back that $(x+y)^2=x^2+y^2$ makes life a lot easier for him also. This argument is really weak, philosophically. $\endgroup$ Aug 21, 2010 at 21:24
  • 33
    $\begingroup$ There's easier, and then there's easier while remaining correct. There's a semi-fake Einstein quotation about this: ‘Everything should be made as simple as possible, but no simpler.’. $\endgroup$ Oct 9, 2011 at 3:46
  • 12
    $\begingroup$ I downvoted for the line starting with "one can take this line too far and try to motivate everything in mathematics from real-world situations". For me, all of mathematics is grounded in reality, and it is often only when I have worked out the simplest most childish connection to reality that I nstart to feel that I can understand some line of research. $\endgroup$ Jan 9, 2012 at 19:09
154
$\begingroup$

Here's one of my favorite axiomatizations of topology!

To make a set $X$ into a topological space, you introduce a relation, "touches," between the elements of $X$ and the subsets of $X$. This relation must have the following properties:

  1. No point touches the empty subset.
  2. If $x$ is an element of $A$, then $x$ touches $A$.
  3. If $x$ touches $A \cup B$, then $x$ touches $A$ or $x$ touches $B$.
  4. If $x$ touches $A$, and every element of $A$ touches $B$, then $x$ touches $B$.

Here, $x$ is an arbitrary element of $X$, and $A$ and $B$ are arbitrary subsets of $X$.

The first three axioms agree very well with my intuitive concept of "touching," and I find the fourth one at least tolerable, if not totally self-evident. If you leave out the fourth axiom, you get the definition of a pretopological space (a set with a Čech closure operator).

In Joshi's Introduction to General Topology, and in most of the literature, this kind of relation is called a nearness relation (page 114). I think one of the first papers on these things was "Nearness--A Better Approach to Continuity and Limits," by P. Cameron, J. G. Hocking, and S. A. Naimpally, which talks about nearness relations on metric spaces.

The definition of continuity in terms of open sets really puzzled me at first. I think the definition in terms of the nearness relation is much clearer!

Let $X$ and $Y$ be topological spaces. A continuous map from $X$ to $Y$ is a map $f$ with the property that if $x$ touches $A$, then $f(x)$ touches $f(A)$.

The definitions of convergence and Hausdorffness are quite pretty, in my opinion, and the definition of connectedness is very intuitive. (WARNING: I'm kinda rusty at this, so these definitions may not be correct.)

The sequence $(x_n)$ converges to the point $x$ if $x$ touches every subsequence of $(x_n)$.

The topological space $X$ is Hausdorff if for any two distinct points $w, x \in X$, there is a subset $A$ of $X$ such that $w$ doesn't touch $A$ and $x$ doesn't touch the complement of $A$.

The topological space $X$ is disconnected if it has two subsets $A$ and $B$, with $A \cup B = X$, such that no point of $A$ touches $B$ and no point of $B$ touches $A$.

$\endgroup$
17
  • 8
    $\begingroup$ this is just the same as kuratowski's closure operator. It is also confusing because there are also the Nearness spaces defined by Herrlich which use covers and are more general than topological spaces.. $\endgroup$
    – faridrb
    Mar 24, 2010 at 3:53
  • 12
    $\begingroup$ @jef: Yes, it's the same as the closure operator, but I find it way easier to motivate. Maybe I'm just weird? Thanks for mentioning Herrlich's nearness spaces---I'd never heard of them before! If anyone is interested in the definition, here's a nice reference: "On Nearness Space," by Seung On Lee and Eun Ai Choi (ccms.or.kr/data/pdfpaper/jcms8_1/8_1_19.pdf). $\endgroup$
    – Vectornaut
    Mar 24, 2010 at 22:51
  • 5
    $\begingroup$ @Vectornaut: Your definition of a continuous map is a nice formalization of informal “a map without breaks” or “a map which preserves infinitesimal distances”. a touches A ↔ a is infinitely close to A. Infinitesimal distance between points does not make sense, so we need to replace one of the points with a set. That was crucial. I went the same way, but directly from the Kuratowski's axioms. a touches A ↔ $a\in cl(A)$. Thank you very much for the references because I did not know even where to start my search or what name it is called. I wonder why point-set topology is not derived this way. $\endgroup$
    – beroal
    Feb 19, 2011 at 14:37
  • 5
    $\begingroup$ Having studied various non-standard topologies, especially ones relating to adeles and profinite groups, I find this axiomatization very useful. $\endgroup$ May 9, 2012 at 9:32
  • 7
    $\begingroup$ @JonaChristopherSahnwaldt, it's not silly at all! Using $\in$ as the "touches" relation gives the discrete topology. $\endgroup$
    – Vectornaut
    Jul 17, 2016 at 18:41
54
$\begingroup$

To me, the concept of an open set is a distillation and abstraction of the properties of open intervals (on the real line) that are critical to defining and working with a continuous function. In my opinion students should never be introduced to the abstract concepts of topology (notably, the concepts of open sets and compactness) unless they have already mastered analysis of functions on the real line and finite-dimensional vector spaces and understand thoroughly the role of open sets and compactness in those settings.

$\endgroup$
4
  • 1
    $\begingroup$ I am with you there Deane. $\endgroup$ Mar 23, 2010 at 23:17
  • $\begingroup$ Charlie, I voted for your answer, too! $\endgroup$
    – Deane Yang
    Mar 23, 2010 at 23:19
  • 26
    $\begingroup$ If this was really the rule, I never would have learned any analysis at all. In the context of analysis there were simply too many details floating around for me to see what was relevant and what wasn't. It made next to no sense to me until I learned a bit of topology, which set aside the parts that don't matter and let me focus on what was actually relevant. At that point I was able to go back and learn the analytic concepts quite easily, but without the chance to study general topology I don't think I could have. $\endgroup$ Feb 5, 2014 at 20:10
  • 32
    $\begingroup$ It's very hard to appreciate compactness as a concept in its own right in a context where it just means "closed and bounded." $\endgroup$ Feb 5, 2014 at 20:11
39
$\begingroup$

After reading the comments on Sigfpe's answer I realized that it would be useful to make a rigorous argument to explain why "rulers" or as like to call them "observable properties" should be open sets. In the process I'd like to explain how we can view general topology as an idealized version of computation by interpreting topological spaces as data types and continuous maps as computable functions.

Computationally an observable property $P$ of a data type $A$ corresponds to a semi-decision procedure. In other words a computable function $\chi_P: A \to Unit$ which returns the unique value $()$ of type $Unit$ if $a \in A$ has the property $P$ and runs forever otherwise. We can interpret $P$ as a subset of $A$ and $\chi_P$ as it's characteristic function. Clearly observable properties pull back under computable functions since if $f:B \to A$ is a computable function $\chi_P \circ f$ is a semi-decision procedure.

Let's translate this into topological language. If we interpret $A$ and $B$ as topological spaces and $f:B \to A$ as a continuous map we have that $f^{-1}(P)$ is observable if $P$ is. Thus it makes sense to interpret observable properties as open sets. We can make this correspondence more precise if we notice that every open set $P$ in $A$ corresponds to a map $\chi_P: A \to \mathbb{S}$ where $\mathbb{S}$ is the Sierpinski space. Thus in our translation $Unit$ corresponds to $\mathbb{S}$, the open point of $\mathbb{S}$ corresponds to $()$, and the closed point $\bot$ of $\mathbb{S}$ corresponds to nontermination.

Now a question remains: why did we choose to represent observable properties by open sets instead of closed sets? The answer lies in the way observable properties behave under intersection and union. Let $P$ and $Q$ be observable properties. The intersection $P \cap Q$ is an observable property we can write a semi-decision procedure $\chi_{P \cap Q}$ by running $\chi_P$ and $\chi_Q$ in succession. Similarly notice that $P \cup Q$ is observable since we can write a semi-decision procedure $\chi_{P \cup Q}$ that runs $\chi_P$ and $\chi_Q$ in parallel and outputs $()$ if one of $\chi_P$ and $\chi_Q$ does. If you have an infinite number of computers it is clear that you can generalize this construction to an infinite union $\bigcup_{i \in I} P_i$ by running all the $\chi_{P_i}$ in parallel. However this will not work for an infinite intersection $\bigcap_{n \in \mathbb{N}} P_n$ because if $\chi_{P_n}$ takes $n$ seconds to terminate, then even running all the $\chi_{P_n}$ in parallel will not help.

