If a function $f(z)$ is continuous on a closed bounded region $D$, then it is bounded in $D$

Question

I tried a proof like this:

Assume the contrary. Then to each $n\in\Bbb N$, there corresponds a $z_n\in D$ such that $|f(z_n)|≥n$ $(*)$. {$z_n$} has a limit point $\xi$. If {$z_n$} has finitely many elements $\xi$ is an element of {$z_n$} and hence belongs in $D$. Otherwise $\xi$ is a cluster point of $D$ and since $D$ is closed, $\xi\in D$.

Due to continuity of $f$ on $D$, $\exists\delta >0$ such that for $|z-\xi|<\delta$ $|f(z)-f(\xi)|<1$ $=>$ $|f(z)|<1+|f(\xi)|<m$, for some positive integer $m$. There is a subsequence {$z_{n_k}$} converging to $\xi$. To $\delta$ there corresponds an $N\in\Bbb N$ such that $\forall k≥N$, $|z_{n_k}-\xi|<\delta$ $=>$ $|f(z_{n_k})|<m$ $(**)$.

$r:=max(N,m)$. Then $(**)$ $=>$ $|f(z_{n_r})|<m$. But $(*)$ $=>$ $|f(z_{n_r})|≥n_r≥r≥m$. Thus we arrive at a contradiction.

Is this proof alright?

Answer 1

I'm going to assume that $f:\mathbb{C}\to\mathbb{C}$ and that $D\subset \mathbb{C}$.

Since $D$ is closed and bounded it is compact by Heine-Borel. Then since $f$ is continuous on a compact set the image $f(D)$ is compact, i.e. closed and bounded.

Answer 2

Your proof is correct. You have just demonstrated the first part of Heine-Borel Theorem (the next part it is proof that $f(D)$ is closed).