Timeline for Why is a topology made up of 'open' sets?
Current License: CC BY-SA 2.5
39 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 9, 2017 at 10:10 | comment | added | Hua | From my understanding of this answer, the main trouble comes from the fact that we can't iterate over "an infinite family" of rulers and do the "matching". Practically it is indeed a problem but mathematically it isn't, is it ? Otherwise, why the definition of sigma-algebra adopts the notion of countable intersection, which is also infinite ? (I admit that countable is nicer than general infinite). Thinking this way, why the definition of topology doesn't adopt the countable intersection ? | |
| Feb 15, 2014 at 17:54 | comment | added | LSpice | @JBL, the kind of structure you describe seems to be a common weakening of a field of sets (en.wikipedia.org/wiki/Field_of_sets) and a topology. It probably has some name that only a set theorist could love …. | |
| Sep 11, 2012 at 1:03 | comment | added | Per Vognsen | 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. | |
| Jan 14, 2012 at 1:47 | comment | added | Toby Bartels | @Changwei: One should not bother saying anything about finiteness in the basic definition. The requirements are: if $U$ and $V$ are open, then $U \cap V$ is open; and the entire space is open. It is then a theorem (to be proved by induction) that the intersection of finitely many open sets is open. But it's the fault of whoever started thinking about this theorem that the natural numbers got introduced; the axioms (if phrased appropriately, which in this case is how they're usually phrased) say nothing about that. | |
| Jan 12, 2012 at 15:41 | comment | added | Kerry | I have a related naive question to ask, namely why we assume the intersection of (finitely) many open sets are open sets. I always feel that condition is ugly since "finite" corresponds more or less to natural numbers. I know this helps to simplify the situation in $\mathbb{R}^{n}$, but I cannot help asking if we can come up with something definition more natural than this one. | |
| Oct 9, 2011 at 4:00 | comment | added | Toby Bartels | @ Kaveh (rather late): No, such a topology need not be $T_0$. If it's not $T_0$, then there are two heights which are different but such that your rulers are incapable of telling you that they are different. Too bad for the people who rely on those rulers. | |
| Jun 27, 2011 at 14:15 | history | made wiki | Post Made Community Wiki | ||
| Dec 19, 2010 at 17:50 | comment | added | Peter LeFanu Lumsdaine | @Kevin, @gowers: “My point is that if you change all the open sets to closed one the analogy seems just as good, but does not work.” To me, that’s not a problem with this analogy! It’s not trying to answer “why are open sets better than closed sets?”; it’s aiming to answer “why are open sets—or, equivalently, closed sets—a natural notion?”. | |
| Jul 15, 2010 at 19:07 | comment | added | Kaveh | I think Kevin and Gowers have a point here. If I remember correctly, Vickers argument is a little bit different: We have rulers of arbitrary but finite precession. With these rulers, we can check if someone is less than 2m tall by finite amount of work (time), we only need to pick a ruler of high enough precision, but we cannot check if he is exactly 2m tall. Observability (affirmablity) in finite time leads to topology, but are not all topologies constructed in this way $T_0$? Computability is a(n even more) special case of topology | |
| Mar 30, 2010 at 18:25 | comment | added | Allison Smith | Rulers feel inherently open to me, too. If I'm trying to decide with certainty whether something is between 1 and 3 cm long (for example) and I have some sort of infinite zoom power, anything in the interval (1,3) will eventually seem clearly in it to me. Anything out of [1,3] will eventually seem clearly out of it. But I will never reach a level of zoom where I can be completely sure about the length of something 3 cm long. Sure within any given tolerance, yes. Completely sure, no. So how could I have a closed ruler? This makes Kevin's new analogy not work for me. | |
| Mar 26, 2010 at 9:37 | comment | added | gowers | I have no such scruples. I basically agree with Kevin and was trying to put his point in a different way. Here's yet another way of putting it. Open sets have a nice stability property. I think it doesn't really add anything to call them rulers instead, since one is then forced to distort one's intuitive picture of what a ruler does. Probably the best one can do is say that there is a more general notion of stability (roughly, one where if a statement is true then it is robustly true) and that in this context it is captured well by open sets. | |
