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.


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?


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.

  • $\begingroup$ Re your last paragraph: ... obviuosly $\endgroup$ Sep 27 '19 at 13:23

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 $\mathbb{C} \setminus E_r$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.