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1. Why is the arbitrary union axiom in the definition of topology necessary?

2. Why is it useful? Why might we expect ("intuitively") that it should be useful?

3. What is the (historical) motivation for the axiom?

Please do not read any of the below unless you are really bored or have a lot of time to kill; the essence of my question is the above.

In what follows, I will follow Dellacherie and Meyer in denoting:

$\underset{f}{\cup}$ = finite unions
$\underset{c}{\cup}$ = countable unions
$\underset{u}{\cup}$ = uncountable unions
$\underset{a}{\cup}$ = arbitrary unions

Similar notation applies for intersections.

In what follows I will take closure under (relative) complementation for granted, simply because almost all set systems which I know of (besides $\pi$ systems) have this property, its motivation is obvious, and because it is involutory it is easily added to the axioms of a set system without changing the included sets very much.

Most spaces which are commonly studied have at most the continuity of the continuum (not that there aren't counterexamples, but at least I don't run into them in my work).

Thus, it seems we could have at most the following types of set systems for a space which has at most the cardinality of the continuum.

  1. $\underset{f}{\cup}$, $\underset{f}{\cap}$ (Boolean) algebra

  2. $\underset{c}{\cup}$, $\underset{f}{\cap}$/$\underset{c}{\cap}$, $\underset{f}{\cup}$ (slightly stronger than a $\lambda$ system)

  3. $\underset{c}{\cup}$, $\underset{c}{\cup}$ $\sigma-$algebra

  4. $\underset{u/a}{\cup}$, $\underset{c}{\cup}$/ $\underset{u/a}{\cap}$, $\underset{c}{\cup}$ ????? Related: Is there a difference between allowing only countable unions/intersections, and allowing arbitrary (possibly uncountable) unions/intersections?

  5. $\underset{u/a}{\cup}$, $\underset{f}{\cap}$/ $\underset{u/a}{\cap}$, $\underset{f}{\cup}$ systems of open sets (topologies)/systems of closed sets resepctively

  6. $\underset{u/a}{\cup}$, $\underset{u/a}{\cap}$ Alexandroff topologies

The motivation for the set axioms of a Boolean algebra are obvious, because they basically constitute "closure under elementary logical operations". For example, all truth functions are well-defined given a Boolean algebra, as is propositional logic.

The motivation for a $\sigma-$algebra is also fairly clear to me -- it is essentially the "countable closure/extension" of a regular algebra, and this property allows us to define countably additive set functions, and along with that obtain closure under pointwise limits of functions as well as the monotone convergence theorem, the two foundations on which integration theory is based.

It is also clear that such additivity could not be extended to the cardinality of the continuum, since the uncountable sum of any non-zero numbers is infinite.

But why do open/closed set systems turn out to be useful? How are these uncountable axioms applied fruitfully in topology?

As far as I am aware, for example with stochastic processes, we usually rely on some separability/first/second-countability properties in order to be able to apply most topologies fruitfully -- i.e. some reduction of the "uncountable" axiom to an essentially countable property. So why not study only metric spaces?

Moreover, the study of limits is only defined/sensible once we have the T1 and T2 separation axioms, so the claim that we study set systems of type 5. in order to study limits seems unjustified to me; at the very least we need that in combination with some separation axioms.

Also, why do we study set systems of type 5., but not of types 4. or 6.? What is the essential property that type 5. set systems capture (in the way that type 1. and type 3. systems capture)?

The best discussion of this topic which I have found so far: https://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets

I agree with the OP of that question however, that the top-voted answer is more of a justification for the notion of metric space (which are always separable in some sense due to the density of the rationals in the reals).

The discussion of how notions of limit really seem to be more of a consequence of a notion of "Hausdorfness" is here: Why not just study the consequences of Hausdorff axiom? What do statements like, "The arbitrary union of open sets is open," gain us? I would also like to point out that the axiomatization of topology in terms of neighborhood systems, plus the Hausdorff axiom, seems to be more relevant to understanding the concept of "limit" than the axioms of a general topological space. And again, I don't see why we would need uncountably many neighborhoods of a given point to define it as the limit of something, since countably many neighborhoods always suffices (at least in the metric space setting).

Motivation:

The motivation for this question came from some rumination about stochastic processes. Namely, the reason why their study can be so difficult is that we are confined to using set systems of type 3., which are only countably closed, whereas we are studying uncountably many random variables.

In order to get any degree of meaningful results, we therefore rely on having some degree of "separability/regularity" properties that essentially allow us to reduce to the countable case where it counts (for instance with regards to the existence of regular conditional distributions).

So as frustrating as it may be trying to study uncountable objects with set systems of type 3., it is clear to me why we can not replace such set systems with any of type 4, 5, or 6, again so that integration and probability are tractable.