| Mar 26, 2010 at 7:33 | comment | added | Kevin Buzzard | 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 | |
| Mar 26, 2010 at 4:34 | comment | added | Mio | @Justin: just read it, nice. Thanks. | |
| Mar 26, 2010 at 3:44 | comment | added | Justin Hilburn | @Mio: I wrote up an answer where I explain exactly why you can't make rulers be closed sets. | |
| Mar 26, 2010 at 3:05 | comment | added | Justin Hilburn | @Kevin, @Gowers: I wrote up a rigorous justification of why the rulers should be open in light of computability theory in another answer. | |
| Mar 25, 2010 at 18:14 | comment | added | Tony Huynh | @Kevin: I think the point is that we should think about the rulers as open sets, because the property of belonging to an open set is more 'stable' than the property belonging to a closed set. For example, it is better to have a (95cm, 105cm) ruler rather than a [95cm, 105cm] ruler because if someone is exactly 95cm tall, then the [95cm, 105cm] ruler will answer yes, which is undesirable because she is actually on the boundary. On the other hand, if the (95cm, 105cm) ruler says yes, then we are more sure of the answer because our set is disjoint from its boundary (open). | |
| Mar 25, 2010 at 17:23 | comment | added | Kevin Buzzard | @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. | |
| Mar 25, 2010 at 2:17 | comment | added | Justin Hilburn | Try this short paper that is much more rigorous (it should be easy to read for anyone who knows topology already): homepages.inf.ed.ac.uk/als/Teaching/MSfS/l3.ps | |
| Mar 25, 2010 at 2:10 | comment | added | Justin Hilburn | Here is a more rigorous source for these ideas: homepages.inf.ed.ac.uk/als/Teaching/MSfS/l3.ps Sigfpe also has two more detailed explanations on his blog: mathoverflow.net/questions/19152/… blog.sigfpe.com/2008/03/what-does-topology-have-to-do-with.html | |
| Mar 24, 2010 at 20:01 | comment | added | gowers | I think the important point is that there is an idealization going on here. A truly real-world ruler would have the property that some points are definitely in, some definitely out, and there's a fuzzy region in the middle. But an open set is defined to be one where if you're in then you're definitely in (whereas if you're on the boundary then it's very hard to tell that you're not in). We could just as well have closed rulers and define topology via closed sets. | |
| Mar 24, 2010 at 17:12 | comment | added | Dan Piponi | @Kevin Topology is a tool to model the situation where you have a system of sets, and a way of demonstrating membership of those sets, but no method of demonstrating non-membership (unless such a demonstration can be constructed from membership of other sets). That's it. I'm not sure how literally you should take the ruler analogy, but think of a topology as a collection of "one-sided" rulers that can be used to demonstrate a value lies in a set, but which can't be used to demonstrate points lie outside it. | |
| Mar 24, 2010 at 15:37 | comment | added | Kevin Buzzard | @Neel: your comments somehow precisely indicate what I don't understand about the analogy! One can perhaps prove that in your setting "verifiability" is a closed condition and non-verifiability is an open condition. That's fine! What I am trying to get to the bottom of is whether the answer above, which has a lot of votes, really does give a conceptually good explanation of the fact that our marked distinguished sets (the open sets) have the property that they're closed under arbitrary unions but only under finite intersections. What stops me switching open to closed and the analogy breaking | |
| Mar 24, 2010 at 14:35 | comment | added | Neel Krishnaswami | Dually, though, nontermination is a "refutable" property and termination is irrefutable. You can refute nontermination by running a program and seeing that it halts. You can't refute a claim a program terminates, because again you can never be sure that it doesn't need just a bit more time. (IOW, there are two ways of equipping a 2-point set with a Sierpinski topology.) | |
| Mar 24, 2010 at 14:31 | comment | added | Neel Krishnaswami | This analogy is a backport from computer science back to geometry, and a bit was lost in the translation. In CS, for open, read "verifiable", and for closed, read "non-verifiable". Termination of blackbox programs is a verifiable property: if someone gives you a program and tells you it halts, then if they're telling the truth, if you wait you'll eventually see the machine stop and know they told you the truth. Nontermination is non-verifiable: no matter how long we wait, we can never be sure that the program won't halt soon, and so we can't verify we were told the truth. | |
