tube lemma

Tube lemma - Let $X$ and $Y$ be topological spaces such that $Y$ is compact. If $N$ is an open set of $X\times Y$ containing a ”slice” $x_0\times Y$, then $N$ contains some ”tube” $W\times Y$, where $W$ is a neighborhood of $x_0$ in $X$.

Proof : $N$ is a union of basis elements $U\times V$, with $U$ and $V$ open sets in $X$ and $Y$ respect. Since $x_0\times Y$ is compact (it is homeomorphic to $Y$), only a finite number $U_1\times V_1,\mathrm{\dots },U_n\times V_n$ of such basis elements cover $x_0\times Y$.

We may assume that each of the basis elements $U_i\times V_i$ actually intersects $x_0\times Y$, since otherwise we could discard it from the finite collection and still have a covering of $x_0\times Y$.

Define $W:=U_1\cap \mathrm{\dots }\cap U_n$. The set $W$ is open and contains $x_0$ because each $U_i\times V_i$ intersects $x_0\times Y$ by the previous remark.

We now claim that $W\times Y\subseteq N$. Let $(x,y)$ be a point in $W\times Y$. The point $(x_0,y)$ is in some $U_i\times V_i$ and so $y\in V_i$. We also know that $x\in W=U_1\cap \mathrm{\dots }\cap U_n\subseteq U_i$.

Therefore $(x,y)\in U_i\times V_i\subseteq N$ as desired. $\mathrm{\square }$