I can't help but list out a few other things to ponder in light of this dictionary:

$\endgroup$
13
  • 8
    $\begingroup$ I can't help expanding on my favourite briefly. The continuity of $\forall_X:(x\rightarrow\mathBB{S})\rightarrow\mathBB{S}$ for compact $X$ gives a way, due to Ulrich Berger, to exhaustively search certain infinite spaces in finite time. It's a beautiful meeting point of CS and topology. Some details here: math.andrej.com/2007/09/28/… $\endgroup$
    – Dan Piponi
    Mar 26, 2010 at 3:27
  • 2
    $\begingroup$ Thanks, I was under the impression that "rulers" or "imprecise measurements" requires a geometrical intuition, but from this post I see that it does not. I am completely happy with sigfpe's answer now. This was something I was trying to get at as well, except for this is much clearer. $\endgroup$
    – abcdxyz
    Mar 26, 2010 at 4:06
  • 5
    $\begingroup$ The list of properties for discrete, Hausdorff, and compact beg for another: that the map $ \exists _ X \colon ( X \to \mathbb S ) \to \mathbb S $ be continuous. For topological spaces, this is always true, but it need not be in ASD. (A space with this property is called overt, a term from constructive locale theory.) $\endgroup$ Jan 14, 2012 at 2:14
  • 3
    $\begingroup$ Oops, that URI up there should be paultaylor.eu/ASD since Taylor is a European individual and not an American university! $\endgroup$ Jan 14, 2012 at 2:15
  • 3
    $\begingroup$ @QiaochuYuan This is embarrassingly ancient, almost five years old now, but I only just saw your question about internal logic of category of compact Hausdorff spaces. My immediate reaction is to say: that category is a pretopos (ncatlab.org/nlab/show/pretopos) since it is extensive and Barr exact; Barr exactness is by monadicity over $Set$. Being a pretopos means one can enact finitary first-order logic inside of it. $\endgroup$
    – Todd Trimble
    Mar 1, 2015 at 16:50
33
$\begingroup$

One of the comments on the original post said that you can define a topology in terms of neighbourhoods. I'd like to amplify on that comment because it's the answer I favour too, if you want to do things as intuitively as possible. In fact, you can do it with basic open neighbourhoods, which is often nicer, for reasons I'll come to in a moment.

The first step would be to axiomatize the notion of a basic open neighbourhood. So it would consist of properties like that if N is a b.o.n. of x then x is an element of N, that if y is also an element of N then there is a b.o.n. N' of y such that N' is a subset of N (and in many systems one would be able to take N'=N), that the intersection of two b.o.n.s of x is another one, and so on. Suppose we've got all that sorted out. Then the rest of the definitions are just like metric space definitions without the need to reformulate those definitions in terms of open sets. To give the most important example, a function $f:X\to Y$ is continuous at x if and only if the following condition holds: for every b.o.n. M of f(x) there exists a b.o.n. N of x such that $f(N)\subset M$. Of course, in a metric space the basic open neighbourhoods of x are the open balls $B_\epsilon(x)$.

In the usual definition of continuity for maps between topological spaces, one never talks about continuity at a point, but it is perfectly possible and natural to do so, as the above shows.

Here's another example: a set F is closed if and only if for every x not in F you can find a basic open neighbourhood N of x that is disjoint from F. Oh, and I should have said that a set U is open if and only if for every x in U you can find a basic open neighbourhood N of x such that $N\subset U$.

The one thing you can't do is reformulate these definitions in terms of sequences, for the simple reason that the sequence reformulations do not generalize to topological spaces (unless you replace them by nets).

Added later: I've just seen some more of the comments on the original post. Much of what I have said is implicit in those comments, but perhaps it is useful to have it spelt out.

$\endgroup$
3
  • $\begingroup$ John Stillwell suggested in the preface to his wonderful-if highly unorthodox-textbook on the subject that to render topology more concrete, that continuity should be removed entirely as it's foundation and it should be expressed entirely in terms of finite elements like simplexes.I think that would really work only in low dimensional spaces that are locally Euclidean. My point is that alternatives to open and closed sets (and the more common "weaker" variant nieborhoods) have been proposed in the past. $\endgroup$ Mar 24, 2010 at 17:34
  • $\begingroup$ > Then the rest of the definitions are just like metric space definitions without the need to reformulate those definitions in terms of open sets. … [A] function $f : X \to Y$ is continuous at $x$ if and only if the following condition holds: for every b.o.n. $M$ of $f(x)$ there exists a b.o.n. $N$ of $x$ such that $f(N) \subset M$. I think that I'm missing something very important here. This seems like you have re-formulated the metric-space definitions, just saying ‘b.o.n.’ everywhere the usual definition would say ‘open containing’. What is the simplification? $\endgroup$
    – LSpice
    Mar 24, 2010 at 21:38
  • 10
    $\begingroup$ Here's an example. Suppose I define the product topology on X={0,1}^N. I won't tell you what an open set is. Instead I'll say that the b.o.n. B_n(x) is the set of all sequences that agree with x up to n. Then I can define the continuity at x of a map from X to X without ever having to say what an open set is. So basic open neighourhoods are playing a role similar to balls of radius epsilon in metric-space theory. To relate this definition to metric spaces I don't have to prove that a map between metric spaces is continuous if and only if the inverse image of an open set is open. $\endgroup$
    – gowers
    Mar 26, 2010 at 9:42
26
$\begingroup$

Not sure if anyone has mentioned this article, but Gregory Moore discussed the development of the notion of open sets vs other historical approaches, in the paper "The emergence of open sets, closed sets, and limit points in analysis and topology" in Historia Mathematica, no. 35, 2008, pages 220-241. Makes for an interesting read.

$\endgroup$
1
  • $\begingroup$ This looks like a very nice article. Thanks very much. $\endgroup$ Jun 28, 2011 at 7:13
25
$\begingroup$

This is an attempt to synthesize ideas that have appeared in other answers, for example sigfpe's and Tim Perutz's. Feel free to edit if you think the ideas can be better expressed.

The idea I want to back is that a topological space is an environment $X$ in which the notion of checking the truth of a statement locally makes sense. In the actual language of topological spaces, we want to be able to talk about statements which are true for a space $X$ if and only if they're true for every open set in an open cover of $X$, and the same should be true for every subspace of $X$. (For example, continuity and differentiability of a function both have this property.)

But whatever an open cover is, it should consist of elements chosen from a distinguished collection of subsets $\mathcal{P}$ of $X$ having certain properties. The empty set and $X$ should both be in $\mathcal{P}$ because checking a statement about $X$ is trivially equivalent to checking it on $X$ and on the empty set. $\mathcal{P}$ should be closed under arbitrary unions because a collection of open sets automatically forms an open cover of its union. $\mathcal{P}$ should be closed under binary intersections because one should be able to build an open cover of a subspace $S$ of $X$ by intersecting an open cover of $X$ with $S$, and if $S$ is itself open, an open cover of $S$ should be extendable to an open cover of $X$.

I don't think I've explained myself very well, though. I also wish I knew enough to say something about the relationship between topology and logic that the above seems to suggest. But one reason I like this perspective is that it suggests certain definitions naturally, such as the definition of compactness or of a manifold.


Some soapboxing: while I can see the pedagogical value of thinking about topological spaces as a natural generalization of metric spaces or even just of $\mathbb{R}$, I think the idea of a topological space is deeper than these roots suggest and I think Minhyong is looking for an answer that reflects this. In other words, I am of the opinion that the definition of a topological space is more natural than the definition of a metric space (or even of $\mathbb{R}$!), so one shouldn't use the latter to motivate the former. But this is just an opinion.

