# Prove $f$ is bounded (Complex Analysis)

Could someone please help me prove the following and tell me their thought process as well if possible?

Let $f$ be a continuous real-valued function on the closed, bounded subset $S$ of the complex plane. Use the Bolzano-Weierstrass Theorem to show that $f$ is bounded.

The hint given was: "If this is not the case then there exists a sequence of points $z_n \in S$ such that $f(z_n)>n$.

Any kind of help will be much appreciated!

• Hint: A closed bounded subset of ${\Bbb C}$ is compact. Every sequence has a convergent subsequence. (IMO: It's not really complex analysis) – H. H. Rugh Aug 6 '16 at 10:50
• That should be $|f(z_n)| > n.$ – zhw. Aug 6 '16 at 22:20

## 1 Answer

Assume that $f$ is not bounded. Then let $z_n$ be the sequence from the hint. As noted by H. H. Rugh we have a convergent subsequence $z_{n_k} \to \tilde z \in S$. Now continuity gives us that $f(z_{n_k}) \to f(\tilde z)< N \in \mathbb N$. Then for sufficiently large $K\in \mathbb N$ we have that $f(z_{n_k})< N+1$, for all $k \geq K$. But this contradicts the hint. Thus our assumption is wrong.

• Thank you so much Martin for your help! Just one thing, how does $f(z_{n_k})<N+1$ contradict $f(z_n)>n$? – Faolan Aug 7 '16 at 1:12
• Because N is a fixed value so some $n_k >N+1$. – Martin Aug 7 '16 at 3:55