Does "compact iff projections are closed" require some form of choice? - MathOverflow

Does "compact iff projections are closed" require some form of choice?

2010-10-14T18:00:09Z

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.

Answer by Mike Shulman for Does "compact iff projections are closed" require some form of choice?

2010-10-14T18:57:11Z

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?

Answer by Andrej Bauer for Does "compact iff projections are closed" require some form of choice?

2010-10-14T20:54:25Z

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

Answer by Todd Trimble for Does "compact iff projections are closed" require some form of choice?

2010-10-15T23:11:37Z

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.

Answer by KP Hart for Does "compact iff projections are closed" require some form of choice?

2010-10-19T14:07:52Z

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 $\pi[G]$.

Answer by Ronnie Brown for Does "compact iff projections are closed" require some form of choice?

2011-10-30T11:47:52Z

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.