$\endgroup$
5
  • 1
    $\begingroup$ Re soapboxing: I also think the idea of a topological space must be deeper than "just" a generalization of metric spaces or of $\mathbb{R}$; this is one of the reasons for this previous question of mine: mathoverflow.net/questions/14634/… $\endgroup$ Mar 24, 2010 at 9:00
  • 1
    $\begingroup$ Qiaochu: I agree with your soapboxing comment. I've indicated my reasons somewhat in the recent edit. $\endgroup$ Mar 24, 2010 at 12:45
  • 1
    $\begingroup$ I think I am missing here why the empty set should be included. It is certainly intuitive that a property that has been checked on $X$ is true on $X$; but, if I were a student presented with this definition, then I'd wonder, I think not unreasonably, why I should check it on $\emptyset$ as well. $\endgroup$
    – LSpice
    Mar 24, 2010 at 21:33
  • 8
    $\begingroup$ I guess another answer is that "arbitrary union" includes the empty union. $\endgroup$ Mar 24, 2010 at 21:36
  • $\begingroup$ I think that it's a mistake to say ‘and on the empty set’. The justification that $\emptyset$ is open (which is given in Qiaochu's last comment) is entirely different from the justification that $X$ is open. $\endgroup$ Jan 14, 2012 at 2:06
24
$\begingroup$

I risk reviving a settled matter.

I aim these sketchy remarks at the expert teacher - not at the neophyte student.

A motivation for open sets in topology might begin with a critique of measurement. Though we often think of measurement in continuous terms, practical measurement really always comes down essentially to answering Boolean questions. Thus the real-valued distance function $d(x,y)$ on a metric space carries the same information as a Boolean-valued function $D(x,y,r)$ where $D(x,y,r)=1$ iff $d(x,y)\leq r$. The Dedekind construction of the reals reflects the sort of divide-and-conquer process of actual measurement, the sort of process that takes us from $D$ to $d$, but not in finite terms.

Now in a non-metric topological space you just allow yourself a richer set of questions than you can index with a variable $r$ running over the reals.

Indeed mathematicians often identity sets with properties - having such and such a property means belonging to the (sub)set of all elements (of a given set) that have that property. Then we can measure the proximity of two things by which properties (that we care about) they share.

At a technical level, the previous paragraph has this reflection: just as a metric allows you to embed a metric space in a product of copies of the positive reals, a topology allows you to embed a general space in a product of copies of the $2$-element Sierpiński space.

Now a student might reasonably challenge the appearance of Sierpiński space in this fundamental role: why the asymmetry? why have just one open point? if binary decisions lie at the heart of the story, why not take as fundamental the $2$-element space discrete?

I say the choice of Sierpiński space mirrors an aspect of practical life. For certain questions, observation may, on the negative side, supply a full refutation, but on the positive side only at best lend support and never full confirmation. For example, we may discover after careful observation that two quantities are not equal, but often we can only amass evidence that they are equal pending more precise measurement. Another example, when we witness a demise we learn that something wasn't eternal, but observation can never confirm that something is eternal.

This concept of decidability motives the axiomatic closure properties of open sets. Intuitively, an open property admits confirmation by a finite amount of evidence. An arbitrary disjunction of open properties (a union) gets confirmed by confirming any one of them and thus also requires only a finite amount of evidence. But a conjunction of open properties requires confirming them all, so we must limit ourselves, at least a priori, to finite conjunctions (intersections).

In summary, the Sierpiński space may seem like a curiosity, if not a monstrosity, but it captures the essence of topology. We have open sets because we care about continuous maps to Sierpiński space, whether consciously or not. We care about continuous maps to Sierpiński space because we care about properties whether they are decidable or not (in the sense of the intuitionists, not in the sense of Turing), i.e., whether or not they disconnect the universe of possibilities. Accepting Sierpiński space commits you to accepting the subspaces of its self-products. The real conceptual hurdle for the topological neophyte lies in contemplating the ubiquity of undecidable properties.

Grothendieck topologies fit very nicely into this point of view (as it leads to topos theory). In essence, Grothendieck challenges the doctrine of identifying properties with subsets. For Grothendieck, a given property may only become visible if one breaks a symmetry or observes some distinction between objects that previously seemed identical. Thus singling out a property might require taking a cover rather than only passing to a subset.

$\endgroup$
5
  • 4
    $\begingroup$ Great answer. I would only note that there is a large area of computer science known as Domain Theory that tries to make the notion of decidability a la Turing the same as the notion of decidability given by considering maps into the Sierpinski space. Here is a short primer on these ideas: homepages.inf.ed.ac.uk/als/Teaching/MSfS/l3.ps $\endgroup$ Dec 20, 2010 at 6:18
  • 4
    $\begingroup$ I believe the correct embedding theorem into powers of the Sierpinski space is the following: a topological space $X$ embeds into a product of copies of the Sierpinski space iff it is Kolmogorov (a.k.a. $T_0$). So the Sierpinski space does not quite have the universal role that you suggest. $\endgroup$ Dec 24, 2010 at 9:30
  • $\begingroup$ @PeteL.Clark it depends on whether you really care about non-$T_0$ spaces. For all I care, every space is Tychonoff (a.k.a. completely regular or $T_{3\frac{1}{2}}$). $\endgroup$ Jun 11, 2014 at 16:15
  • 2
    $\begingroup$ @David: I find your remark curious. If you only care about Tychonoff spaces, it follows that you don't care about the Sierpinski space. So why comment on it? $\endgroup$ Jun 22, 2014 at 22:31
  • $\begingroup$ In c. 1996 Ī discovered this formulation of general topology for myself and was astonished that it’s not found in textbooks. BTW, this space is also attributed to P.S. Alexandrov. $\endgroup$ Mar 22, 2015 at 15:52
22
$\begingroup$

In this answer I will combine ideas of sigfpe's answer, sigfpe's blog, the book by Vickers, Kevin's questions and Neel's answers adding nothing really new until the last four paragraphs, in which I'll attempt to settle things about the open vs. closed ruler affair.


DISCLAIMER: I see that some of us are answering a question that is complementary of the original, since we are trying to motivate the structure of a topology, instead of adressing the question of which of the many equivalent ways to define a topology should be used, which is what the question literally asks for. In the topology course that I attended, it was given to us in the first class as an exercise to prove that a topology can be defined by its open sets, its neighbourhoods, its closure operator or its interior operator. We later saw that it can also be stated in terms of convergence of nets. Having made clear these equivalence of languages, its okay that anyone chooses for each exposition the language that seems more convinient without further discussion. However, I will mantain my non-answer since many readers have found the non-question interesting.


Imagine there's a set X of things that have certain properties. For each subset of S there is the property of belonging to S, and in fact each property is the property of belonging to an adequate S. Also, there are ways to prove that things have properties.

Let T be the family of properties with the following trait: whenever a thing has the property, you can prove it. Let's call this properties affirmative (following Vickers).

For example, if you are a merchant, your products may have many properties but you only want to advertise exactly those properties that you can show. Or if you are a physicist, you may want to talk about properties that you can make evident by experiment. Or if you predicate mathematical properties about abstract objects, you may want to talk about things that you can prove.

It is clear that if an arbitrary family of properties is affirmative, the property of having at least one of the properties (think about the disjunction of the properties, or the union of the sets that satisfy them) is affirmative: if a thing has at least one of the properties, you can prove that it has at least one of the properties by proving that property that it has.

It is also clear that if there is a finite family of affirmative properties, the property of having all of them is affirmative. If a thing has all the properties, you produce proofs for each, one after the other (assuming that a finite concatenation of proofs is a proof).

For example, if we sell batteries, the property P(x)="x is rechargeable" can be proved by putting x in a charger until it is recharged, but the property Q(x)="x is ever-lasting" can't be proved. It's easy to see that the negation of an affirmative property is not necessarily an affirmative property.

Let's say that the open sets are the sets whose characteristic property is affirmative. We see that the family T of open sets satisfies the axioms of a topology on X. Let's confuse each property with the set of things that satisfy it (and open with affirmative, union with disjunction, etc.).

Interior, neighbourhood and closure: If a property P is not affirmative, we can derive an affirmative property in a canonical way: let Q(x)="x certainly satisfies P". That is, a thing will have the property Q if it can be proved that it has the property P. It is clear that Q is affirmative and implies P. Also, Q is the union of the open sets contained in P. Then, it is the interior of P, which is the set of points for which Q is a neighbourhood. A neighbourhood of a point x is a set such that it can be proved that x belongs to it. The closure of P is the set of things that can't be proved not to satisfy P.