| Mar 24, 2010 at 13:30 | comment | added | Kevin Buzzard | 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. | |
| Mar 24, 2010 at 9:10 | comment | added | Neel Krishnaswami | One other thing about this explanation: if you let people break a ruler into two pieces to get two smaller rulers, then you can explain the axioms for uniform spaces (which I always found even more mysterious than the ones for topological spaces). | |
| Mar 24, 2010 at 8:05 | comment | added | Mio | I mean I get where I was wrong in my reasoning, but not why having the magical power of being able to deal with uncountably many sets of one sort but not the other is any less arbitrary than the customary definition of topological spaces. | |
| Mar 24, 2010 at 7:19 | comment | added | Justin Hilburn | The complement of an open set isn't necessarily open and only open sets correspond to measuring devices. Thus Merlin can't tell you whether an element is in a uncountable intersection of open sets by looking at the uncountable union of their complements. This is supposed to mirror the behavior of semi-decidable propositions in computer science and logic. A good reference is the paper Synthetic Topology of Data Types and Synthetic Spaces by Martin Escardo. | |
| Mar 24, 2010 at 3:37 | comment | added | Mio | Suppose I ask Merlin to tell me whether an element belongs to a union of a collection of sets, then he can give me a set containing that element in case of a positive answer, but I must also accept an empty collection of sets from him in case of a negative answer. The question of whether an element is in an intersection of a collection of sets is equivalent to me asking him to tell me whether it is in the union of their complements, and if he comes back with no sets then I know that the element belongs to the intersection. In both cases the collections can have any number of sets. | |
| Mar 24, 2010 at 2:33 | comment | added | JBL | Think of Merlin trying to convince Arthur that a certain length falls into a certain collection (some union or intersection). The question is, can Merlin do this? In the case of an arbitrary union, if Merlin is able to figure out for himself that it does, then he can easily convey a proof of this fact to Arthur. In the case of an arbitrary intersection, this is not true. Maybe your question really is, "What do we get if we change the axioms of topological spaces so that we only require finite unions of open sets to be open?" (I don't know the answer to this question.) | |
| Mar 24, 2010 at 1:59 | comment | added | Mio | Right, I was thinking about the time it may take to find the one out of uncountably many. I still don't understand why I shouldn't think about it that way when I read your explanation, but never mind... | |
| Mar 24, 2010 at 0:48 | comment | added | Dan Piponi | Interesting bit of trivia. As well as writing on the connections between logic, computation and topology, Steve Vickers wrote most of the ZX Spectrum ROM. | |
| Mar 24, 2010 at 0:43 | comment | added | Dan Piponi | Don't think of it in terms of how long it takes to find a demonstration, but how long the demonstration would be once you've found it. If $l$ lies in $\bigcup_i U_i$ it might be hard to find a single $U_i$ it lies in. But once you have, the demonstration that $l\in\bigcup_i U_i$ reduce to showing that $l\in U_i$. Actually, with a little work you can tweak this into a question of how long it takes to compute, but that's another story. | |
| Mar 24, 2010 at 0:40 | comment | added | JBL | No: you just need to show the one member of the union to which it belongs. | |
| Mar 23, 2010 at 23:49 | comment | added | Mio | Question for sigfpe (sorry, I don't have enough reputation to post comments, feel free to delete this). It can also take an infinite time to show that something belongs to a union of, i.e. one of, uncountably many sets, no? | |
| Mar 23, 2010 at 23:34 | comment | added | Neel Krishnaswami | I first saw this kind of explanation in Steve Vickers' Topology via Logic, which is an excellent little book. | |
| Mar 23, 2010 at 23:18 | comment | added | Dan Piponi | @Qiaochu I wish I knew who to credit with the idea. It's based on what I've picked up from reading a bunch of computer science literature where there is a deep connection between topology and computability. But it's one of those things that's "in the air" so to speak, rather than something I could say I'd read from a specific paper. | |
| Mar 23, 2010 at 23:12 | comment | added | Qiaochu Yuan | 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. | |
| Mar 23, 2010 at 23:04 | history | answered | Dan Piponi | CC BY-SA 2.5 |