Intuition for Kuratowski-Mrówka characterization of compactness

Question

Fact. A space $X$ is compact iff for every space $Y$, the projection $X\times Y\rightarrow Y$ is a closed map.

The finite subcover definition of compactness seems reasonably intuitive: finite covers mean it can't be too "spread out". This fits in with the characterization by nets, which are just there because sequences may have too few points to measure problems.

I have no intuition whatsoever for this characterization. How should one visualize it, why should one expect it, etc?

Answer 1

What does the closedness of the projection $\pi : X \times Y \to Y$ really mean?

That the compactness of $X$ implies that every projection $\pi : X \times Y \to Y$ is closed really stems from the Tube Lemma, and a characterisation of closed (continous) maps.

• Tube Lemma. (See Lemma 26.8 from Munkres's Topology.) If $X$ is compact, $Y$ is any space, $y_0 \in Y$, and $W \subseteq X \times Y$ is open such that $X \times \{ y_0 \} \subseteq W$, then there is an open neighbourhood $V$ of $y_0$ such that $X \times V \subseteq W$.

• A continuous function $f : X \to Y$ is closed iff for each $y \in Y$ and each open $U \subseteq X$ with $f^{-1} [\{ y \}] \subseteq U$ there is an open neighbourhood $V$ of $y$ such that $f^{-1} [V] \subseteq U$.

Restating the latter in the special case of the projection $\pi : X \times Y \to Y$ we get

• $\pi : X \times Y \to Y$ is closed iff for each $y \in Y$ and each open $W \subseteq X \times Y$ with $X \times \{ y \} = \pi^{-1} [\{y\}] \subseteq W$ there is an open neighbourhood $V$ of $y_0$ such that $X \times V = \pi^{-1} [ V ] \subseteq W$.

So the statement that the projection $\pi : X \times Y \to Y$ is closed for every $Y$ is really just a restatement of the Tube Lemma for $X$.

"Intuition" for the sufficiency of the statement

The Tube Lemma is probably easier to gain an intuition for. It basically says that if $W \subseteq X \times Y$ is open and contains a segment $X \times \{ y_0 \}$ (which is homeomorphic to $X$), then it can't be too "erratic" or become arbitrarily "thin" around that segment.

• If $X$ is compact, then since $\langle x,y_0 \rangle \in W$ there are open $U_x \subseteq X$ and $V_x \subseteq Y$ such that $\langle x,y_0 \rangle \in U_x \times V_x \subseteq W$. If $X$ is compact, we only need finitely many of the $U_x$s to cover $X$, and the intersection of the corresponding $V_x$s will yield $V$.

• If $X$ is not compact, you can imagine that the $U_x$s form an open cover with no finite subcover, and so when you intersect the corresponding $V_x$s for a family of the $U_x$s which do cover $X$, you might not get an open set as desired.

The trouble then is picking out an appropriate space to witness the latter for each non-compact space.

Encoding information about an open cover in an auxiliary space

That the closedness of each projection $\pi : X \times Y \to Y$ implies compactness rests on constructing an auxiliary topological space $Y$ for a given family $\mathcal{U}$ of open subsets of $X$ in such a way that properties of space $Y$ and the projection $\pi : X \times Y \to Y$ encode facts about the family $\mathcal{U}$. In particular, $\pi$ being closed will imply that either $\mathcal{U}$ does not cover $X$, or that a finite subfamily covers $X$.

The actual construction of $Y$ is a bit synthetic, and may not yield to intuition. But we can try to pick apart some ideas.

Consider the general case of a topological space $X$, and a family $\mathcal{U}$ of open subsets of $X$. We consider the set $Y = X \cup \{ \mathord{*} \}$ where $\mathord{*} \notin X$ with the topology generated by the basis consisting of

• $\{ x \}$ for each $x \in X$; and
• all sets of the form $\{ \mathord{*} \} \cup ( X \setminus ( U_1 \cup \cdots \cup U_n ) )$ where $U_1 , \ldots , U_n \in \mathcal{U}$.

Central Fact. For $A \subseteq Y$, $\mathord{*} \in \overline{A}$ iff $A$ cannot be covered by finitely many sets from $\mathcal{U}$.