Axioms of separation: If T is not T0, there are x, y that can't be distinguished by proofs and if it is not T1, there are x, y such that x can't be distinguished from y (we can think that they are apparently identical batteries, but x is built in such a way that it will never overheat. So if it overheats, then it's y, but if it doesn't, you can't tell).

Base of a topology: Consider a family of experiments performable over a set X of objects. For each experiment E we know a set S of objects of X over which it yields a positive result (nothing is assumed about the outcome over objects that do not belong to S). If you consider the properties that can be proved by a finite sequence of experiments, the sets S are affirmative and the topology generated by them is the family of all the affirmative properties.

Compactness: I don't know how to interpret it, but I think that some people know, and it would be nice if they posted it. (Searchable spaces?)

Measurements: A measurement in a set X is an experiment that can be performed on each element of X returning a result from a finite set of possible ones. It may be a function or not (it is not a function if there is at least one element for which the result is variable). The experiment is rendered useful if we know for each possible result r a set T_r of elements for which the experiment certainly renders r and/or a set F_r for which it certainly doesn't, so let's add this information to the definition of measurement. An example is the measurement of a length with a ruler. If the length corresponds exactly with a mark on the ruler, the experimenter will see it and inform it. If the length fits almost exactly, the experimenter may think that it fits a mark or may see that it doesn't. If the length clearly doesn't fit any mark (because he can see that it lies between two marks, or because the length is out of range), he will inform it. It is sufficient to study measurements that have only a positive outcome and a negative outcome, a set T for which the outcome is certainly positive and a set F for which the outcome is certainly negative.

Imprecise measurements on a metric space: If X is a metric space, we say that a measurement in X is imprecise if there isn't a sequence x_n contained in F that converges to a point x contained in T. Suppose that there is a set of imprecise measurements available to be performed on the metric space. Suppose that, at least, for each x in X we have experiments that reveal its identity with arbitrary precision, that is, for each e>0 there is an experiment that, when applied to a point y, yields positive if y=x and doesn't yield positive if d(y,x)>e. Combining these experiments we are allowed to prove things. What are the affirmative sets generated by this method of proof? Let S be a subset of X. If x is in the (metric) interior of S, then there is a ball of some radius e>0 centered at x and contained in S. It is easy to find an experiment that proves that x belongs to S. If x is in S but not in the interior (i.e, it is in the boundary), we don't have a procedure to prove that x is in S, since it would involve precise measurement. Therefore, the affirmative sets are those that coincide with its metric interior. So, the imprecise measurements of arbitrary precision induce the metric topology.

Experimental sciences: In an experimental science, you make a model that consists of a set of things that could conceivably happen, and then make a theory that states that the things that actually happen are the ones that have certain properties. Not all statements of this kind are completely meaningful, but only the refutative ones, that is, those that can be proved wrong if they are wrong. A statements is refutative iff its negation is affirmative. By applying the closure operator to a non refutative statement we obtain a statement that retains the same meaning of the original, and doesn't make any unmeaningful claim.

An example from classical physics: Assume that the space-time W is the product of Euclidean space and an affine real line (time). It can be given the structure of a four-dimensional real normed space. Newton's first law of motion states that all the events of the trajectory of a free particle are collinear in space-time. To prove it false, we must find a free particle that incides in three non-collinear events. This is an open condition predicated over the space W^3 of 3-uples of events, since a small perturbation of a counterexample is also a counterexample. Assuming that imprecise measurements of arbitrary precision can be made, it is an affirmative property. I think that classical physicists, by assuming that these kind of measurements can be done, give exact laws like Newton's an affirmative set of situations in which the law is proved false. I also suspect (but this has more philosophical/physical than mathematical sense) that the mathematical properties of space-time (i.e. that it is a normed space over an Archimedean field) are deduced from the kind of experiments that can be done on it, so there could be a vicious circle in this explanation.

$\endgroup$
14
  • 2
    $\begingroup$ Axioms of separability: It seems hard to tell the difference between your $ T _ 0 $ and your $ T _ 1 $. So to be precise: a space is $ T _ 0 $ iff, whenever $ x \ne y $, there is a verifiable property that can be proved of $ x $ but not of $ y $ or there is a verifiable property that can be proved of $ y $ but not of $ x $. The space is $ T _ 1 $ iff, whenever $ x \ne y $, there is a verifiable property that can be proved of $ x $ but not of $ y $ and there is a verifiable property that can be proved of $ y $ but not of $ x $. $\endgroup$ Jan 14, 2012 at 4:05
  • 2
    $\begingroup$ In Sierpiński space, there are two points, $ \top $ and $ \bot $, and one nontrivial verifiable property, which can be proved of $ \top $ but not of $ \bot $. So it is $ T _ 0 $. But there is no verifiable property that can be proved of $ \bot $ but not of $ \top $, so it is not $ T _ 1 $. We may take this as set of Turing machines modulo halting status, so $ \top $ is the (equivalence class of) machines that halt and $ \bot $ is the (equivalence class of) machines that run forever; the nontrivial verifiable property is that the machine halts. $\endgroup$ Jan 14, 2012 at 4:09
  • 2
    $\begingroup$ Another separation axiom: (not ‘separability’, I just noticed): a space is $ T _ 2 $ iff, whenever $ x \ne y $, there are two mutually contradictory verifiable properties, one of which holds of $ x $ and one of which holds of $ y $. Equivalently, in the product space, that $ x \ne y $ is itself verifiable. (In Sierpiński space, it's not: you'll never verify that $ \top \ne \bot $, even after you see $ \top $ halt.) The other separation axioms can't be stated in such elementary terms, I think. $\endgroup$ Jan 14, 2012 at 4:39
  • 1
    $\begingroup$ Compactness (as requested): If a verifiable property holds of at least one value, then this property of the property can be verified (by checking that value). If a property holds of all values, then this property of the property might be hard to verify (since a priori one would need to check all values). But for a compact space, this can be verified. Of course, a finite space is compact, since we can simply check each value in turn. $\endgroup$ Jan 14, 2012 at 4:48
  • 2
    $\begingroup$ Compactness: As far as affirmative properties are concerned, any disjunction of cases boils down to a finite one. $\endgroup$
    – user56097
    Oct 26, 2017 at 21:07
17
$\begingroup$

There are several interpretations of the original question, but one is, why focus on open sets rather than closed sets? I have an unusual answer.

Suppose you want to do constructive mathematics. (Don't ask me why, you just do.) So you abstract the properties of open and closed subsets from the real line. Then you see that open subsets are closed under arbitrary union but only finitary intersection, OK. Dually, you see that closed sets are closed under arbitrary intersection but … not under finitary union! For example, the union of $ [ 0 , 1 ] $ and $ [ 1 , 2 ] $ cannot be proved to be closed. (The closure of the union is $ [ 0 , 2 ] $, but to prove that the union itself is all of $ [ 0 , 2 ] $ requires the lesser limited principle of omniscience. Or less formally, there is no definite method to decide whether a number near $ 1 $ is in $ [ 0 , 1 ] $ or in $ [ 1 , 2 ] $.) So open sets are better behaved and naturally you prefer to axiomatise them.

But as you continue with constructive topology, more advanced things fail, such as the Tychonoff Theorem (which implies the axiom of choice and thus excluded middle). Then you learn that this stuff works in locale theory, so you abandon traditional topological spaces for locales. And here the duality between open and closed is restored; a locale's frame of opens can just as well be interpreted as a coframe of closeds, and only tradition tells us to do the first. (In the locale of real numbers, the union of the closed sublocales $ [ 0 , 1 ] $ and $ [ 1 , 2 ] $ is the closed sublocale $ [ 0 , 2 ] $, and the thing that you can't prove constructively is that every point in this union belongs to at least one of its addends.)

So this answer only works in a very unusual frame of mind: setting off down an unusual path but not going all the way.

$\endgroup$
0
15
$\begingroup$

Without specifying a precise answer to the question, I am surprised that there has been so little emphasis on continuity as the motivating concept for topology - topological spaces seem to me to have been designed, so to speak, to capture the notion of continuity in as much generality as seemed possible at the time, and particularly in non-metric contexts - and incidentally clarifying some proofs by throwing away the metric structure.

