Let $E$ be a closed, bounded set and let $f(z)$ be a continuous complex function in $E$. Prove that $f(z)$ is bounded in $E$.

I began the argument the same way that the boundness theorem is addressed in Real Analysis (believing that the argument shouldn't change much).

Assume $f(z)$ is not bounded on $E$. Then $\forall n \in \mathbb{N}, \space\space\exists z_n \in E \space\space$ s.t. $\space |f(z_n)| > n$. Construct the sequence $(z_n)_{n=1}^\infty \subset E \space$ from these $z_n$.

Note that $(z_n)_{n=1}^\infty$ is bounded, as $E$ is bounded.

Then by Bolzano-Weierstrass, $(z_n)_{n=1}^\infty$ has a limit point $L$, and so there exists a subsequence $(z_{n_k})_{k=1}^\infty$ which converges to $L$. $\space$ Moreover, $\space L \in E \space$ since $E \space$ is a closed set.

This implies $\lim_{k \to \infty} f(z_{n_k}) = f(L)$, $\space$ and so $\lim_{k \to \infty} |f(z_{n_k})| = |f(L)| \space$ because $f$ is continuous on $E$, and $f(z)$ continuous implies $|f(z)|$ is continuous.

Now, in real analysis proofs, we say this is a contradiction, as $\lim_{k \to \infty} |f(z_{n_k})| = \infty \space$ by assumption. However, we have introduced the extended complex numbers in my complex analysis class, and I don't see why $|f(L)| \ne \infty$ would be necessary, which is the crux of the contradiction.

So my question is:

If working in the extended complex numbers, would this proof method fail as I believe it would, or am I missing something? If it does fail, does this also mean that the theorem would not be true, or just that this is an insufficient way to prove it?

  • $\begingroup$ I guess you do not call a set bounded if it contains $\infty$? $\endgroup$ – user99914 Sep 15 '17 at 0:58
  • $\begingroup$ @JohnMa: I am making the assumption that $L \in E$ is finite, and that $f(L) = \infty$. As I understand, the values in the image of $f$ do not necessarily relate to $E$ $\endgroup$ – infinitylord Sep 15 '17 at 1:02
  • $\begingroup$ you do not need to make the assumption the f(L) is infinite in the complex sense of infinity..just see my answer $\endgroup$ – Marios Gretsas Sep 15 '17 at 1:04

Your proof is fine because the absolute value of a complex number is a non-negative real number.

So indeed $|f(z_{n_k})| \to +\infty$

Note that if $f(z_{n_k}) \to f(L)$ then $|f(z_{n_k})| \to |f(L)|$

thus $|f(L)|$ must be finite because of the sequential continuity of $f$.

Now you can prove your statement using Heine-Borel Theorem.

Because from this theorem in $\mathbb{R}^n$ and $\mathbb{C}$ every closed and bounded subset of these spaces is compact and the converse.

Also $f(E)$ is compact because continuous functions preserve compactness so $f(E)$ is bounded.

  • $\begingroup$ You're absolutely right! I forgot about the very simple result that convergent sequences are necessarily bounded, which implies $|f(L)|$ is bounded. I debated using Heine-Borel Theorem as well, but I had just learned it today, and didn't quite feel comfortable applying it so readily. $\endgroup$ – infinitylord Sep 15 '17 at 1:10
  • $\begingroup$ @infinitylord also keep in mind that in general if $a_n \to a$ then $|a_n| \to |a|$ in the complex plane and in $\mathbb{R}^n$ under the euclideian norm $\endgroup$ – Marios Gretsas Sep 15 '17 at 1:14

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