Tube lemma

Remarks

For each $y \in Y$, there is a basic open box $U_y \times V_y \subseteq N$ that contains $(x_0, y)$. The sets $V_y$ cover the compact space $Y$, so finitely many $V_{y_1}, \ldots, V_{y_n}$ cover it. Then $W = U_{y_1} \cap \cdots \cap U_{y_n}$ is open, contains $x_0$, and satisfies $W \times Y \subseteq N$.

The lemma gives the proof that the product of two compact spaces is compact. It also shows that the projection $X \times Y \to X$ is a closed map when $Y$ is compact.

Compactness of $Y$ is necessary. Let $X = Y = \R$, $x_0 = 0$, and $N = \set{(x, y) : \abs{xy} < 1}$. Then $\set{0} \times \R \subseteq N$, but for any $\varepsilon > 0$ the point $(\varepsilon / 2, 4 / \varepsilon)$ lies in $(-\varepsilon, \varepsilon) \times \R$ and not in $N$.

Going deeper

For a student

The tube lemma is the key step in the proof that a finite product of compact spaces is compact. A more general form says that if $A \subseteq X$ and $B \subseteq Y$ are compact and $A \times B \subseteq N$ with $N$ open, then there are open $U \supseteq A$ and $V \supseteq B$ with $U \times V \subseteq N$. This general form is often used to show that compact sets in Hausdorff spaces can be separated by open sets.