If $K_\epsilon=\bigcup_{z\in K}\overline{D_\epsilon (z)}$, $K_\epsilon$ is compact.

Question

Let $\Omega$ be open and $K\subseteq\Omega$ be compact. If we define $K_\epsilon=\bigcup_{z\in K}\overline{D_\epsilon (z)}$ such that $K_\epsilon \subseteq \Omega$, then prove that $K_\epsilon$ is compact.

Answer 1

Assume $\Omega\subseteq\mathbb{C}$, so compactness is equivalent to being closed and bounded. In particular, $\overline{D_\varepsilon(0)}\subset\mathbb{C}$ is compact. Now $K_\epsilon$ is the image of $K\times\overline{D_\epsilon(0)}$ under the continuous map $+\colon\mathbb{C}\times\mathbb{C}\to\mathbb{C}$. Clearly $K\times\overline{D_\epsilon(0)}$ is compact (being the Cartesian product of two compacts), so $K_\epsilon$, as the continuous image of a compact set, is compact.

Answer 2

Your question is not wrong, it is unclear. You didn't say what is $\Omega$. The "yes" answer posted by @user10354138 is correct, assuming you meant that $\Omega$ is a subset of the complex plane (which seems a reasonable assumption, just because $\Omega$ and $z$ are likely to be used in a complex analysis text). Without this assumption, the answer need not be "yes".

Let $\mathbb Q$ denote the rational numbers with the usual metric and topology (inherited from the reals). Assuming that $\Omega\subseteq\mathbb Q$, then the answer is no, even if $K=\{q\}$ is a single point, since $\overline{D_\epsilon (q)}$ is never compact.

Answer 3

Let $L=\bigcup_{z\in K}\overline{D_\varepsilon}(z)$ and $\{w_n\}\subset L$. We need to show that $\{w_n\}$ possesses a convergent subsequence, with limit in $L$.

At first, there exists a sequence $\{z_n\}\subset K$, such that $w_n\in \overline{D_\varepsilon}(z_n)$, and hence $$ w_n=z_n+\zeta_n, \quad\text{for some $|\zeta_n|\le \varepsilon$.} $$ As $K$ is compact, $\{z_n\}$ possesses a convergent subsequence, say $\{z_{n_k}\}$, with $z_{n_k}\to z_0\in K$. Also, since $\{\zeta_{n_k}\}\subset \overline{D}_\varepsilon(0)$ and since $\overline{D}_\varepsilon(0)$ is also compact, then $\{\zeta_{n_k}\}$ also possesses a convergent subsequence, say $\{\zeta_{n_{k_j}}\}$, with $\zeta_{n_{k_j}}\to \zeta_0\in \overline{D}_\varepsilon(0)$, and hence $|\zeta_0|\le \varepsilon$.

Altogether $$ w_{n_{k_j}}=z_{n_{k_j}}+\zeta_{n_{k_j}}\to z_0+\zeta_0\in \overline{D}_\varepsilon(z_0)\subset L. $$ Indeed, every sequence in $L$ possesses a convergent subsequence, with limit in $L$.