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

compactnesscomplex-analysisgeneral-topologysequences-and-series

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?

Best Answer

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 $\Omega$.
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