So then I thought "oh so we must use topologies because we need uncountability for some reason". But then I thought about how almost all topologies which I am aware of/use are in some way "separable" or "first/second-countable", which basically constitutes reducing back to the countable case, the same way we do with stochastic processes. I.e. there is no fundamental gain in generality, since the only cases of uncountable objects we can/do study are those special cases which can be reduced back to the countable case in some meaningful way.

I'm really at a loss as to why an uncountable closure axiom would be at all useful, and why we would want only finite closure for the complementary operation, rather than some type of infinite closure.

I feel like it has to have something to do with the distinction between "discrete" and "continuous" spaces, which leads us back again to the study of stochastic processes, and the distinction between "discrete" and "continuous" time.

Namely, for a set with countable cardinality, like the natural numbers, the power set coincides with the reasonable choices for sigma algebra and topology.

In other words, both concepts are clearly only needed when moving to the uncountable case, where they don't necessarily coincide with the power set, i.e. we can gain a concept of "continuousness", rather than the "discreteness" which we are necessarily confined to with set systems of countable sets. Again, the purpose of $\sigma-$algebra in this case is clear to me, since it is actually only closed under countable set operations, but now I really don't understand why we are able to use the uncountable set axiom profitably in open/closed set systems, especially since in all cases which I am aware of we have some sort of "separability" which reduces the concept of topology to a countable one. So why not define the concept which we actually use, which is countable, not uncountable in nature?

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Chill2Macht
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    I think the first nonmetrizable space that we encounter as analysts is the weak star topology on an infinite dimensional Banach space which is the dual of some other infinite dimensional Banach space. For example, $L^p(I)$ with $1<p<\infty$ and $I$ an interval in $\mathbb{R}$. We wind up needing this basically because the strong topology doesn't have enough compact sets for us to be able to extract convergent subsequences from certain approximation procedures. – Ian Jun 09 '16 at 17:30
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    This also comes up in stochastic process theory (there are whole books about proving things in probability theory using weak convergence methods; for example I've looked at one that looks at large deviation theory from this perspective). Is this enough motivation? – Ian Jun 09 '16 at 17:33
  • Not really to be honest; the weak-topology is still separable, correct? Since it is a space of real/complex functionals? Metric spaces were just one example of topological spaces whose study "reduces to the countable case". Also it is unclear to me where the uncountable union axiom comes into play when studying that topology (maybe something with cylinder sets)? But anyway we are mostly interested as analysts inweak-limits, correct? Since as I argued earlier, that isn't really encapsulated in the full generality offered by general topological spaces. – Chill2Macht Jun 09 '16 at 19:04
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    Well, my point was that as you've said, the uncountable union axiom basically doesn't come up in separable metric spaces, which is to say essentially all metric spaces of interest in analysis. So I went for the first nonmetrizable space I could think of. As it happens that $L^p$ space in particular is separable, and the dual of a separable space is separable in its weak star topology, so you are right. – Ian Jun 09 '16 at 19:17
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    I guess this is another illustration case: if open sets only allow countable unions, then closed sets must only allow countable intersections. This could be bad: for example, if you have a mapping that sends each point $x$ in the Cantor set to a closed interval $I(x)$, the intersection of ${ f \in C[0,1] : f(x) \in I(x) }$ should be closed (if a sequence of functions is in that intersection and converges uniformly, then the limit is in that intersection, certainly), but you would not be able to infer that from the topology axioms. – Ian Jun 09 '16 at 19:17
  • Ah so we need set systems of type 5 for pointwise convergence of functions on uncountable spaces, even if not for points in the spaces themselves? So in some sense then we need to have topology to define a useful integration theory, so we actually need topological sigma algebras, and not just arbitrary sigma algebras for closure under pointwise convergence of functions? – Chill2Macht Jun 09 '16 at 19:31
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    I think that's right; I'm not sure whether you can write such things down in a "countable" way by exploiting separability, though. It seems like you should not be able to do that in my example because all points in the Cantor set are isolated (which is why I used the Cantor set rather than, say, the irrationals in $[0,1]$). – Ian Jun 09 '16 at 19:36
  • hmm that's an intersting example because it also addresses the issue of the distinction between "discrete" and "continuous spaces", since the Cantor space, although not discrete itself, is the product of discrete spaces (using the weak topology generated by the coordinate mappings, ie cylinder sets). – Chill2Macht Jun 09 '16 at 19:53
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    Aside: you do have arbitrary meets; it's just that they're not given by intersection. $$\bigwedge_{i \in I} U_i = \text{interior}\left( \bigcap_{i \in I} U_i \right) $$ (also, finite meets distribute over arbitrary unions, but infinite meets don't have to) –  Jun 12 '16 at 08:33
  • hm that point about distributivity is very interesting and something which I hadn't thought about before -- that really might be it actually – Chill2Macht Jun 12 '16 at 18:06

1 Answers1

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Here are three characterizations that have nothing to do with arbitrary unions

A base on $X$ is a collection $N \subseteq \mathcal{P}(X)$ of neighborhoods satisfying

  • Every point is contained in at least one neighborhood
  • If $A$ and $B$ are neighborhoods and $P \in A \cap B$, then there is a neighborhood $C$ such that $P \in C \subseteq A \cap B$.

