nLab tube lemma

Idea

The tube lemma may refer to one of several fundamental lemmas in topology (point-set topology to be more exact¹) that underlie many arguments involving compact spaces. That is to say: much like the Yoneda lemma of pure category theory, there are various closely related statements that can be said to fall under the rubric “tube lemma”; we consider several examples here.

Statements and proofs

Proposition

(closed-projection characterization of compactness)

If $X$ is any space and $Y$ is a compact, then the projection map $p: X \times Y \to X$ out of the product topological space is a closed map, i.e., under direct image (the left adjoint $\exists_p: P (X \times Y) \to P(X)$ to the inverse image map $p^\ast: P(X) \to P(X \times Y)$), closed sets in $X \times Y$ are mapped to closed sets in $X$.

Assuming the principle of excluded middle, the statement of prop. is equivalent to an alternative formulation:

Proposition

let $\forall_p: P(X \times Y) \to P(X)$ be the right adjoint to $p^\ast: P(X) \to P(X \times Y)$. (In classical logic, $\forall_p = \neg \exists_p \neg$.) Then

• If $Y$ is compact, then $\forall_p: P(X \times Y) \to P(X)$ takes open sets in $X \times Y$ to open sets in $X$. (Compare overt space.)

Proof

This alternative may be proven directly as follows. Let $W \subseteq X \times Y$ be an open set, and suppose $\{x\} \subseteq \forall_p(W)$. This means precisely that $p^\ast\{x\} = \{x\} \times Y \subseteq W$. We are required to show that there is a neighborhood $O$ of $x$ such that $O \subseteq \forall_p(W)$, which is to say $O \times Y \subseteq W$.

This $O \times Y$ may be pictured as a little open “tube” around $\{x\} \times Y$ which fits inside $W$, thus explaining the name of the eponymous lemma.

Lemma

(Tube lemma)

Let $X$ be a topological space and let $Y$ be a compact topological space. For a point $x \in X$ and $W$ open in $X \times Y$, if $\{x\} \times Y \subseteq W$, then there is an open neighborhood $O$ of $x$ such that the “tube” $O \times Y$ around $\{x \} \times Y$ is still contained: $O \times Y \subseteq W$.

Proof

Let $\mathcal{O}_x$ denote the set of open neighborhoods of $x$, and for $U \in \mathcal{O}_x$, let $U\multimap W$ be the largest open $V$ of $Y$ such that $U \times V \subseteq W$. The collection $\{U \multimap W: U \in \mathcal{O}_x\}$ is an open cover of $Y$; since $Y$ is compact, there is a finite subcover $U_1 \multimap W, U_2 \multimap W, \ldots, U_n \multimap W$. Then $O = U_1 \cap \ldots \cap U_n \in \mathcal{O}_x$ is the desired open. For, $O \subseteq U_i$ implies $(U_i \multimap W) \subseteq (O \multimap W)$, so that

$$Y = \bigcup_{i=1}^n (U_i \multimap W) \subseteq O \multimap W$$

and this implies $O \times Y \subseteq O \times (O \multimap W) \subseteq W$.

Proof

(of the tube lemma via closed-projection characterization of compactness)

Let

$$C \coloneqq (X \times Y) \backslash W$$

be the complement of $W$. Since this is closed, by prop. also its projection $p_X(C) \subset X$ is closed.

Now

$$\begin{aligned} \{x\} \times Y \subset W & \;\Leftrightarrow\; \{x\} \times Y \; \cap \; C = \emptyset \\ & \;\Rightarrow\; \{x\} \cap p_X(C) = \emptyset \end{aligned}$$

and hence by the closure of $p_X(C)$ there is (by this lemma) an open neighbourhood $U_x \supset \{x\}$ with

$$U_x \cap p_X(C) = \emptyset \,.$$

This means equivalently that $U_x \times Y \cap C = \emptyset$, hence that $U_x \times Y \subset W$.

Consequences

Tychnoff theorem

Corollary

(binary Tychonoff theorem)

The product topological space $X \times Y$ of two compact topological spaces $X, Y$ is itself compact.

Proof

Let $\mathcal{U}$ be any open cover of $X \times Y$, and let $\mathcal{B}$ be the collection of opens of $X \times Y$ that are unions of finitely many elements of $\mathcal{U}$. For each $x \in X$, we have that $\{x\} \times Y$ is compact since $Y$ is, so there is $B \in \mathcal{B}$ such that $\{x\} \times Y \subseteq B$, whence there is $U \in \mathcal{O}_x$ with $U \times Y \subseteq B$ by the tube lemma.

It follows that the collection

$$\{U\; open\; in\; X: U \times Y \subseteq B \; for\; some\; B \in \mathcal{B}\}$$

covers $X$, whence by compactness of $X$ there is a finite subcover $U_1, \ldots, U_n$ for which $U_i \times Y \subseteq B_i$ for some $B_i \in \mathcal{B}$, and then $B = B_1 \cup \ldots \cup B_n$ belongs to $\mathcal{B}$ and is all of $X \times Y$.