What can we recover from the epsilon-delta formulation of continuity if we don't allow measurement? It is possible that this question is more readily answered by reference to closed sets than to open ones.

Clearly the concept then takes off in all sorts of directions, where intuitions motivated by metrics are confounded (as mine was initially with the Zariski Topology).

$\endgroup$
11
$\begingroup$

Two platitudes:

(1) On a metric space, $\mathbb{R}$-valued functions which are continuous in the ($\epsilon$-$\delta$)-sense are the same as those for which the preimage of an open is open. So one can achieve the aim of discussing continuity by using open sets.

(2) The standard open sets in a metric space satisfy the axioms for a topology.

However, the open sets in a metric space satisfy many other properties too (Hausdorff, etc.).

So - as a former colleague of mine pointed out - to motivate our definition we ought to say why we can't reasonably drop one of the axioms for a topology - the intersection axiom, say. After all, our examples will still satisfy the axioms, and we'll still be able to prove some standard lemmas about spaces and continuous functions.

The answer, I think, is that continuity really ought to be local: a function is continuous if it's continuous when restricted to each of the sets making up an open cover. In proving this, we use both the union and intersection axioms.

$\endgroup$
5
  • $\begingroup$ Your answer is tempting me to say that I put a topology on a set X by choosing a subring of the ring of bounded real functions on X, and calling the functions in that subring continuous. If I went down this dangerous road, would I always end up with a unique topology on X? Would I have to be careful about which subring I chose? (The empty subring wouldn't work, because constant functions are always continuous, but that isn't much of a restriction.) I'd like to say I'm using the Banach-Stone theorem in reverse, but I'm not, because I can get non-Hausdorff topologies, like the indiscrete one. $\endgroup$
    – Vectornaut
    Mar 24, 2010 at 23:35
  • $\begingroup$ @Vectornaut. Surely maps to the reals can't be meaningful in studying general topological spaces (think, say, of affine varieties of positive characteristic...)? My point was rather that topological spaces are designed to facilitate a discussion of continuous maps, generalising classical continuous functions by asking that the preimage of opens are open. If we demand that continuity be determined locally, and that constant maps are continuous, the topological space axioms are essentially forced on us. $\endgroup$
    – Tim Perutz
    Mar 25, 2010 at 1:42
  • 2
    $\begingroup$ Regarding (1) of the answer above: I'm sorry, but I do not fully understand why the fact that a mapping between metric spaces is continuous if and only if the preimage of an open set is open should justify the predominace of open sets in topology vs. closed sets. It is also true that a mapping between metric spaces is continuous if and only if the preimage of a closed set is closed. So this does not yield a pedagogic motivation for using open sets in my opinion. $\endgroup$ Jun 15, 2010 at 21:53
  • 1
    $\begingroup$ The open sets determine and are determined by the closed sets, so one just has to make a choice about which to axiomatize. Axioms for one give complementary axioms for the other. My point was that the conventional axioms for open sets (or their complementary axioms for closed sets) appear to be minimal ones under which continuity is locally determined and constant functions are continuous. $\endgroup$
    – Tim Perutz
    Jun 15, 2010 at 22:36
  • $\begingroup$ Apologies for commenting on such an old post, and apologies if I'm making some silly error, but where is the intersection axiom used? If we have an open cover $\{U_a\}_{a\in A}$, a function $f:X\to Y$, and $f_a = f|_{U_a}: U_a \to Y$ is continuous, then for $x\in f^{-1}(V)$ (where $V$ open in $Y$), $x$ in some $U_a$ so $f_a^{-1}(V)$ is a neighborhood of $x$ in $f^{-1}(V)$. Thanks for any clarification you can provide. $\endgroup$
    – D.R.
    Nov 11, 2020 at 7:52
11
$\begingroup$

Topology can be defined directly, without open sets, as the study of "metric spaces without the metric", i.e., modulo metric deformations or homeomorphism. This matches reasonably well the intuition of a qualitative geometry, insensitive to stretching and bending.

One can then prove that the structure of open sets is a complete invariant (since we start from metrizable spaces), and one can observe, with some experience, that reasoning about the topology of metric spaces (i.e., proofs of properties that are invariant under deformation or homeomorphism) can be formulated directly in terms of this invariant. In other words, not only the topologically invariant maps but the constructions of those maps descend to the category of topological spaces in terms of open sets. This means that we can work natively in manifestly topologically invariant terms provided that the invariant thing --- the structure of open sets --- is taken as the object of study. This is a rare case of a total or near-total success of the Erlangen program, where thinking in terms of that which is invariant really suffices for the original purposes of the subject.

(I say near-total, but don't know of any example where topologically invariant properties of a metric spaces are most easily proved using one or more metrics.)

Once topology is set up in terms of open sets one can look at examples beyond the motivating intuition, such as Zariski topology, the long line or pathological spaces. As far as those extensions start to challenge the adequacy of the open-set formalism it is because they are based on phenomena different from the stretching and bending ideas abstracted from picturesque low-dimensional situations.

$\endgroup$
1
  • 4
    $\begingroup$ But of course there are other structures on a metric space that are complete topological invariants, for example the convergence structure (specifying which sequences converge to which points). Abstracting this gives the category of convergence spaces (well, sequential convergence spaces without some motivation to generalise from sequences to nets), of which $ Top $ is a full subcategory. So while this explanation is historically accurate, it doesn't explain why topological spaces precisely are so important (if they really are). $\endgroup$ Jan 14, 2012 at 5:11
9
$\begingroup$

Hello Minhyong. I think other people have already given several excellent answers about how to motivate topology, and I'm not sure I have much to add. But there are a couple of other parts to your question. I had the impression that the notion of a topological space was introduced by Hausdorff in his 1914 book on set theory (Mengenlehre). But my history is somewhat shaky, and it would be nice if someone else could confirm this. It certainly seems that by the early 20th century there was a radical shift in the way mathematics was done.

Jumping ahead to your added comment, I agree that a Grothendieck topology is a very natural extension of this idea. However, for Grothendieck this was secondary; the associated category of sheaves or topos was the important thing. Of course, you know that, but perhaps not everyone does.

$\endgroup$
9
$\begingroup$

I have really quite enjoyed reading this thread, although I must confess I don't quite understand all of it. It has been many years since I have studied topology of any sort, and not only am I rusty, my mind is aging, and not as agile as it was in my youth. Anyway, for what it's worth, here is my take on basic definitions of topology. My first introduction to formal topology (rather than the specialized versions of it in real and complex analysis) was in simple homotopy theory, and the textbook I used gave the open sets axiomatically as a starting point. Although I saw (with some difficulty) that this was a generalization of the properties of the real vector spaces I was used to, and although it was in an algebraic style, I struggled with it's non-intuitive presentation. I believe (although I am hardly an expert) that "nearness" is a better approach to topology. It makes sense, even to a non-mathematician. For example, the statement of continuity in such a setting is simplicity itself: f:D->R is continuous at a point x in D, when: x is near a set A implies f(x) is near f(A). Furthermore, the axiomatic presentation of neighborhoods is simpler than the axioms for open sets in one important respect, you only need to consider the meet (for sets, intersection) of two neighborhoods, and the (possibly partial) order of the collection of the neighborhoods (for sets, the natural ordering on the power set, containment).

In a metric space, which automatically comes with a rich topological structure, you can easily resolve these notions to the traditional definitions. You lose the purely geometric flavor of topology in the process, though. Topology is concerned with matters in the small, and in the large, the nature of what are defined to be neighborhoods to a large part determines how intricate the spatial structure is.

As far as the aesthetic of why unions can be infinite, but intersections have to be finite, I have 2 thoughts: first, open and closed sets are somewhat dual notions, it is entirely possible to begin with the notion of a closed set, in which case you can allow only finite unions, but infinite intersections. In fact, this approach makes more sense for practical applications, since our physical tools for dealing with calculations (and our brains) are in fact finite. The second thought I have, is that when you take the union of two sets, you always get something "bigger", but when you take the intersection, you may get a null result. It seems natural to restrict the basic study to finite intersections, because infinite intersections can behave qualitatively different than infinte unions (similar to how, in whole numbers, substraction isn't always possible, but addition is).

