# A theorem of compact set.

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.

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.
• Why does $x \in \partial A$? Commented Oct 20, 2017 at 13:12
• @ArnabChatterjee. What is important is that it is in the boundary of $\Omega$. To see that that is true, note that every open ball containing $x$ also contains infinitely many $x_n$ by definition. Then taking open balls of radius small enough gives the result. Commented Oct 20, 2017 at 14:05
• What I thought is that if $x \in A$ then clearly $x \in \Omega$. Then there exists $\delta>0$ such that $B(x;\delta) \subset \Omega$. Now if $\{x_{k_n} \}$ be the subsequence converging to $x$ then there exists $p \in \mathbb N$ such that for all $n \geq p$, $d(x_{k_n},x) < \delta$. Take $q \geq p$ such that $\frac {1} {k_q} < \delta$. Then if I take any $y \in B({x_{k_q}}, {\frac {1} {k_q}}) \cap \Omega^{c}$ then by triangle inequality $d(x,y) < 2\delta$ and it is true for any $\delta' < \delta$. Hence $x=y$. But then $x \in \Omega^c$!! Commented Oct 20, 2017 at 15:36
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 $A \subset \cup_{x \in A} B(x,\epsilon) \subset \Omega$.