There are many equivalent ways of defining the notion of compact space, but some require some kind of choice principle to prove their equivalence. For example, a classical result is that for $X$ to be compact, it is necessary and sufficient that every ultrafilter on $X$ converge to a point in $X$. The necessity is easy to prove, but the sufficiency requires a choice principle to the effect that every filter can be extended to an ultrafilter.

Some years ago I heard from a very good categorical topologist that many, perhaps most of the useful properties of compact spaces $X$ readily flow from the fact that for every space $Y$, the projection map $\pi: X \times Y \to Y$ is closed. Of course that is a very classical consequence of compactness which can be left as an exercise to beginners in topology, and I was struck by the topologist's assertion that you could in fact use this as a definition of compactness, and that this is a very good definition for doing categorical topology. (I am still not sure what he really meant by this, but that's not my question.)

My own proof that this condition implies compactness goes as follows. Let $Y$ be the space of ultrafilters on the set $X$ with its usual compact Hausdorff topology, and suppose the projection $\pi: X \times Y \to Y$ is a closed map. Let $R \subseteq X \times Y$ be the set of pairs $(x, U)$ where the ultrafilter $U$ converges to the point $x$. One may show that $R$ is a closed subset, so the image $\pi(R)$ is closed in $Y$. But every principal ultrafilter (one generated by a point) converges to the point that generates it, so every principal ultrafilter belongs to $\pi(R)$. Now principal ultrafilters are dense in the space of all ultrafilters, so $\pi(R)$ is both closed and dense, and therefore is all of $Y$. This is the same as saying that every ultrafilter on $X$ converges to some point of $X$, and therefore $X$ is compact.

I was at first happy with this proof, but later began to wonder if it's overkill. Certainly it uses heavily the choice principle mentioned above, and my question is whether the implication I just proved above really requires some form of choice like that.

  • $\begingroup$ What are the conditions on $Y$ when we say "for every space $Y$"? Your argument suggests that (at least in the presence of choice) it's enough to take $Y$ to be a particular compact Hausdorff space depending on $X$ $\endgroup$
    – Yemon Choi
    Oct 14, 2010 at 18:08
  • $\begingroup$ No conditions are put on $Y$. As you observe, the a priori weaker implication we get by strengthening the hypothesis to say (e.g.) "if $Y$ is compact Hausdorff" is nevertheless sufficient to prove compactness, at least if choice is assumed. $\endgroup$
    – Todd Trimble
    Oct 14, 2010 at 18:16
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    $\begingroup$ Incidentally, this view-point on compactness is the basis for the notion of properness in algebraic geometry. Recall (in topology) that a continuous map $f: X \to S$ is called proper if the preimage of compact sets are compact. If $S$ is a point, then the canonical map $X \to S$ is proper if and only if $X$ is compact, so properness is a relative version of compactness. One can check that $X \to S$ is proper if and only if for any $T \to S$, the fibre product $X\times_S T \to T$ is a closed map. (This generalizes your statement. And, as a caveat, we may need to restrict to locally ... $\endgroup$
    – Emerton
    Oct 14, 2010 at 20:36
  • $\begingroup$ ... compact space for the asserted equivalence to hold.) In algebraic geometry, working with the Zariski topology, it turns out that using with the "inverse image of compact is compact" definition doesn't give the desired analogue, but "all pull-backs are closed maps" does. Anyway, all this is just to remark that this point of view on compactness is not just used by categorical topologists. $\endgroup$
    – Emerton
    Oct 14, 2010 at 20:38
  • $\begingroup$ Thanks Emerton -- I was sort of aware of this, but the comment surely deserves to be made. $\endgroup$
    – Todd Trimble
    Oct 14, 2010 at 20:50

5 Answers 5


Martin Escardó wrote a very nice note "Intersections of compactly many open sets are open" which you might want to read.

