# The Tube Lemma

The Tube Lemma is a useful tool in working with Cartesian products of finitely many compact spaces. A general discussion is followed by three applications of the lemma.

Let $X$ and $Y$ be topological spaces. A slice in the Cartesian product $X \times Y$ is a subspace of the form $\left\{x\right\} \times Y$ or $X \times \left\{y\right\}$ where $x \in X$ and $y \in Y$. A tube is an open subset of the Cartesian product that is of the form $G \times Y$ or $X \times H$ where $G$ is open in $X$ and $H$ is open in $Y$. In the Euclidean plane, a slice would be either a vertical line or a horizontal line and open strips (vertical or horizontal) are examples of tubes.

Tubes are one type of open subsets of the Cartesian product $X \times Y$. The Tube Lemma is applicable when one of the factors is compact. Let $Y$ be the factor that is compact. A good way of thinking about the lemma is that when you consider the slices $\left\{x\right\} \times Y$ as “points”, the tubes $G \times Y$, where $x \in G$, behave like a base. The following is a statement of the lemma.

The Tube Lemma
Let $X$ be a space and let $Y$ be a compact space. For each $x \in X$, and for each open set $U$ of $X \times Y$ such that $\left\{x\right\} \times Y \subset U$, there is an open set $O \subset X$ such that $\left\{x\right\} \times Y \subset O \times Y \subset U$.

Proof. Let $x \in X$ and let $U \subset X \times Y$ be open in the product space such that $\left\{x\right\} \times Y \subset U$. For each $y \in Y$, choose open sets $A_y \subset X$ and $B_y \subset Y$ such that $(x,y) \in A_y \times B_y \subset U$. Since $Y$ is compact, we can find finitely many $B_y$ whose union is $Y$, say, $Y=B_{y(1)} \cup \cdots \cup B_{y(n)}$. Let $O=A_{y(1)} \cap \cdots \cap A_{y(n)}$. It follows that $\left\{x\right\} \times Y \subset O \times Y \subset U$. $\blacksquare$

Remarks
The lemma is not true when none of the factors is compact. Let $X=\mathbb{R}$ and $Y=\mathbb{R}$ with the usual topology. Let $U$ be defined by:

$U=\left\{(x,y):x \ne 0,y <\frac{1}{\lvert x \lvert}\right\} \cup \left\{(0,y):y \in \mathbb{R}\right\}$

The open set $U$ contains the slice $\left\{0\right\} \times \mathbb{R}$. But no tube can be situated between this slice and $U$.

The Tube Lemma can be used in proving that the product of two compact spaces is compact. By induction, it follows that the product of finitely many compact spaces is compact. However, the lemma cannot be used in proving the compactness of product space with infinitely many compact factors (the Tychonoff Theorem).

The Tube Lemma also shows that both the Lindelof property and paracompactness are preserved in taking two-factor Cartesian product as long as one of the factors is compact. As a corollary, the product of two Lindelof spaces is Lindelof if one of the factors is $\sigma$-compact. We have the following theorems.

Theorem 1
Let $X$ and $Y$ be compact spaces. Then $X \times Y$ is compact.

Proof. Let $\mathcal{U}$ be an open cover of $X \times Y$. For each $x \in X$, let $\mathcal{U}_x$ be a finite subcollection of $\mathcal{U}$ such that $\mathcal{U}_x$ is a cover of $\left\{x\right\} \times Y$. By the Tube Lemma, there is an open set $O_x \subset X$ such that $\left\{x\right\} \times Y \subset O_x \times Y \subset \cup \mathcal{U}_x$. Since $X$ is compact, there are finitely many $x_1,x_2,\cdots,x_n \in X$ such that $X=O_{x_1} \cup \cdots \cup O_{x_n}$. It follows that $\mathcal{U}_{x_1} \cup \cdots \cup \mathcal{U}_{x_n}$ covers $X \times Y$. $\blacksquare$

Theorem 2
Let $X$ be a Lindelof space and $Y$ be any compact space. Then $X \times Y$ is Lindelof.

Proof. The same proof in Theorem 1 applies except that there are countably many $O_x$ that cover $X$, leading to a countable subcover of the original open cover. $\blacksquare$

Corollary 3
Let $X$ be a Lindelof space and $Y$ be any $\sigma$-compact space. Then $X \times Y$ is Lindelof.

Theorem 4
Let $X$ be a paracompact space and $Y$ be any compact space. Then $X \times Y$ is paracompact.

Proof. The proof begins just as in Theorem 1. Let $\mathcal{U}$ be an open cover of $X \times Y$. For each $x \in X$, let $\mathcal{U}_x$ be a finite subcollection of $\mathcal{U}$ such that $\mathcal{U}_x$ is a cover of $\left\{x\right\} \times Y$. By the Tube Lemma, there is an open set $O_x \subset X$ such that $\left\{x\right\} \times Y \subset O_x \times Y \subset \cup \mathcal{U}_x$. Since $X$ is paracompact, let $\left\{V_x: x \in X\right\}$ be a locally finite open refinement of $\left\{O_x: x \in X\right\}$ such that $V_x \subset O_x$ for each $x \in X$.

Let $\mathcal{W}=\left\{(V_x \times Y) \cap U: x \in X,U \in \mathcal{U}_x\right\}$. It can be shown that $\mathcal{W}$ is a cover of $X \times Y$, is a refinement of $\mathcal{U}$, and is locally finite. The first two points are clear. To show that it is a locally finite collection of sets, let $(a,b) \in X \times Y$. There is some open $V \subset X$ such that $a \in V$ and $V$ meets only finitely many $V_x$. Then $V \times Y$ meets only finitely many sets in $\mathcal{W}$. $\blacksquare$