You can develop topology in terms of bases. e.g.

$f : X \to Y$ is continuous if and only if, for every neighborhood $V \subseteq Y$ and every point $P \in X$ such that $f(P) \in V$, there is a neighborhood $U \subseteq X$ such that $P \in U$ and $f(P) \subseteq V$.

Let $O(N)$ denote the set of all arbitrary unions of neighborhoods from $N$. Then $O(N)$ satisfies the open set axioms. Furthermore:

If $N$ and $N'$ are two different bases on $X$, then the identity map defines a homeomorphism $(X,N) \to (X,N')$ if and only if $O(N) = O(N')$.

Conversely, the collection of open sets in a topological space satisfies the neighborhood axioms.

A closure operation on $X$ is an operation on subsets of $X$ satisfying

  • $\overline{\varnothing} = \varnothing$
  • $A \subseteq \overline{A}$
  • $\overline{A \cup B} = \overline{A} \cup \overline{B}$
  • $\overline{\overline{A}} = \overline{A}$

The collection of sets satisfying $A = \overline{A}$ satisfies the closed set axioms.

Conversely, in a topological space, closure is a closure operation.

A nearness relation on $X$ is a relation between elements of $X$ and subsets of $X$

  • No point is near $\varnothing$
  • If $P \in A$, then $P$ is near $A$
  • $P$ is near $A \cup B$ if and only if $P$ is near $A$ or $P$ is near $B$ (or both)
  • If $P$ is near $A$ and every point of $A$ is near $B$, then $P$ is near $B$

The set of near points to $A$ defines a closure relation, so we again get a topological space.

Conversely, $P \in \bar{A}$ is a nearness relation.

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    The axiomatization of "nearness" is mine and maybe has errors, although I'm sure I've seen it elsewhere. Also, I didn't look up the phrasing of continuity for bases, so it may have an error too. The other two formulations are well-known. –  Jun 12 '16 at 08:30
  • I guess what I am confused about is why we need the arbitrary union of base sets to be open, when in all cases I know of, countable unions suffice. Do you know a (non-discrete) case where arbitrary unions of base sets are necessary as opposed to just countable unions? – Chill2Macht Jun 12 '16 at 14:08
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    @William: If every arbitrary union can be written as a countable union, then you still have arbitrary unions. If you don't have that property, then all three of the characterizations I gave imply that requiring arbitrary unions is the "right" definition and merely countable unions is "wrong". –  Jun 12 '16 at 14:18
  • I know, I just don't know any examples where the stronger property is necessary, so it's hard to understand why it is "just right" and not "too strong" if that makes sense. For example one can justify countable but not arbitrary unions for sigma algebras by noting that we want countable additivity of measures but that uncountable additivity is impossible. It is unclear to me, for lack of a good example, why we don't encounter similar problems with arbitrary unions for topologies – Chill2Macht Jun 12 '16 at 14:27
  • Thank you again by the way so much for all of your help; I really appreciate it. I already upvoted your answer; if you can add an example where arbitrary meets are necessary and not just sufficient I will accept it. I thought the Zariski topology might be an example, but that it is most often used in the context of Noetherian rings suggests otherwise, although admittedly I am unsure. – Chill2Macht Jun 12 '16 at 14:30
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    Two simple counterexamples I can think of are the discrete topology on an uncountable set (take a base of singletons) and the long line (take a base consisting of covers of the individual intervals constructing it). Maybe there are things involving function spaces? I don't know this sort of detail about those. –  Jun 12 '16 at 14:34
  • However, I think it might be a good idea to ask a separate question specifically about counterexamples. –  Jun 12 '16 at 14:34
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    @William: Oh, I should also add that typical arguments use arbitrary unions (e.g. for every point having some property choosing a neighborhood with some property); so even if you had a space where arbitrary unions could be written as countable unions, you would still prefer to have theorems formulated in terms of the arbitrary unions. Otherwise, arguments would have to take unnecessary detours through working with countable dense subsets or countable subcovers. –  Jun 12 '16 at 14:48