Exponentiability of the compact-open topology

The tube lemma may be used to show that the mapping space from a locally compact topological space to any topology spaces equipped with the compact-open topology is an exponential object in the category Top of topological spaces. See there.

Converse of the tube lemma

As explained above, the three statements

1. Tube lemma, Lemma

2. $\forall_p: P(X \times Y) \to P(X)$ preserves openness if $X$ is compact

3. $\exists_p: P(X \times Y) \to P(X)$ preserves closedness if $X$ is compact (prop. )

are virtually tautologically equivalent, and in this sense any one of these forms may be referred to as the tube lemma. In this section, we indicate that the converse of the tube lemma holds, stated as follows.

Theorem

(closed-projection characterization of compactness) If for all spaces $X$ the projection $p: X \times Y \to X$ is closed (form 3), then $Y$ is compact.

Various proofs may be given. If one likes the characterization of compactness that says every ultrafilter converges, then a neat conceptual proof runs as follows: given a space $Y$ satisfying the hypothesis, take $X$ to be the Stone-Cech compactification $\beta Y$ of the discrete space $Y_d$ on the underlying set of $Y$, and let $C \subseteq \beta Y \times Y$ be the convergence relation (i.e., $(U, y)$ belongs to $C$ iff the ultrafilter $U$ converges to $y$). One checks that $C$ is closed. Then the direct image $p(C)$ is closed by hypothesis. Notice also that $p(C)$ contains the set of principal ultrafilters $prin(y)$ ($prin(y)$ converges to $y$ after all), and this set may be identified with the dense set $Y_d \subset \beta Y$. Being closed and dense, $p(C)$ is all of $\beta Y$, but this simply says that every ultrafilter on $Y$ converges, so that $Y$ is compact.

The characterization of compactness via ultrafilter convergence has a slight drawback of relying on the ultrafilter theorem, a kind of choice principle. If one doesn’t like this, there are various workarounds that avoid it, but let it be said that all proofs (some of which are collected here) have basically the same intuitive character. Given $Y$, one forms a space $X$ by adjoining one or more ideal points to the discrete space $Y_d$, where the open sets around an ideal point correspond to elements of some filter $F$ that we want to show clusters around, or converges to, some point $y$ (in order to show compactness of $Y$). Just as for the convergence relation in the proof sketched above, one passes to the closure $C$ of the diagonal $Y \to Y_d \times Y \hookrightarrow X \times Y$. The image $p(C)$ is closed and contains the dense subset $Y_d$, and so contains any ideal point $p$, and thus $(p, y) \in C$ for some $y$. This turns out to mean $F$ converges or clusters to $y$, as desired.

Here is a more precise enactment of one such proof.

Proof

Let $Y$ satisfy the hypothesis of Theorem ; to prove $Y$ is compact, suppose $\mathcal{C}$ is some collection of closed sets such that every finite intersection of elements of $\mathcal{C}$ is inhabited. Let $X = Y_d \sqcup \{\infty\}$, formed by adjoining an ideal point $\infty$ to the discrete space $Y_d$ on the underlying set of $Y$, and stipulating that whenever $K \in \mathcal{C}$, the set $K \sqcup \{\infty\}$ is an open neighborhood of $\infty$. The topology thus generated consists of arbitrary subsets $U \subseteq Y$ together with sets $F \sqcup \{\infty\}$ where $F$ belongs to the filter generated by $\mathcal{C}$. Note that the finite intersection property of $\mathcal{C}$ guarantees that the filter is proper, meaning in particular $\infty$ is not an open point, equivalently that the closure of $Y_d$ in $X$ is all of $X$.

We have a diagonal embedding $Y \stackrel{\Delta}{\to} Y_d \times Y \hookrightarrow X \times Y$; let $C$ be the closure of $Y$ in $X \times Y$. Of course $p(C) \subseteq X$ contains $Y_d$, and is closed by hypothesis, so as we just observed, $p(C) = X$. So $\infty \in p(C)$; this means that $(\infty, y) \in C$ for some $y \in Y$. By closure of $C$, for each $K \in \mathcal{C}$ and any neighborhood $U$ of $y$, the neighborhood $(K \sqcup \{\infty\}) \times U$ of $(\infty, y)$ meets $\Delta(Y) \subseteq X \times Y$. This just says every open $U \in \mathcal{O}_y$ intersects $K$; since $K$ is closed in $Y$, this means $y \in K$. Thus

$$y \in \bigcap_{K \in \mathcal{C}} K$$

and the non-emptiness of this intersection proves that $Y$ is compact.

References

• Wikipedia, Tube lemma

• Todd Trimble, Characterizations of compactness, nLab link.

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1. The use of the phrase “point-set topology” signals that the tube lemma proper does refer to points, as opposed to the “point-free” or “pointless” topology as developed in the theory of locales. Point-free analogues of the main consequences of the tube lemma, such as the fact that the product of two compact topological spaces is again compact, are quite nontrivial and require a significantly different approach. ↩