TUBE LEMMA

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 $X\; imes\; Y$ of the set $\{x\}\; imes\; Y$ for some $x\; in\; X$, then there exists a neighborhood U of x such that C also covers $U\; imes\; Y$.

Contents

Proof

Proof

Consider the projection $C\text{'}\; =\; pi\_Y\; C\; =\; \{\{y\; in\; Y\; |\; exists\; x\; .\; (x,y)in\; V\},\; |,\; V\; subseteq\; X\; imes\; Y,\; V\; in\; C\}$ 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 $C\text{'}\_\{sub\}\; subseteq\; C\text{'}$. This subcover must have arisen as the projection of open sets back in the original open cover of $X\; imes\; Y$; that is, there is a subset $C\_\{sub\}\; subseteq\; C$ such that $pi\_Y\; C\_\{sub\}\; =\; C\text{'}\_\{sub\}$.
Now $C\_\{sub\}$ is a finite open cover of a slice over Y, and by considering the projection $pi\_X\; C\_\{sub\}$ of $C\_\{sub\}$ 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.