Lemma 3.10.7 (Tube Lemma).label Let $X$ be a topological space and $Y$ be a compact space, then
- (1)
For any $x\in X$ and $U\in\cali{N}^{o}_{X\ti Y}(\curl{x}\ti Y)$, there exists $V\in\cali{N}_{X}(x)$ such that $V\ti Y\suf U$.
- (2)
For any $A\suf X$ and $U\in\cali{N}^{o}_{X\ti Y}(A\ti Y)$, there exists $V\in\cali{N}_{X}(A)$ such that $V\ti Y\suf U$.
Proof.
- (1)
For each $y\in Y$, let $U_{y}\in\cali{N}_{X}(x)$ and $V_{y}\in \cali{N}_{Y}(y)$ such that $(x,y)\in U_{y}\ti V_{y}\suf U$, then $\curl{V_y:y\in Y}$ is an open cover of $Y$. By compactness, there exists $\curl{y_j}_{1}^{n}$ such that $\cups{n}{j=1}V_{y_j}=Y$. Let $V=\caps{n}{j=1}U_{y_j}$, then $V\in\cali{N}_{X}(x)$ and $V\ti Y\suf U$.
- (2)
For each $x\in A$, there exists $V_{x}\in\cali{N}_{X}(x)$ such that $V_{x}\ti Y\suf U$. Let $V=\cups{}{x\in A}V_{x}$, then $V\in\cali{N}_{X}(A)$ and $V\ti Y\suf U$.
$\square$
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