MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The usual axiomatization of a topological space (in the sense of Bourbaki) goes by declaring certain subsets as being open and such that a few axioms are fulfilled by the family of open subsets.

It is trivial to get an equivalent axiomatization of a topological space in terms of declaring certain subsets as being closed (by choosing the complements of the open subsets and dualizing the axioms).

My question is whether there is a good axiomatization in use of what a locally compact Hausdorff space is in terms of what its compact subspaces will be.

share|cite|improve this question
up vote 1 down vote accepted

This is as of now probably more of a really long comment than a complete answer.

There is a one-to-one correspondence between compact Hausdorff spaces with distinguished dense subspaces and structures called proximity spaces. Similarly, there is a more general correspondence between locally compact Hausdorff spaces with distinguished dense subspaces and structures called local proximity spaces. In essence, we obtain the notion of a compact (locally compact) Hausdorff space with a distinguished dense subspace from structures that appear at first glance to have nothing to do with compactness.

Intuitively, a proximity space is a set $X$ along with a binary relation $\mathcal{\delta}$ on $P(X)$ such that $\delta$ gives one the notion of whether two subsets of $X$ touch each other in some sense. For example, if $(X,d)$ is a metric space, then $(X,d)$ becomes a proximity space where $A\delta B$ if and only if $d(A,B)=0$. Every proximity space $(X,\delta)$ induces a completely regular topology on $X$ and with each proximity space $(X,\delta)$, we associate a compactification of $X$ known as the Smirnov compactification of $X$. From a proximity space and its Smirnov compactification, we obtain an equivalence between the category of proximity spaces and the category consisting of all pairs $(X,C)$ where $X$ is a proximity space and $C$ is a compactification of $X$.

Furthermore, one can generalize the notion of a proximity space to the notion of a local proximity space to obtain a similar representation for locally compact spaces. Intuitively, a local proximity space is a set $X$ along with a notion of which subsets of $X$ are bounded and whether subsets of $X$ touch each other. The category of local proximity spaces is equivalent to the category of locally compact Hausdorff spaces along with a distinguished dense subset.

The reader is referred to the book [1] for more information on proximity spaces and local proximity spaces.

[1] Naimpally, S. A., and B. D. Warrack. Proximity Spaces. Cambridge [Eng.: University, 1970.

share|cite|improve this answer
This answer has been helping me most as this gives me more than just rewriting the usual axiom (which is a basic exercise). – Marc Nieper-Wißkirchen Dec 21 '13 at 9:23

Based on Qiaochu's answer, we can explicitly work out the axioms, and hopefully reduce the messiness somewhat:

The empty set is compact.

Any intersection of compact sets is compact.

A finite union of compact sets is compact. (These replace the axioms of a topological space)

If the empty set is the intersection of infinitely many compact sets, then it is the intersection of a finite subcollection of them. (This says that our compact sets are actually-compact-in-the-new-topology.)

For any point $P$, there are compact sets $A$ containing $P$, $B$ not containing $P$, such that for any compact set $C$, $C- (A \cap C ) \cup B$ is compact. (This replaces local compactness, with $A - (A \cap B)$ the open neighborhood)

For any pair of points $P_1,P_2$, we can find $A_1,A_2,B_1,B_2$ satisfying the conditions of the previous axiom such that $A_1 \cap A_2 \subset B_1 \cup B_2$. (This replaces Hausdorff.)

We do not need a condition to prove that all the sets that are-actually-compact-in-the-new-topology are compact, because the local compactness axiom gives us an open cover such that anything covered by finitely many open sets is covered by finitely many compact sets, so if it's closed, it's compact.

It's possible that some of these axioms can be simplified further.

share|cite|improve this answer

Claim: A subspace $S$ of a locally compact Hausdorff space $X$ is closed iff $S \cap C$ is compact for each compact subspace $C$.

Proof. The implication $\Rightarrow$ is clear. For the implication $\Leftarrow$, let $x \not \in S$ and let $U$ be a neighborhood of $x$ with compact closure. Then by hypothesis $S \cap \bar{U}$ is compact, hence closed, hence $S^c \cap \bar{U}$ is open in $\bar{U}$, hence $S^c \cap U$ is open in $U$, so there is a neighborhood $x \in V \subseteq U$ disjoint from $S$. $\Box$

For a Hausdorff space the above property is equivalent to being compactly generated.

From here you can rewrite all of the relevant axioms solely in terms of compact subspaces but it seems messy and you end up basically restating all of the usual conditions anyway.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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