My question is motivated by the previous discussion 'Why is a topology made of open sets?'. While the axioms for arbitrary unions and finite intersections are without doubt essential to the concept of a topological space, the 1st axiom (the set itself and the empty set are open) seems rather technical. So, do we really need these conditions in order to build most (if not all) of the point-set-topology without significant changes? In other words, if we leave out the 1st axiom, can point-set-topology stil remain as useful and powerful in terms of what we actually need to do analysis and geometry?

EDIT: Expanded the topic name in accordance with my question.

closed as no longer relevant by Pete L. Clark, Tom Leinster, Mariano Suárez-Álvarez, Harry Gindi, Yemon Choi Mar 26 '10 at 3:42

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    The empty set is the empty union, and the entire set is the empty intersection. In other words, from a categorical perspective we want to keep both of them to ensure that the obvious category built from open sets has an initial and terminal object. – Qiaochu Yuan Mar 24 '10 at 22:45
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    I think a better justification for using the empty set as an open set is one which is not about the empty set as a space in its own right, but rather one which refers to examples of open sets in spaces that people are really interested in: nonempty topological spaces (Euclidean space, etc.). The question is not about why the empty set is a topological space, but why the empty set should be an open subset of a topological space. Incidentally, the one-point set would be a final object. – KConrad Mar 24 '10 at 22:55
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    @KConrad: I think Qiaochu was talking about the category of open subsets of a fixed topological space with morphisms given by inclusions. Then the empty set is initial and the whole space is terminal. – Rasmus Mar 24 '10 at 23:09
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    Rasmus: I see your point, but still I don't think this is an issue which should be settled by category theory. It's sort of like justifying the zero ring because it serves as an initial/final object in the category of rings. While that is a "high-level" argument in favor of allowing the zero ring, there is a justification for the zero ring that makes sense at a simpler level: I would like to create a ring $R/I$ for any ideal $I$, and when $I = R$ I must allow the zero ring. OK, now I see where that is going: why let $R$ be an ideal in $R$? Oy. – KConrad Mar 24 '10 at 23:21
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    I have voted to close. Yes, you could dispense with saying that the total space and the empty subspace are open at the cost of having to say "except for the total space / empty subspace" in many definitions and constructions. There's nothing really lost, because we do not, after all, really have a choice in the matter: if we want a clean definition, then it is forced on us by the axioms. Otherwise we get a sloppy definition, but not one with any hidden flaws. So nothing essential is gained or lost either way (except cleanliness). Four answers is enough, I think. – Pete L. Clark Mar 25 '10 at 0:01
up vote 19 down vote accepted

Here's a boring reason, and it may or may not convince you: any function $f : X \to Y$ between topological spaces has the property that the preimage of the entire space $Y$ is the entire space $X$, and the preimage of the empty subset of $Y$ is the empty subset of $X$. So if you allow topological spaces in which either the entire set or the empty set is not open, there are no continuous functions from these spaces to "classical" topological spaces! Given that you agree with me that this is undesirable behavior, I think you are forced to make the entire set and/or the empty set either always open or always not open, and I think if you pick the second option then nothing changes except that, as KConrad says, it becomes unnecessarily harder to say things.

Actually, the situation is even worse: if the empty set isn't allowed to be open in $X$ then the continuous functions $X \to Y$ cannot miss any open set in $Y$, and if the entire set isn't allowed to be open in $X$ then the continuous functions $X \to Y$ cannot take values entirely in a proper open subset of $Y$. I think these are both much more unnatural than allowing the entire set and the empty set to be open. This is assuming you agree that the standard definition of continuity is natural.

  • These are indeed some significant changes if we leave out the 1st axiom. Thanks for pointing these out! I would not call them indesirable or unnatural though. It just has some different implications to continuous maps. – M.G. Mar 25 '10 at 15:10

If the empty set and the whole space are not open, then many statements you would like to make about open sets need qualifying remarks. It really can happen that two open sets are disjoint (two open balls that are far apart) or their union is the whole space (an appropriate pair of open half-planes that overlap). If the empty set were not open then we would have to say that any finite intersection of open sets is open or is empty. You'd have to tack on "or empty" in a lot of statements (e.g., the complement of a closed set is open or is empty... I assume you would like to call the whole space closed?). It is easier to allow the empty set as an open set to avoid a profusion of "or empty" qualifiers in theorems.

