Prove $E_r = \{z \in \mathbb{C} \text{ }|\text{ } |z - w| \leq r \text{ for some } w \in E \}$ is compact, where $E$ is compact.

Question

I am reading a complex analysis book.

There is the following proposition in the book.

Let $E$ be a compact set in the complex plane $\mathbb{C}$.
Let $r$ be a positive real number.
Let $E_r = \{z \in \mathbb{C} \text{ }|\text{ } |z - w| \leq r \text{ for some } w \in E \}$.

Then, $E_r$ is a compact set.

Proof:

Since $E$ is compact, there exists $R > 0$ such that $E \subset D(0, R) := \{z \in \mathbb{C} | |z - 0| < R\}$.
Since $E_r \subset D(0, R + r)$, $E_r$ is bounded.
If $z \in \mathbb{C} - E_r$, then $|z - w| > r$ for all $w \in E$.
So, $\mathbb{C} - E_r$ is open.
So, $E_r$ is closed.

So, $E_r$ is compact.

But I don't think it is obvious that $\mathbb{C} - E_r$ is open.

So I proved that $\mathbb{C} - E_r$ is open as follows.

1. Let $E$ be a compact set in $\mathbb{C}$.
   $E \ni z \to |z - a| \in \mathbb{R}$ is continuous.

2. Let $\operatorname{dist}(z, E) := \min \{|z - w| | w \in E\}$.
   $\mathbb{C} \ni z \to \operatorname{dist}(z, E) \in \mathbb{R}$ is continuous.

3. Let $z_0 \in \mathbb{C} - E_r$.
   Then, $\operatorname{dist}(z_0, E) > r$.
   Since $\operatorname{dist}(z, E)$ is continuous, $|z - z_0| < \delta \implies r < \operatorname{dist}(z, E)$ for some posivitve real number $\delta$.
   So, $|z - z_0| < \delta \implies z \in \mathbb{C} - E_r$ for some posivitve real number $\delta$.

I don't think it is obvious that $\mathbb{C} - E_r$ is open.

Is this fact really obvious? We don't need to prove this fact?

Answer 1

The set $E_r$ is equal to $\{z\in\mathbb C\mid d(z,E)\leqslant r\}$. Since the map $z\mapsto d(z,E)$ is continuous and $E_r$ is the inverse image of $[0,r]$ by that map, $E_r$ is a closed set.

Of course, I must add that being obvious is subjective.

Answer 2

The criterium used is (probably) the following : a set $U$ is open iff for all $x\in U$, one can find an open neighborhood $V$ of $x$ that's contained in $U$.

So, $\mathbb{C} \setminus E_r$ is open since for any $z \in \mathbb{C} \setminus E_r$, the open ball of center $z$ and radius $r$ is contained in $\mathbb{C} \setminus E_r$.