  • $\begingroup$ This looks like it probably answers my question (and then some!), but I'll need to look it over more carefully when I find some time. Thank you! $\endgroup$
    – Todd Trimble
    Oct 14, 2010 at 21:11
  • $\begingroup$ Now that I've stared at this reference a bit longer, I can see it definitely answers my question. This is a very elegant development (which I'll probably write up on my personal web at the nLab, for possible exportation to the nLab proper), and I thank you again Andrej for bringing it to my attention. $\endgroup$
    – Todd Trimble
    Oct 14, 2010 at 22:40
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    $\begingroup$ Oh, silly me, I should have announced: no, choice principles are not required! $\endgroup$
    – Todd Trimble
    Oct 14, 2010 at 22:43
  • $\begingroup$ That's a very nice way of thinking about the closed-projection chararcterization of compactness. $\endgroup$ Oct 15, 2010 at 18:46

As a statement about locales, or even toposes, the equivalence is true without any choice and even without excluded middle. A clever construction of an appopriate locale Y can be found in the proof of C3.2.8 in Sketches of an Elephant; the proof for toposes is the culmination of chapter C3.2. Since compactness and closedness of spaces and maps are detected by their underlying locales and locale maps, this would imply the corresponding result for spaces if (1) the resulting Y is spatial and (2) the locale product X×Y agrees with the space product. However, at the moment I don't see any reason why either of those conditions should hold for a general X; the construction of Y is very frame-theoretic.

(Depending on the meaning of "categorical topology," one reason this might be a good definition is that a general notion of "closed map" might be easier to come by, for instance in a category equipped with a closure operator on subobjects. Another nice thing about it is that it generalizes very well to proper maps: a map is proper iff any pullback of it is closed.)

Regarding an actual proof, one thought would be to use filters instead of ultrafilters. With LEM but without any additional choice, compactness is equivalent to "every filter base has a cluster point." So perhaps one could replace your space Y of ultrafilters by a space of filters and your relation R of convergence by a relation of clustering?

  • $\begingroup$ As usual, you are very helpful, Mike -- these look like very nice leads! I'll take a look at the Elephant. Someone somewhere else also suggested I consider nets/filters, so this angle is probably worth another look as well. $\endgroup$
    – Todd Trimble
    Oct 14, 2010 at 19:37

This can be done choiceless and quite elementarily as follows: - if $F$ is closed in $X\times Y$ and $y\notin\pi[F]$ then, as $F$ is closed its complement is the union of a family of basic open sets, $\mathcal{B}$, say this has a finite subfamily that covers $X\times \lbrace y\rbrace$, say $\lbrace U_i\times V_i:i\le n\rbrace$, where we may as well we assume that $y\in V_i$ for all $i$; then $\bigcap_{i\le n}V_i$ is a open set that contains $y$ and it is disjoint from $\pi[F]$. This shows "$X$ compact" imples "projections parallel to $X$ are closed." - conversely if $X$ is not compact then there is a family $\mathcal{F}$ of closed sets with the finite intersection property but with empty intersection, which we may assume to be closed under finite intersections. Let $Y=X\cup\lbrace\mathcal{F}\rbrace$, topologized by making all points of $X$ isolated and the sets $F\cup\lbrace\mathcal{F}\rbrace$ ($F\in\mathcal{F}$) form a local base at $\mathcal{F}$. Now let $G$ be the closure, in $X\times Y$, of the diagonal of $X$, the fact that $\bigcap\mathcal{F}=\emptyset$ implies that $\mathcal{F}\notin\pi[G]$, the fact that no $F\in\mathcal{F}$ is empty implies that $\mathcal{F}$ belongs to the closure of $\pi[G]$.

  • $\begingroup$ Let me elaborate on a point which was confusing me, before you wrote this up. Here are two definitions of the topology on a product $X \times Y$. Definition 1: A set $U \subseteq X \times $ is open if, for every $(x,y)$ in $U$, there is are open sets $V \subseteq X$ and $W \subseteq Y$ such that $x \in V$, $y \in W$ and $X \times Y \subseteq U$. Definition 2: A set $U \subseteq X \times Y$ is open if there is a set $A$ and families $(V_a)_{a \in A}$ and $(W_a)_{a \in A}$ of open subsets of $X$ and $Y$ such that $U = \bigcup_{a \in A} V_a \times W_a$. (continued) $\endgroup$ Oct 19, 2010 at 14:34
  • $\begingroup$ I was thinking in terms of definition 1, and worried that I needed choice to pass to definition 2 -- thinking that I had to choose some particular $V$ and $W$ for each $(x,y)$. But I was wrong. If $U$ is open according to definition 1, then just take $A$ to be the set of all ordered pairs $(V,W)$ such that $V$ is open in $X$, $W$ is open in $Y$ and $V \times W \subseteq U$. No choice needed! $\endgroup$ Oct 19, 2010 at 14:35
  • $\begingroup$ And, once you are working with definition 2, it is very easy to give a choice free proof, as you explain very nicely. $\endgroup$ Oct 19, 2010 at 14:36
  • $\begingroup$ Thank you, KP Hart. I discovered a few days ago that this is the proof in Bourbaki (and it is also the point-set version of the localic proof alluded to by Mike Shulman in his answer, given by Johnstone in the Elephant). $\endgroup$
    – Todd Trimble
    Oct 19, 2010 at 15:33

In case anyone is interested, this is a rendition of the proof I was looking for, contained in the article pointed out by Andrej Bauer. (I'm not looking for upvotes; this is just to round out the discussion.)