If, as has been suggested in a comment, the issue being raised is whether or not that first axiom about topologies is simply redundant, it isn't. Without that axiom we could consider any single subset of a space as a topology on the space: that one set is closed under arbitrary unions and finite intersections of itself. In that setting the concept of an open cover loses its meaning, so it really seems like a dead end.

Edit: Without the whole space being allowed as open (which can happen for "topologies" without that first axiom), there need not be open coverings, and then the usefulness of point-set topology is seriously damaged.

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    Similarly, it can happen that two open sets have union the entire set, and we don't want to say that the union of open sets is open or the entire set... – Qiaochu Yuan Mar 24 '10 at 22:55
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    I don't think he wants to require that the empty set is never open, but rather to lift the restriction that it must be open. For instance, in algebraic geometry - say the theory of algebraic curves - a lot of statements might become simpler if we said that the closed sets are exactly the finite sets of points. This would be consistent, since the intersection of any two nonempty open sets on an irreducible topological space is nonempty. – zeb Mar 24 '10 at 23:06
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    I also think this does not answer the question. You are just saying that for a space admitting two disjoint open sets, this axiom follows from the intersection axiom. So only irreducible topological spaces remain. For these, calling the empty set open or not more is a matter of conventions. But I guess this could change the notion of continuos function for non surjective functions. – Andrea Ferretti Mar 24 '10 at 23:18
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    The empty set is vacuously a finite set by any definition of finite... – Harry Gindi Mar 24 '10 at 23:19
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    Andrea, I was giving as an example one subset by itself. That would fit the conditions of a "topology" without the first axiom, but then the space doesn't admit an open covering in that "topology". It's hard to do much at all without open coverings. – KConrad Mar 24 '10 at 23:25

If one wants constant functions to always be continuous, then one must necessarily have the empty set and the whole space to be open.

From a category theory perspective, it is the continuous functions that are the more fundamental building block of topology, than the open or closed sets. I believe that there is some equivalent way to axiomatise topology via continuous functions using the machinery of sheaves, which is in some ways more "natural" than the simple but somewhat arbitrary-looking axioms for open sets, but I am not an expert on these matters.

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    Right. But as with Jonas Meyer's commment above, you can work around this in a silly way if you want to: say a map between topological spaces is continuous iff the preimage of every open set is either open, or the empty set, or the whole space. (Note that you don't need to also check the preimages of the empty set and the whole space: there can be no surprises there.) – Pete L. Clark Mar 25 '10 at 3:27
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    The first sentence is not 100% accurate; any function into a space Y is continuous if Y has no open sets! – Qiaochu Yuan Mar 25 '10 at 5:35
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    It's not clear to me how much of a problem it is sometimes to have discontinuous constant functions, if the openness of the empty set follows whenever you have two disjoint open sets (and the openness of the whole space follows if every point has a neighbourhood). One could then define a space to be T_{-3} if the empty set and the whole space happen to be open and comment that all reasonable spaces are T_{-3}. But non-T_{-3} spaces are just too silly to be worth considering, so one doesn't do this. – gowers Mar 25 '10 at 9:25

I see a little more in the question now. It seems that the OP is proposing to eliminate the axioms that the empty space and the total space are open but maintain the axioms that arbitrary (nonempty!) unions and finite (nonempty!) intersections of open sets are open.

In this case there is a little content here, because you can try to figure out whether or not $\emptyset$ and $X$ will then be open or not.

Note one disturbing fact: with the elimination of the above axiom, there is nothing to imply the existence of any open sets at all! Whether this is good or bad, if it happens there is nothing further to say, so let's assume that there is at least one open set.

Claim: If $X$ is a Hausdorff with more than one point, then the empty set and the total space are open.

Proof: Indeed, the Hausdorff axiom asserts that any two points have disjoint open neighborhooods, so the intersection of these is empty. The same axiom says in particular that every point has at least one open neighborhood (!) which is clearly equivalent to $X$ being open.

Claim: If $X$ is T1 with more than one point, then the total space is open, but the empty set need not be.

Proof: T1 means that the singleton sets are closed. If there are least two of them, take their intersection: this makes the empty set closed, hence the total space open. On the other hand, the cofinite topology on an infinite set is T1 and the empty set is not open (if we do not force it to be).

