Question on compact sets in $\mathbb{R}^2$ Let $\Omega$ be an open set in $\mathbb{R}^2$, $K \subseteq \Omega$ and K be compact.
Prove that there exists $r>0$ such that $E=\bigcup_{z \in K} \bar{D}(z,r)$ is a compact subset of $\Omega$, where $\bar{D}(z,r)$ is the closed disc centred at $z$ with radius $r$.
I found Rudin using this fact in one of the proofs in his 'Real and Complex Analysis', however I am having difficult proving this fact is true.
My attempt: Let $2r$ be the minimum distance from the boundary of $\Omega$ to K, then $D(z,r)$ will all lie in $\Omega$. Since the union of $D(z,r)$ covers K, we can reduce the collection into a finite covering of K, say $D(z_1,r),D(z_2,r),...,D(z_N,r)$ Then $\bigcup_{n=1}^{N} \bar{D}(z_n,r)$ will be a compact subset of $\Omega$. However this set is not E, and it is necessary for the result to work for E for the proof to work.
Any help is appreciated!
 A: Define $f(z)= d(z,K).$ Then $f$ is continuous on $\mathbb {R}^2.$ Note that for every $z \in \mathbb {R}^2,$ there exists a $w\in K$ such that $f(z) = d(z,w)$ (because a real continuous function on a compact set attains a minimum value). From this you can verify that
$$E_r = \cup _{z\in K}\overline {D(z,r)} =\{z \in \mathbb {R}^2: d(z,K)\le r\}.$$ 
This shows each $E_r,$ which equals $f^{-1}([0,r]),$ is closed. It's also bounded (by $[\sup_{z\in K} |z|]+r$). So each $E_r$ is compact. To finish, note that the compact set $K$ and the closed set $\Omega ^c$ are disjoint. Hence (standard exercise), $d(K,\Omega ^c) = s$ for some $s>0.$ Setting $r=s/2,$ we see $E_r$ has the desired properties.
A: Choosing $r$ as you did, we clearly have $E\subset \Omega$. It remains to see that $E$ is compact. In view of the Heine-Borel-Lebesgue theorem, we need to see that $E$ is closed and bounded.
Since $K$ is compact, it is bounded, say $K \subset D_R(0)$. Then we have $\overline{D}_r(z) \subset D_{R+r}(0)$ for all $z\in K$, whence $E\subset D_{R+r}(0)$, so $E$ is bounded.
We have $w \in \overline{D}_r(z) \iff \lvert w-z\rvert \leqslant r \iff z \in \overline{D}_r(w)$, so
$$\mathbb{C}\setminus E = \{ w : K \cap \overline{D}_r(w) = \varnothing\}.$$
If $K\cap \overline{D}_r(w) = \varnothing$, then $s := \min \{ \lvert z-w\rvert : z \in K\} > r$. By the triangle inequality, we thus have $K\cap \overline{D}_r(u) = \varnothing$ for all $u \in D_{s-r}(w)$, i.e. $D_{s-r}(w)\subset \mathbb{C}\setminus E$. So $\mathbb{C}\setminus E$ is open, and the proof that $E$ is compact is complete.
Along the lines you started on, you generally can't write $E$ as a finite union of compact disks, but you can cover it by a finite family of compact disks of radius $s > r$. You start by covering $K$ by a finite family of (open) disks of radius $s-r$, and then consider the finite family of disks with radius $s$ and the same centres $x_k$. By the triangle inequality, that family covers $E$: If $z \in K$, there is a $k$ such that $\lvert z- x_k\rvert < s-r$, and hence $\overline{D}_r(z) \subset D_s(x_k)$. That shows that $E$ is relatively compact in $\Omega$, but to see that it is closed, you still need an argument like above.
A: Construct the set $E$ as in your attempt and consider arbitrary sequence $\{e_1,e_2,\ldots\}$ of points from $E$. Then, $e_i=z_i+\rho_ie^{\alpha_i}$ with $z_i\in K$, $\rho_i\in[0,r]$ and $\alpha_i\in[0,2\pi]$. By compactness, there is a subsequence $i_t$ such that $z_{i_t}\rightarrow z\in K$, $\rho_{i_t}\rightarrow \rho\in [0,r]$ and $\alpha_{i_t}\rightarrow \alpha\in [0,2\pi]$. It remains to note that $e=z+\rho e^{\alpha}\in E$ is a limit point of $\{e_1,e_2,\ldots\}$.
A: I think we can adapt the Lebesgue Number Lemma here:
$\Omega$ is open so for each $x\in K$ there is an open ball $B(x,r_{x})$ containing $x$ and contained in $\Omega$. 
Since $K$ is compact, we can find a finite subcover of $K$, say $\left \{ B(x_{i},r_{i}) \right \}^{n}_{i=1}$. 
Define $h:K\to \mathbb R$ by $h(x)=\frac{1}{n}\sum_{i=1}^{n}d(x,(K \setminus B(x_{i},r_{i})))$. Then $h>0$ and $h$ is continuous so it has a minimum value $\delta >0$.
Now let $E=\bigcup_{z\in K}\overline B(z,\delta )$ and consider any $\overline B(z_{0},\delta )\subseteq E$. 
Then 
$\delta \leq h(z_{0})\leq d(z_{0},(K \setminus B(x_{j},r_{j})))$, 
where $d(z_{0},(K \setminus B(x_{j},r_{j})))$ is the largest of the numbers $\left \{ d(z_{0},(K \setminus B(x_{i},r_{i}))) \right \}^{n}_{i=1}$ 
Thus,
$\overline B(z_{0},\delta )\subseteq B(x_{i},r_{i})\subset \Omega$ 
so that $E$ has the required property. 