The elegant proof of the implication I was after (see Martin Escardo's article) is perfectly constructive. So this answers completely my question: no choice is involved.

Major edit: In a previous version of this post, I had made a rash claim "above the fold" that the implication "$X$ is compact implies $\pi: X \times Y \to Y$ is a closed map for every space $Y$" seems to involve the axiom of choice. But Mike Shulman sent me an email which set me straight: if you do it right, AC is not required for this direction either.

Start with the following easy observation (which follows by playing around with complements): $\pi: X \times Y \to Y$ is a closed map precisely when the set

$$\{y \in Y: X \times \{y\} \subseteq U\}$$

is open whenever $U \subseteq X \times Y$ is open.

Now, suppose $\pi: X \times Y \to Y$ is a closed map for every $Y$. To show $X$ is compact, we show that $X$ belongs to any open cover of $X$ that is closed under finite unions. Let $\Sigma$ be such an open cover.

Construct a space $Y$ as follows: the points of $Y$ are open sets of $X$, and the open sets of $Y$ are subsets $W \subseteq Y$ such

  • $W$ is upward-closed: if $U \in W$ and $U \subseteq V$ for $V \in Y$, then $V \in W$, and

  • $\Sigma \cap W$ is nonempty (unless $W$ is empty).

It is straightforward to check this defines a topology on $Y$ (that the intersection of two opens $W$, $W'$ of $Y$ is again open uses the fact that $\Sigma$ is closed under finite unions).

Observe that if $U$ belongs to $\Sigma$ and $U' \subseteq U$, the principal upward-closed set $prin(U') = \{V \in Y: U' \subseteq V\}$ is open in $Y$.

Now consider the set $E = \{(x, U) \in X \times Y: x \in U\}$. Claim: this is open in $X \times Y$. Proof: for every $(x, U) \in E$, the set $U \times prin(U)$ is an open set which contains $(x, U)$, and $U \times prin(U) \subseteq E$ because for every $(y, V) \in U \times prin(U)$, we have $y \in V$. Given $(x, U) \in E$, there exists $U' \in \Sigma$ containing $x$ (since $\Sigma$ is a cover), and then for $U'' = U \cap U'$, the set $U'' \times prin(U'')$ is an open that contains $(x, U)$, and this is included in $E$ because $y \in V$ for every $(y, V) \in U'' \times prin(U'')$.

By the open-set reformulation of the closed map condition, the set

$$\{V \in Y: X \times \{V\} \subseteq E\}$$

is open in $Y$, is nonempty (because $X$ belongs to it), and so this set intersects $\Sigma$ by definition of the topology of $Y$. Thus $X \times \{V\} \subseteq E$ for some $V \in \Sigma$. But then $V$ is all of $X$! So $X \in \Sigma$ for any open cover $\Sigma$ closed under finite unions; therefore $X$ is compact.

  • 2
    $\begingroup$ Confusion: Why is $U \times prin(U)$ open? We don't seem to have assumed $U\in\Sigma$... $\endgroup$ Oct 16, 2010 at 16:32
  • $\begingroup$ My bad, and thanks for catching that Harry. The responsible thing would be to cross out the bogus proof and insert the correction, but I can't seem to figure out how to cross out. Anyway, I'll fix this. $\endgroup$
    – Todd Trimble
    Oct 16, 2010 at 17:16

This question is related to the notion of proper map, of which there is quite a lot in Bourbaki and also in my book `Topology and groupoids'. Note the elegant Bourbaki definition: a map $f: X \to Y$ is proper if for all spaces $Z$ the map $$f \times 1_Z: X \times Z \to Y \times Z$$ is a closed map.

Once I was teaching a second year course in analysis and realised how nice were the proofs involving sequences. So I started looking at questions like `what is the one point sequential compactification?' (add an extra point to which the non convergent sequences converge!). It all worked out quite well and was published as

[12]. ``Sequentially proper maps and a sequential compactification'', J. London Math Soc. (2) 7 (1973) 515-522.


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