Suppose $A$ is compact and $U$ is open in $(X,d)$ such that $A\subset U$. Show that there exists $\epsilon >0$ such that $A\subseteq \bigcup_{a\in A }B_{\epsilon}(a)\subseteq U$ and $\{x\in X \mid d(x,A)<\epsilon\}\subseteq U$
Now $U^c$ is closed with $U^c \cap A =\phi$ so we must have $d(U^c,A)>0$
So $\exists \epsilon>0$ such that $d(x,y)\geq\epsilon \; \forall x\in U^c, y\in A$
Hence $\forall a\in A \rightarrow B_{\epsilon}(a)\cap U^c =\phi$
so that $A\subseteq \bigcup_{a\in A}B_{\epsilon}(a)\subseteq U$ [IS THIS CORRECT? Refer to this statement my $(1)$]
I think $\{x\in X\mid d(x,A)<\epsilon\}\subseteq U$ is an obvious implication of above statement? If not, please tell me.