Tube lemma

Statement
Let $$X$$ be a compact space and $$A$$ any topological space. Consider $$X \times A$$ endowed with the product topology. Suppose $$a \in A$$ and $$U$$ is an open subset of $$X \times A$$ containing the entire slice $$X \times \{ a \}$$. Then, we can find an open subset $$V$$ of $$A$$ such that:

$$a \in V$$, and $$X \times V \subseteq U$$

In other words, any open subset containing a slice contains an open cylinder that contains the slice.

Proof
Given: A compact space $$X$$, a topological space $$A$$. $$a \in A$$, and $$U$$ is an open subset of $$X \times A$$ containing the slice $$X \times \{ a \}$$.

To prove: There exists an open subset $$V$$ of $$A$$ such that $$a \in V$$ $$X \times V$$ is contained in $$U$$.

Proof:


 * 1) A collection of open subsets inside $$U$$ whose union contains $$X \times \{ a \}$$: For each $$x \in X$$, we have $$(x,a) \in U$$, so by the definition of openness in the product topology, there exists a basis open subset $$M_x \times N_x \subseteq U$$ containing $$(x,a)$$. In particular, we get a collection $$M_x \times N_x, x \in X$$ of open subsets contained in $$U$$, whose union contains $$X \times \{ a \}$$.
 * 2) This collection yields a point-indexed open cover for $$X$$: Note that since $$M_x \times N_x$$ is a basis open set containing $$(x,a)$$, $$M_x$$ is an open subset of $$X$$ containing $$X$$, so the $$M_x, x \in X$$, form an open cover of $$X$$.
 * 3) (Given data used: $$X$$ is compact): This cover has a finite subcover: Indeed, since $$X$$ is compact, we can choose a finite collection of points $$\{ x_1, x_2, \dots, x_n \} \subseteq X$$ such that $$X$$ is the union of the $$M_{x_i}$$s.
 * 4) If $$V$$ is the intersection of the corresponding $$N_{x_i}$$s, then $$V$$ is open in $$A$$, $$a \in V$$, and $$X \times V \subseteq U$$: First, $$V$$ is open since it is an intersection of finitely many open subsets of $$Y$$. Second, each $$N_{x_i}$$ contains $$a$$, so $$a \in V$$. Third, if $$(x,v) \in X \times V$$, then there exists $$x_i$$ such that $$x \in M_{x_i}$$. By definition, $$v \in N_{x_i}$$, so $$(x,v) \in M_{x_i} \times N_{x_i} \subseteq V$$. Thus, $$(x,v) \in U$$.

Textbook references

 * , Page 168, Lemma 26.8, Chapter 3, Section 26 (the proof is given before the theorem, as Step 1 of the proof of Theorem 26.7 on page 167)