Getting extra stuff is quite common in mathematics, you extend a field, or create a semi-direct product, or consider generated objects. But all this "extra stuff" is rather meaningless without some core thing that has some intrinsic behavior. In topology, I believe nearness should be that core thing. Much of topology's development was motivated by the idea of getting a handle on what a limit (and convergence) ought to mean, and these are notions which have their roots in approximation.

So, naively, one could argue, in real (one-dimensional) analysis we use open intervals as the basic building block, since we are often concerned about local behavior on very small intervals, and using that to extend to larger sets. extending this to a collection of open sets, or to a collection of neighborhoods can be shown to be equivalent constructions; however, using open sets focuses more on the "boundarylessness" of these sets (and thus emphasizes the density of the real numbers), whereas using neighborhoods focuses more on the "localness" of these intervals, lending itself more easily to abstraction while keeping some intuitive spatial idea.

$\endgroup$
1
  • $\begingroup$ I agree that nieghborhoods gives quite a bit more flexibilty when working with more subtle aspects of topology and analysis then merely open sets.By the way,David-I'm 35 and applying to PHD programs.Age is just a number when it comes to the human mind as long as it's healthy. $\endgroup$ Jul 1, 2010 at 21:30
8
$\begingroup$

This may be a naive answer, but for me the concept captured by a topological space is when a point is infinitely close to a set. This happens when the point is in the closure of the set (or, equivalently, when every neighborhood of the point intersects the set). The definition by open sets may obscure this.

$\endgroup$
1
  • 2
    $\begingroup$ For the definition of "infinitely close" (or the way I would define "touches"), I prefer continuous functions: wherever the set goes, that point goes with it. (So a closed set is one that can go to a point without dragging more points with it--but that requires a tautological definition of continuity again.) $\endgroup$ Mar 25, 2010 at 0:20
8
$\begingroup$

I would say that topology is defined in terms of open sets and closed sets. I think it is motivated by two theorems from the Calculus. The Bolzano-Weierstrass theorem and the intermediate value theorem.

In its simplest form, the Bolzano-Weierstrass theorem says that an infinite subset of a closed bounded interval $[a,b]$ of the real numbers has a limit point. You find that limit point as follows. As there are infinitely many points in the set, then there are infinitely many in the left half or the right half of $[a,b]$. Say the left half. Divide that interval in half and there are infinitely many in one half or the other. Proceed this way to produce a sequence of closed intervals $I_{n+1}\subset I_n$ with the length of the $n$th interval equal to $\frac{b-a}{2^n}$. By Cantor's theorem $\cap_n I_n$ is nonempty, and the point in there is your limit point.

It didn't really make a difference if you broke the interval into two pieces or 10 pieces. This leads to the notion of compactness, by saying that every open cover has a finite subcover. The fact that the intersection of the closed intervals $I_n$ is nonempty is the complementary notion that a collection of closed subsets with the finite intersection property has nonempty intersection. The proof of the Bolzano-Weierstrass theorem leads you to think of open and closed sets.

A similar analysis of the proof of the intermediate value theorem leads likewise to open sets and closed sets.

Really, the concept of a topology was an incredible creative leap, that allowed people to take ideas from the Calculus and apply them in other places. Similar leaps to me, are the notion of sigma algebra, distribution (in the PDE sense), and the construction of homological algebra. :)

$\endgroup$
1
  • $\begingroup$ It's worth remembering that all of these notions have a long history: great minds wrestled with the notions for a long time before the present ones were settled on, for all reasons given in the many excellent answers above. The compactness notion I understand was particularly long in coming after the Heine-Borel theorem was originally formulated and proven. I seem to recall that modern topology wasn't settled on until the 1930s? $\endgroup$ Apr 21, 2011 at 2:54
8
$\begingroup$

I don't think that Grothendieck topologies should be viewed as analogous to ordinary topologies. It is true that a topology on a set induces various Grothendieck topologies on various categories, but so does every system of basic open neighborhoods. In my opinion, it is more fruitful to think of a Grothendieck topology as the analogue of a system of basic open neighborhoods and a topos as the analogue of a topological space.

Let me try to answer the following question, which may or may not be the question that is actually being asked: Why do we prefer topological spaces to systems of basic open neighborhoods, and topoi to Grothendieck topologies? I think that the answer has to do with morphisms.

To give a morphism of spaces with basic open neighborhoods, one must give a function that respects those neighborhoods. One can't, however, require that the pre-image of a basic open neighborhood be open. Instead, one has to require that each point contained in the pre-image of a basic open neighborhood have a basic open neighborhood inside the pre-image.

Not only is this definition more complicated than the one for topological spaces (and the extension to Grothendieck topologies is by no means obvious!), but there are multiple distinct but isomorphic systems of open neighborhoods on the same space. A topology is a maximal system of open neighborhoods in a given isomorphism class, which makes it a "best model" for a particular notion of nearness.

Another interpretation that makes this model appealing is that given a system of basic open neighborhoods, the open sets are the "local properties" (those that hold at a point if and only if they hold at all sufficiently nearby points). (If one believes a slogan like "there are two -1-categories: TRUE and FALSE" then open sets are "-1-sheaves"; this completes the formal analogy with Grothendieck topologies and their associated topoi, which are their associated categories of "0-sheaves".)

$\endgroup$
2
  • $\begingroup$ Dear Jonathan, following that logic, why do we study top. spaces instead of their sheaf toposes? $\endgroup$ Jul 1, 2010 at 23:29
  • 2
    $\begingroup$ Or, for that matter, why not study the higher topoi of stacks and higher stacks? If you're only interested in spaces, those theories add extra overhead without giving you anything new. If you're interested in something like the etale topology then you don't have the option of restricting your attention to topological spaces. $\endgroup$ Jul 2, 2010 at 0:39
7
$\begingroup$

At least for me, the first time I learned about a metric space, was to discuss when sequences converge. I am not a math-history buff, but the concept of metric seems to stem from the need to formalize the concept of convergence. However, everything about of convergence depends only on the topology the metric generates. Hence one only needs to understand the open sets.

Topological spaces generalize metric spaces in the sense that every metric space gives rise to one and all concepts of convergence are captured by this topological space. Even more, the category of metric spaces and continuous maps sits inside the category of topological spaces as a full subcategory (this is just saying that a map between metric spaces is continuous if and only if the preimage of an open is open). However, you may object, and justifiably, that you can embed metric spaces into several categories, so, why is topological spaces the right generalization?

There are many "practical answers", for instance, the wide range of examples of abstract spaces which are not metric spaces, e.g., Spec(R) of a ring. However, the core of the matter is that topological spaces correctly axiomatize the notion of convergence. What I mean by this is, a topological space is completely determined by the convergence of the ultrafilters on its underlying set. This is seem most notably for compact Hausdorff spaces; a topological space is compact Hausdorff if and only iff every ultrafilter has a unique limit, and in fact, compact Hausdorff spaces are precisely the algebras of the ultrafilter monad. However, we are interested in non-compact examples, since we study unbounded metric spaces for example- but even here we can have ultrafilters with no limit- so, a complete generalization should take this into account. Furthermore, the space Spec(R) is very often non-Hasudorff, which means, ultrafilters which have a limit point, may have more than one. So, to understand convergence is to understand the set of limits of each ultrafilter. If X is a space and BX is its set of ultrafilters, we get a map BX->P(X) which sends each ultrafilter to its set of limit points (possibly empty). This corresponds to a relation $R \subset X \times BX$, which made be seen as a map BX->X in the bicategory of sets and relations. More precisely, we get a "relational algebra" for the ultrafilter monad. The converse is true as well: the category of topological spaces is equivalent to the category of relational algebras for the ultrafilter monad. This a theorem of Barr. The upshot is, there is a bijection between topologies on a set and "convergence systems for ultrafilters" on that set.

Anyhow, this probably goes way beyond what you can explain to most undergraduates.