If the projection $\pi : X \times Y \to Y$ is closed, then in particular the implication $$\mathord{*} \in \overline{ \pi[\Delta] } \Rightarrow \mathord{*} \in \pi[ \overline{\Delta} ]$$ holds, where $\Delta = \{ \langle x,x \rangle : x \in X \} \subseteq X \times Y$.

• As $\pi[\Delta] = X$, we have that $\mathord{*} \in \overline{\pi[\Delta]}$ iff $X$ cannot be covered by finitely many sets from $\mathcal{U}$.

• Now $\mathord{*} \in \pi [ \overline{\Delta} ]$ iff there is an $x \in X$ such that $\langle x , \mathord{*} \rangle \in \overline{\Delta}$.

  For $x \in X$, if $x \in \bigcup \mathcal{U}$, then take $U \in \mathcal{U}$ containing $x$. Clearly $U$ is an open neighbourhood of $x$ in $X$, and $\{ \mathord{*} \} \cup ( X \setminus U )$ is an open neighbourhood of $\mathord{*}$ in $Y$, however $( U \times ( \{ \mathord{*} \} \cup ( X \setminus U ) ) ) \cap \Delta = \varnothing$, so $\langle x , \mathord{*} \rangle \notin \overline{\Delta}$. On the other hand, if $x \notin \bigcup \mathcal{U}$, then as every open neighbourhood of $\mathord{*}$ contains $x$ it follows that $\langle x , \mathord{*} \rangle \in \overline{\Delta}$.

  So for $x \in X$ we have that $\langle x,\mathord{*} \rangle \in \overline{\Delta}$ iff $x \notin \bigcup \mathcal{U}$, and so $\mathord{*} \in \pi[\overline{\Delta}]$ iff $\bigcup \mathcal{U} \neq X$.

Putting this together, in order for $\pi$ to be closed it must be the case that either $X$ can be covered by finitely many sets from $\mathcal{U}$, or that $\mathcal{U}$ does not cover $X$.

We can use the same space $Y$ to investigate how the Tube Lemma for $X$ implies the compactness of $X$. Note that $W = ( X \times Y ) \setminus \overline{ \Delta }$ is an open subset of $X \times Y$. For the Tube Lemma for $X$ to hold, it must be that the implication

$X \times \{ \mathord{*} \} \subseteq W$ ⇒ $X \times V \subseteq W$ for some open neighbourhood $V$ of $\mathord{*}$

holds.

• One can show that $X \times \{ \mathord{*} \} \subseteq W$ iff $\bigcup \mathcal{U} = X$. (In particular, as above, $\langle x, \mathord{*} \rangle \in \overline{\Delta}$ iff $x \notin \bigcup \mathcal{U}$.)
• If $V$ is an open neighbourhood of $\mathord{*}$ such that $X \times V \subseteq W$, then without loss of generality there are $U_1 , \ldots , U_n \in \mathcal{U}$ such that $V = \{ \mathord{*} \} \cup ( X \setminus ( U_1 \cup \cdots \cup U_n ) )$. It then follows that $U_1 \cup \cdots \cup U_n = X$.

Answer 2

Some concrete examples might be helpful. Let's choose $X = Y = \mathbb{R}$. Now the graph of $\arctan: \mathbb{R} \to (-\pi/2, \pi/2)$ is a closed subset of $X \times Y$. However, the projection $X \times Y \to Y$ is not a closed map since the projection of the graph is $(-\pi/2, \pi/2)$. This demonstrates that $\mathbb{R}$ is not compact.

So if $X$ is not compact, the "shadow" of a closed set in $X \times Y$ might not be closed.

There are many more examples in $\mathbb{R}^2$ that help. If we have some horizontal line, and a closed subset of $\mathbb{R}^2$ that asymptotically approaches this horizontal line, then the projection to the $y$-axis does not contain the point where this line intersects the $y$-axis, but contains everything below (or above) it. Thus the projection is not closed. But if $X$ is some compact subset of the $x$-axis, then a closed subset of $X \times \mathbb{R}$ can't asymptotically approach a vertical line.