I am coming around to agree with K. Conrad in that having $X$ be open in itself may not be just a formality. A topological space in which some point has no neighborhood sounds like trouble...I guess I was thinking that if you get into real trouble, you can just throw $X$ back in as an open set! If you want to put it that way, this is some kind of completion functor from the OP's generalized topological spaces to honest topological spaces.

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    Anything you can say in topology applies only to the union of all of the open sets in X. If there are points in X not in this union, they might as well not be there; in other words, I see no conceivable reason to exclude the entire set from being open. – Qiaochu Yuan Mar 25 '10 at 1:38
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    I just noticed something amusing in the above rather trivial claims. If you take T_1 to mean (as I did) that points are closed, then in the above world of non-automatic openness of $\emptyset$ and $X$, Hausdorff does not imply T_1: consider a one point space with non-open empty set! – Pete L. Clark Mar 25 '10 at 3:30

This question may be better answered by going to the intuition.

Let X be a set with some sort of "local structure"; a metric space will do, but any reasonable sort of "closeness" is fine. We want to say a set UX is open if, whenever U contains a point p, it also contains all those points qX which are "near p".

Given this, U = X is clearly open. If pU, we want all the points of X "near p" to be in U; but every point of X is in U, so this is trivially true.

∅ is also clearly open. For this to be false, ∅ would have to contain a point p but not all the points near p; but as ∅ contains no points, this is impossible.

Also consider the following: Let f : X → {0,1} be the constant function sending every pX to 0. We want constant functions to be continuous, no matter how fine the topology on {0,1}. So f -1[{0}] = X and f -1[{1}] = ∅ both must be open (and both must be closed).

To address the question in a somewhat less-categorical way, I would point out that you can in fact do all of topology without using the expression "open set", by instead refering to the filter of neighborhoods of every point --- then a function $f:X\rightarrow Y$ is continuous iff for all $x\in X$ and every neighborhood $V$ of $f(x)$ there is a neighborhood $U$ of $x$ such that $f(U)\subset V$. You'll remember this as the "epsilon-delta"-style of continuity criterion from calculus, only without mentioning $ε$ or $δ$. The viewpoints are equivalent in the sense that one can define an open set as being a neighborhood of all its points, or contrariwise define a neighborhood of $x$ as including some open set containing $x$.

There's another alternative, to study presentations of topologies; usually a base for a topology is given. These have the advantage of being available as raw data, because the closure requirements for a base are much less stringent from a set-theoretic point of view than those of the whole system of open sets; one consequence is that you can study the topological space $(X,\langle B\rangle)$ in any universe including $X$, $B$ and $(X,B)$. This notion of a presented topology then lets you compare how the properties of a space-given-the-base change after a forcing extension of the universe --- if you're into that sort of thing.

The moral of this story is that open sets are not really the object of study in topology, and changing what you mean by "open set" is mostly going to distract you; continuous functions are one object of study --- as others have pointed out --- and there are many ways to define those.

  • Even if you just use the notion of the neighbourhood filter, you still have to decide for each point whether the entire space is a neighbourhood of it, or, equivalently, whether that point even has a neighbourhood. (You do get around concerns about emptiness, of course.) – LSpice Dec 6 '17 at 13:24

One can define a topological space in terms of closed sets to make this easy to see.

A topological space $T$ satisfies:
- The intersection of any sets $X_i \in T$ is closed (i.e., in $T$)
- The union of finitely many sets $X_i \in T$ is closed.

Then, if we take, e.g., $T=\mathbb{R}$, by the first property above, $(0,1) \cap (10,20) = \emptyset \in T$. So we should consider the empty set to be closed.

The complement of $\emptyset$ is $\mathbb{R}$, so the whole space is closed as well (there was clearly nothing particular about using the reals here so this holds for arbitrary $T$). So we should consider $T$ to be closed as well. [No, you have used the existence of disjoint open sets, which does not hold in a general topological space -- PLC.]

By De Morgan's laws, one can check that the above characterizations of a topological space are equivalent to the usual definition in terms of open sets, so for consistency we should require $\emptyset$ and $T$ to be both open and closed in addition to the above two properties.

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    This is not true. – Harry Gindi Mar 24 '10 at 23:22
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    To elaborate, all that you've done is show that the empty set must be closed, from which it follows that the whole space should be open! A tautology! – Harry Gindi Mar 24 '10 at 23:25

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