$\endgroup$
4
  • $\begingroup$ By my last sentence I mean, to undergraduates who have not yet seen what a topological space is! $\endgroup$ Mar 23, 2010 at 23:56
  • 1
    $\begingroup$ David: That's exactly right. What you write is interesting, but I think it's clear that an answer of the form `The axioms for a topological space are equivalent to these other ones.' is not quite satifactory. $\endgroup$ Mar 24, 2010 at 0:12
  • $\begingroup$ I like this answer. $\endgroup$
    – abcdxyz
    Mar 25, 2010 at 8:40
  • $\begingroup$ "the category of topological spaces is equivalent to the category of relational algebras for the ultrafilter monad." -- they are isomorphic, I guess. $\endgroup$ Mar 30, 2015 at 9:21
7
$\begingroup$

It is indeed appropriate to ask "why open sets in topology ?" However, the answer is not so simple precisely since the concept of topology proves to be far more deep and complex than Hausdorff, Kuratowski or Bourbaki have ever imagined, and which may call the HKB topology. For starters, it turned out decades ago that the usual, open set based, that is, HKB topology leads to a category which is not Cartesian closed. And this creates serious difficulties when dealing with topologies on function spaces, and in particular, in duality theory for locally convex spaces. More simply, and without categories, there are most important topological type processes in mathematics which simply cannot be described by HKB topologies. Such are, for instance, in measure theory and partially ordered spaces. As a consequence, various more general concepts of pseudo-topologies have been suggested. What happened, however, is that the doors thus opened up proved to be too large for those who tried to pursue them, or would have liked to use them ... In other words, topologies beyond HKB are a far less cozy venture than usually customary in mathematics ... Some details about the above and relevant references may be found in

arXiv:1001.1866 [pdf, ps, other] Title: Beyond Topologies, Part I Authors: Elemer E Rosinger, Jan Harm van der Walt Subjects: General Mathematics (math.GM)

arXiv:1005.1243 [pdf, ps, other] Title: Rigid and Non-Rigid Mathematical Theories: the Ring $\mathbb{Z}$ Is Nearly Rigid Authors: Elemer E. Rosinger Subjects: General Mathematics (math.GM)

$\endgroup$
7
$\begingroup$

Here's how I like to think about it.

We can all agree that a topological space should be a set $X$ together with some extra structure encoding how the points of $X$ fit together. It seems pretty reasonable ask that this structure is sophisticated enough to answer the following question whenever $x \in S \subset X$:

Given any choice of "direction" is there freedom to nudge $x$ some small "amount" in that direction without bumping into any points of $X \setminus S$?

We say that $x$ is an interior point of $S$ if the answer to the above question is "yes". I would say the following assertions about interior points are completely reasonable.

  1. Any $x \in X$ is an interior point of $X$.
  2. If $S \subset T \subset X$ and $x$ is an interior point of $S$, then $x$ is an interior point of $T$.
  3. If $S,T \subset X$ and $x$ is an interior point of both $S$ and $T$, then $x$ is an interior point of $S \cap T$.

For instance, (1) holds because there are no points in $X \setminus X$ to concern ourselves about bumping into. (3) holds because, if I specify a direction, then I can move $x$ an amount $a_s$ (in this direction) without hitting points from $X \setminus S$ and an amount $a_t$ without hitting points from $X \setminus T$, so if I move $x$ the smaller of these two amounts, I won't hit anything in $X \setminus (S \cap T)$.

If we take the above as axioms for a machine that tells us which points are interior to which sets, and then define an open set to be a set each of whose points is an interior point, then it is simple to recover the standard axioms for open sets:

  1. $\varnothing$ and $X$ are open.
  2. The union of arbitrarily many open sets is open.
  3. The intersection of two open sets is open.

The only issue I can see with this approach is that one might be able to convince oneself that interior points should satisfy more axioms. For instance, if $X = \{0,1\}$ and $1$ can be moved a little bit in any direction without bumping into $0$, then shouldn't it be possible to move $0$ a little bit in any direction without bumping into $1$? This would seem to preclude the existence of the Sierpiński topology $\{\varnothing, \{1\} ,X\}$. Or perhaps this is merely an invitation to be more imaginitive about the geometry of the situation? For instance, maybe there is a little round bowl with $1$ at the bottom and $0$ is sitting on the rim. If I give a $0$ a little push in the direction of $1$, no matter how small, $0$ will roll into the bowl and bump into $1$.

$\endgroup$
6
  • 1
    $\begingroup$ I don't think you have to worry about your example. It would simply be the situation where you can't wiggle 0 about for some specific example. In general you have to allow for this to happen in your view otherwise you have no idea which points are interior or not. I do like this way of motivating the axioms. $\endgroup$ Dec 3, 2011 at 16:08
  • 1
    $\begingroup$ I do like this. But notice that specifying an ‘interior point’ condition gives more information than specifying the open sets! You have (using also Axiom 0, that an interior point of $ S $ must be a member of $ S $) written down the definition of a pretopological space. It is more standard (or at least that's what I've seen) to say that $ S $ is a neighbourhood of $ x $ instead of that $ x $ is an interior point of $ S $. One reason is that there are two notions of interior, and that is the crux of the matter. $\endgroup$ Jan 14, 2012 at 5:57
  • $\begingroup$ Define the preinterior $ S ^ \circ $ of $ S $ to be the set of all points of which $ S $ is a neighbourhood. Call $ S $ open if its preinterior is all of $ S $ (which is how you defined this). Then let the interior $ I S $ of $ S $ be the union of all of the open subsets of $ S $. Nothing in your axioms proves that $ I S $ is all of $ S ^ \circ $. Instead we have $ S \supseteq S ^ \circ \supseteq S ^ { \circ \circ } \supseteq \cdots I S = I I S $. (In principle one could transfinitely iterate the preinterior, but I don't know how long this sequence can get or if it must converge.) $\endgroup$ Jan 14, 2012 at 6:08
  • 1
    $\begingroup$ To fix this, one may add the axiom of that a preinterior must be open, or in your terms, every interior point of $ S $ must be an interior point of the set of interior points of $ S $. But this is significantly more complicated than the other axioms, and suggests that topological spaces are just a special case of pretopological spaces, much as Hausdorff spaces are just a special case of topological spaces. $\endgroup$ Jan 14, 2012 at 6:13
  • 3
    $\begingroup$ @TobyBartels: The sequences of preinteriors (or dually preclosures) can become arbitrarly long. Consider any ordinal number $\alpha$ and define a closure operator by $A \mapsto [0,\sup(A)+1]\cap\alpha$. The sequence of closures of $\{0\}$ becomes stationary only at the $\alpha$-th step. And of course this is as bad as it gets: If $X$ has cardinality $\kappa$, then the chain will become stationary before the $\kappa^+$-th term. $\endgroup$ Apr 3, 2015 at 22:03
7
$\begingroup$

I have long understood that the initial ideas of topology arose from the notion of "neighbourhood" and were then found to be equivalent to the definition in terms of open sets. One advantage of the neighbourhood concept was that the definition of continuity using that is nearer to the $\varepsilon$-$\delta$ definition used in analysis.

The neighbourhood definition is more easily motivated than that in terms of open sets, but one then shows the equivalence. However one finds difficulties with the neighbourhood definition in defining, say, identification spaces, and this illustrates nicely a feature of mathematics, that equivalent concepts may have their best uses in different areas. Horses for courses!

Einstein wrote in 1915:

"Concepts which have proved useful for ordering things easily assume so great an authority over us, that we forget their terrestrial origin and accept them as unalterable facts. They then become labelled as conceptual necessities, a priori situations, etc. The road of scientific progress is frequently blocked for long periods by such errors. It is therefore not just an idle game to exercise our ability to analyse familiar concepts, and to demonstrate the conditions on which their justification and usefulness depend, and the way in which these developed, little by little... "

Thus Grothendieck in his 1984 "Esquisse d'un programme" Section 5, argues that the notion of topological space is motivated from analysis rather than geometry, and the latter requires spaces with more structure, in particular what he calls stratified spaces. I have found filtered spaces important in basic homotopical algebraic topology.

$\endgroup$
6
$\begingroup$

I want to complete something said by Andrew Stacey above: like him I think that the only reason to motivate the use of the open sets it's because they are more easy to use. Topology is the study of property preserved by invertible continuous transformation (following the Erlangen program): this definition clearly need the notion of continuity, I've always thought of continuity as the relation of proximity of points, so the first thing to do topology is to define the notion of proximity and neighbourhood are most natural way to do so (at least for me). Anyway dealing with neighbourhoods is more complex than working with open set, for example the definition of topology with neighbourhoods require five axioms while classical definition with open sets require just three axioms. So while it seems more natural the study of topology via neighbourhoods it is more convenient dealing with open sets which allow to simplify the proofs.

