Tube lemma

Lemma 26.8 (The tube lemma). Consider the product space $X × Y$, where $Y$ is compact.
If $N$ is an open set of $X × Y$ containing the slice $x_0 × Y$ of $X × Y$, then $N$ contains some tube $W × Y$ about $x_0 × Y$, where $W$ is a neighborhood of $x_0$ in $X$.

Proof

First let us cover $x_0 × Y$ by basis elements $U × V$ (for the topology of $X × Y$) lying in $N$. The space $x_0 × Y$ is compact, being homeomorphic to $Y$. Therefore, we can cover $x_0 × Y$ by finitely many such basis elements $$ U_1 × V_1, …, U_n × V_n $$ (We assume that each of the basis elements $U_i × V_i$ actually intersects $x_0 × Y$, since otherwise that basis element would be superfluous; we could discard it from the finite collection and still have a covering of $x_0 × Y$.) Define $$ W=U_1 ∩ ⋯ ∩ U_n $$ The set $W$ is open, and it contains $x_0$ because each set $U_i × V_i$ intersects $x_0 × Y$.
We assert that the sets $U_i × V_i$, which were chosen to cover the slice $x_0 × Y$, actually cover the tube $W × Y$. Let $x × y$ be a point of $W × Y$. Consider the point $x_0 × y$ of the slice $x_0 × Y$ having the same $y$-coordinate as this point. Now $x_0 × y$ belongs to $U_i × V_i$ for some $i$, so that $y ∈ V_i$. But $x ∈ U_j$ for every $j$ (because $x ∈ W$). Therefore, we have $x × y ∈ U_i × V_i$, as desired.Since all the sets $U_i × V_i$ lie in $N$, and since they cover $W × Y$, the tube $W × Y$ lies in $N$ also. See Figure 26.2.