Tube lemma

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In mathematics, in the field of topology, the tube lemma is a result which states that if $X$ and $Y$ are topological spaces with $Y$ compact, then in the product space, any open cover of a slice over $Y$ also covers a tube about that slice. More formally, if there is an open cover C of open sets in $X \times Y$ of the set $\{x\} \times Y$ for some $x \in X$ , then there exists a neighborhood U of x such that C also covers $U \times Y$ .

[edit] Proof

Consider the projection $C' = \pi_Y C = \{\{y \in Y | \exists x . (x,y)\in V\}\, |\, V \subseteq X \times Y, V \in C\}$ of C to an open cover of open sets of Y. Evidently C' actually covers Y. By assumption Y is compact, so there is a finite subcover $C'_{sub} \subseteq C'$ . This subcover must have arisen as the projection of open sets back in the original open cover of $X \times Y$ ; that is, there is a subset $C_{sub} \subseteq C$ such that $π Y C s u b = C' s u b$ .

Now $C s u b$ is a finite open cover of a slice over Y, and by considering the projection $π X C s u b$ of $C s u b$ down to X, and taking the (finite!) intersection of all open sets in it, one obtains the open neighborhood U of x that meets the requirements of the lemma.