I hope this answer my help.

$\endgroup$
4
$\begingroup$

It might be a futile attempt to add anything worthwhile to this long list of interesting answers, but let me add my own pedestrian $0.02:

One may think of topology as a set of rules about what's close to what. In other words, it tells me that if I pick a point in the space, then there are several rules (i.e., open sets) that tell me that with respect to some question this set of points is "close" to my chosen point. Considering many rules (i.e., intersecting the open sets) gives me better and better approximation of which points are "really close" to the chosen one. It seems clear that then the union and intersection of these rules would have to belong to the rules.

If we were in a Euclidean space, then we might agree that one way to measure what's close is to put a small (open) ball around a point. If we can't measure, we can't do this, so we need to do something more general and a single open set will not be enough (it's not enough even in a Euclidean space as the radius of the ball that defines "closeness" would certainly depend on the way we want to measure closeness).

So far both open and closed sets would qualify for these rules, but I feel that open sets work better: A rule of "closeness" should be independent of any single point. In other words, a rule should behave the same with respect to any point it applies to (i.e., any point contained in the corresponding set). This clearly picks open sets over closed sets.

I suppose one might say that none of this explains what happens with infinitely many rules/sets. I suppose we could say that if we take an infinite set of rules that define closeness, then on one hand we might still say that satisfying any one of the rules is still a reasonable rule while satisfying all the rules is a little bit too much to ask. If you feel this part of my argument is a little shaky, then we agree. I don't have a very good explanation for the behavior of infinite unions and intersections. If I was indeed trying to explain this to undergrads, then at this point I would probably switch over to see what happens in a Euclidean space with all this non-sense about rules of "closeness" and come to the conclusion that a good way to define rules is to say that their corresponding sets contain little balls around every point in them. Then deduce the axioms of open sets in a topology and then say that we should see what these give us if we forget that we were in a Euclidean space.

$\endgroup$
3
$\begingroup$

Topological spaces are good abstract spaces to study limit and continuity, just as vector spaces are good abstract spaces to study linear combinations. Like many abstractions, proofs are studied in less abstract settings (e.g., $\mathbb{R}$) to see what makes them tick.

So why open sets?

Topology is defined in terms of open sets because that formulation was introduced (by Hausdorff?) at just the right time to become popular and drive out any competing formulations. There are quite a few equivalent formulations: closed sets; neighborhoods; operation of taking interiors; closure operation; predicate that says when a point is a limit point of a set; and so forth.

$\endgroup$
3
$\begingroup$

I'm going to take a finite, "combinatorial geometry" approach, thinking about, for instance, convex polytopes.

Suppose we have a finite set of things. For instance, a tetrahedron is made up of 4 points, 6 lines, 4 faces, and one solid. We notice, first, that some of these things are adjacent to each other, and some things are not adjacent to each other. So we have a graph.

Then we further note that this graph can be directed. If two things are adjacent, one of them must be bigger. If they have the same dimension, we have forgotten about the boundary that, in fact, separates them. This is a consequence of including, in some sense, all the shapes you can, to maximally clarify the geometry of your space.

So we have a directed graph. It is easy to verify that this graph should satisfy the axioms of a partially ordered set. If $A \geq B$ ($B$ is an edge of $A$) and $B \geq C$ ($C$ is a vertex of $B$) then $A\geq C$ ($C$ is a vertex of $A$).

Now, finite partially-ordered sets are just finite T0 topological spaces. The open sets contain everything $\geq$ their elements, and the closed sets contain everything $\leq$ their elements. It is not obvious in this context why these are natural objects of study, although they are fairly easy to define, and thus must be useful for something.

To determine the difference between finite and infinite union, we must break out of the finite world. We're going to do that, however, only by breaking it up into smaller finite pieces. A face might turn into 4 faces, 4 edges, and 1 vertex, for instance. A set on the unbroken space becomes a set on the broken space. Its closure is preserved, while the smallest open set containing it gets smaller.

Thus, though in the finite case we can consider the smallest closed set containing something or the smallest open set containing it, only the first notion is preserved as we increase the number of objects in our space and decrease their size, bringing us closer to infinite, continuous mathematics. Thus, we abandon the notion of a smallest open set containing something, which means we must abandon infinite intersection of open spaces, and therefore, infinite union of closed spaces.

$\endgroup$
2
$\begingroup$

First, note that a mapping between metric spaces is continuous if and only if the inverse image of an open set is always open. There are various concepts for metric spaces that you can likewise find equivalent formulations for in terms of open (and closed) sets, for example compactness. Convergence of a sequence to a point can be rephrased in terms of neighbourhoods of the point, with no reference to any ε. Then you could, for example, notice how you can talk about pointwise convergence of functions, but there is no corresponding metric. So you need a more general framework for talking about different kinds of convergence, and soon enough, topological spaces won't seem so strange anymore.

$\endgroup$
2
  • 1
    $\begingroup$ I'm sorry, but I do not fully understand why the fact that a "mapping between metric spaces is continuous if and only if the inverse image of an open set is open" should justify the predominace of open sets in topology vs. closed sets. It is also true that a mapping between metric spaces is continuous if and only if the inverse image of a closed set is closed. So this does not yield a pedagogic motivation for using open sets in my opinion. $\endgroup$ Jun 15, 2010 at 21:49
  • $\begingroup$ The short answer is that I did not read the original question as “why is a topology made up of open sets and not of closed sets?”. If that were the question, of course my answer would not be much help. But I don't think it was. $\endgroup$ Jun 15, 2010 at 22:56
2
$\begingroup$

Disclaimer: There are many topologists here and they may not like the philosophical flavour of my answer :-)

I think it all starts with the end... actually with the notion of "end". Take an open interval (a,b). It is bounded, yet you cannot reach its ends! At first this may look weird, but then one realizes this weirdness is to be attributed to the mathematically exact observation of this object (mathematicians can distinguish between so many things like point, set of points, boundedness, boundary etc.). Encountering such "weirdness" only shows the strong need for an exact and abstract formulation of "having no end". The lack of ends we then call "openness". If we are looking for the best generalization of this concept, the first approach would be of course a set-theoretical one. And in fact, in the case of intervals it turns out that the property "having no ends" is inhereted into arbitrary unions and finite intersections. Any attemps to expand this lead either to contradictions to our basic example or to unjustified reduction in generality.

$\endgroup$
4
  • $\begingroup$ I think that this is a little dangerous—it could make it difficult for students to understand later how, say, $[a, b]$ can be open in itself. I'd also have trouble motivating why, say, arbitrary unions, but only finite intersections, of ‘endless’ sets are again ‘endless’. $\endgroup$
    – LSpice
    Mar 24, 2010 at 21:40
  • 2
    $\begingroup$ I have intentionally left out the postulate that the empty set and the whole set are also defined to be open. It seems to me rather a technical axiom and maybe there is a way to do most of the point-set-topology without it... For arbitrary unions, it seems intuitive to me that expanding an "endless set" with other "endless things" should be "endless" too (the key point is expanding). As for intersections, (my) intuition reaches only so far as to say that if A and B are "endless", then $A\cap B$ is "endless", in other words "endless" is a "shareable" property. $\endgroup$
    – M.G.
    Mar 24, 2010 at 22:16
  • $\begingroup$ As always, the binary version of the statement does not imply in general the infinite version of the statement. In fact, since intersection is kind of refinement, it would be rather contra-intuitive to assume that the statement should be valid for infinite inttersections. But then again, of course, all this is pretty vague, and what might be intuitive for some, might look less intuitive or contra-intuitive or even wrong to others. $\endgroup$
    – M.G.
    Mar 24, 2010 at 22:21
  • $\begingroup$ In fact, if binary intersections exist, then all finite interesections of nonempty collections exists. But not neccessarily the whole space, i.e. the intersection of the empty collection. $\endgroup$
    – user2529
    Jan 10, 2013 at 15:28

Not the answer you're looking for? Browse other questions tagged or ask your own question.