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Let $U\subseteq \mathbb{R}^2$ be open, and $C\subset U$ be compact. Show there exists $V$ open and $D$ compact such that $C\subset V\subset D\subset U$.

My attempt : For each $x\in U$ consider balls $B(x,\epsilon_x)\subset U$. Now $\displaystyle C\subset \bigcup_{x\in U} B(x,\epsilon_x)$ so there is a finite subcollection $\{x_1,\dots ,x_k\}\subset U$ such that $\displaystyle C\subset \bigcup_{i=1}^{k} B(x_i,\epsilon_{x_i})=V$ say. Now since $C$ is compact in $\mathbb{R}^2$ it's closed and bounded. So $C\subseteq B(0,M)$ for some $M>0$. Now we can replace $U$ by $U\cap B(0,M)$ wlog. So we may assume $U$ is bounded. So if we can show $\bar{V}\subseteq U$ then we are done. Then we can take $D=\bar{V}$ and we have $D$ is compact. But I can't do this. Some help? Thanks.

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You can't prove that $\overline{V} \subset U$, because the identity $V = U$ might hold.

You need to modify how you choose your sets $B(x, \epsilon_x)$. Simply choose $\epsilon_x$ small enough, so that $x \in B(x, \epsilon_x) \subset \overline{B(x, \epsilon_x)} \subset U$ holds. The remainder of the proof stays the same. Observe that $\bigcup \limits_{i = 1}^n \overline{B(x_i, \epsilon_{x_i})} \subset U$ is a finite union of compact sets and thus compact.

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