The set $\Omega \subset \mathbb{C}$ is compact iff every sequence $\{ z_n \} \subset \Omega$ has a subsequence that converges to a point in $\Omega$

Question

My textbook Complex Analysis by Stein and Shakarchi presents the following theorem:

The set $\Omega \subset \mathbb{C}$ is compact if and only if every sequence $\{ z_n \} \subset \Omega$ has a subsequence that converges to a point in $\Omega$.

After many hours of unsuccessfully trying to prove this, I found the following proof here:

Every sequence in a closed and bounded subset is bounded, so it has a convergent subsequence, which converges to a point in the set, because the set is closed.

Conversely, every bounded sequence is in a closed and bounded set, so it has a convergent subsequence.

The problem is that I don't see how it is possible to conceive of the same proof using only the information my textbook had given me:

The interior of $\Omega$ consists of all its interior points. Finally, a set $\Omega$ is open if every point in that set is an interior point of $\Omega$.

A set is closed if its complement $\Omega^c = \mathbb{C} - \Omega$ is open. This property can be reformulated in terms of limit points. A point $z \in \mathbb{C}$ is said to be a limit point of the set $\Omega$ if there exists a sequence of points $z_n \in \Omega$ such that $z_n \not= z$ and $\lim_{n \to \infty} z_n = z$. The reader can now check that a set is closed if and only if it contains all its limit points. The closure of any set $\Omega$ is the union of $\Omega$ and its limit points, and is often denoted by $\bar{\Omega}$.

Finally, the boundary of a set $\Omega$ is equal to its closure minus its interior, and is often denoted by $\partial \Omega$.

A set $\Omega$ is bounded if there exists $M > 0$ such that $|z| < M$ whenever $z \in \Omega$. In other words, the set $\Omega$ is contained in some large disc. If $\Omega$ is bounded, we define its diameter by

$$\text{diam}(\Omega) = \sup_{z, w \in \Omega} |z - w|.$$

A set $\Omega$ is said to be compact if it is closed and bounded.

Using this information, and despite many hours of effort, I was unable to prove the aforementioned theorem. Keep in mind that I have just begun learning complex analysis, so the information in this textbook is the full extent of my understanding.

I was wondering if someone could please show me how the aforementioned theorem could be proved using the information from my textbook. Or is the information in my textbook insufficient?

Answer 1

Let $\Omega$ be a compact set and let $(z_n)_{n\in\mathbb N}$ be a sequence of elements of $\Omega$. Since $\Omega$ is bounded, the sequence $(z_n)_{n\in\mathbb N}$ is bounded and so the sequence $\bigl(\operatorname{Re}(z_n)\bigr)_{n\in\mathbb N}$ has a convergent subsequence $\bigl(\operatorname{Re}(z_{n_k})\bigr)_{k\in\mathbb N}$, by the Bolzano-Weierstrass theorem; let $x$ be its limit. By the same reason, the sequence $\bigl(\operatorname{Im}(z_{n_k})\bigr)_{k\in\mathbb N}$ has a convergent subsequence; let $y$ be its limit. But then $(z_n)_{n\in\mathbb N}$ has a subsequence which converges to $x+yi$. Since $\Omega$ is closed, $x+yi\in\Omega$.

Now, suppose that $\Omega$ is not compact. There are two possibilities:

1. $\Omega$ is not closed. Then take $z\in\overline\Omega\setminus\Omega$. Since $z\in\overline\Omega$, it is the limit of a sequence $(z_n)_{n\in\mathbb N}$ of elements of $\Omega$. Then every subsequence of $(z_n)_{n\in\mathbb N}$ converges to $z$ and therefore no subsequence of $(z_n)_{n\in\mathbb N}$ converges to an element of $\Omega$.
2. $\Omega$ is unbounded. Then, for each $n\in\mathbb N$, take $z_n\in\Omega$ such that $\lvert z\rvert\geqslant n$. Then no subsequence of $(z_n)_{n\in\mathbb N}$ converges and, in particular, no subsequence of $(z_n)_{n\in\mathbb N}$ converges to an element of $\Omega$.