Proposition 3.10.6.label Let $X$ be a Hausdorff space, $A,B\suf X$ be compact such that $A\cap B=\emp$ then there exists $U\in\cali{N}(A)$ and $V\in\cali{N}(B)$ such that $U\cap V=\emp$. In particular, if $X$ is also compact then $X$ is normal.
Proof. For each $x\in A,y\in B$, there exists $U_{x,y}\in \cali{N}(x),V_{x,y}\in\cali{N}(y)$ such that $U_{x,y}\cap V_{x,y}=\emp$. By compactness of $B$ there exists $B_{x}\suf B$ finite such that $\cups{}{y\in B_x}V_{x,y}\in\cali{N}(B)$. Let $U_{x}\define \caps{}{y\in B_x}U_{x,y}\in\cali{N}(x)$ and $V_{x}\define \cups{}{y\in B_x}V_{x,y}\in\cali{N}(B)$, then $U_{x}\cap V_{x}=\emp$. By compactness of $A$, there exists $A_{0}\suf A$ finite such that $\cups{}{x\in A_0}U_{x}\in\cali{N}(A)$. Then we can take $U=\cups{}{x\in A_0}U_{x}\in\cali{N}(A)$ and $V=\caps{}{x\in A_0}V_{x}\in\cali{N}(B)$ since
$\square$
Post a Comment