Let $A$ be a compact set contained in a open set $\Omega$. Then there exists $\epsilon > 0$ such that $A \subset \cup_{x \in A} B(x,\epsilon) \subset \Omega$.
How can I prove it? I have tried it by proving compactness implies total boundedness. But that doesn't work properly.Please help me.
Thank you in advance.
Assume this isn't true. Then for every $n\in \mathbb{N} $ there exists some $x_n\in A$ such that $B (x_n,\frac{1}{n})\cap \Omega^c\neq \emptyset $. Consider the sequence $x_n $. By compactness of $A $, there exists a convergent subsequence of $x_n $. Call the limit $x $. Then $x\in A $, but $x $ is a boundary point of $\Omega $ and hence not an element of $\Omega $, so we have our contradiction.
Let $f(x)= d(x,\partial \Omega)$ defined on $A$. Since $A$ is compact and $f$ is continuous on $A$, $f$ has a minimum value $\epsilon_0\geq 0$. If $\epsilon_0 = 0$, there exists $x_0\in A$ such that $x_0 \in \partial \Omega$, which cannot be true by hypothesis.
Now, we set $\epsilon = \frac{\epsilon_0}{2}$, and we have $A \subset \cup_{x \in A} B(x,\epsilon) \subset \Omega$.
