§1.1 Topological Preliminaries

1. We first prove that closedness implies the inclusion of accumulation points. Let $𝑋$ be a closed set, and assume that some accumulation point ${𝑧}_0$ of $𝑋$ satisfies ${𝑧}_0\notin 𝑋$. Since $𝑋$ is closed, ${ℂ}^{𝑛}\setminus 𝑋$ is open, and there exists an open ball ${𝐵}^{𝑛}\left({𝑧}_0,𝜀\right)$ that is fully contained in ${ℂ}^{𝑛}\setminus 𝑋$. However, this contradicts the definition of accumulation point, since ${𝐵}^{𝑛}\left({𝑧}_0,𝜀\right)\setminus \left\{{𝑧}_0\right\}$ does not intersect with $𝑋$.
2. Assume $𝑋$ is an arbitrary set which includes all its accumulation points. We will show that ${ℂ}^{𝑛}\setminus 𝑋$ is open, which implies that $𝑋$ is closed. Let ${𝑧}_0\in {ℂ}^{𝑛}\setminus 𝑋$. Since ${𝑧}_0$ is not an accumulation point of $𝑋$ (as otherwise ${𝑧}_0\in 𝑋$), there exists an open ball ${𝐵}^{𝑛}\left({𝑧}_0,𝜀\right)$ such that $\left({𝐵}^{𝑛}\left({𝑧}_0,𝜀\right)\setminus \left\{{𝑧}_0\right\}\right)\cap 𝑋=⌀$. Then, ${𝐵}^{𝑛}\left({𝑧}_0,𝜀\right)\subset {ℂ}^{𝑛}\setminus 𝑋$, and hence ${ℂ}^{𝑛}\setminus 𝑋$ is open. $\square$

We will first show that any compact set is closed and bounded.

First we will show that $𝑋$ is bounded. Suppose that $\forall 𝑅>0$, $\exists 𝑧\in 𝑋$ where $\Vert 𝑧\Vert >𝑅$. Consider the collection of open sets

$\mathrm{𝒰︀}$ forms an open cover of $𝑋$. Then by the assumption that $𝑋$ is compact, there exists a finite subcover in $\mathrm{𝒰︀}$, namely $\left\{{𝐵}^{𝑛}\left(0,{𝑘}_1\right),\dots ,{𝐵}^{𝑛}\left(0,{𝑘}_{𝑚}\right)\right\}$ which covers $𝑋$. Then,

By contradiction, $𝑋$ must be bounded.

$𝑋$ must also be a closed set. For the sake of contradiction, assume that there exists a point ${𝑧}_0\in \overline{𝑋}\setminus 𝑋$. Since ${𝑧}_0\notin 𝑋$, the following open collection of sets covers $𝑋$:

There then exists a finite subcover $\mathrm{𝒞︀}=\left\{{ℂ}^{𝑛}\setminus \overline{{𝐵}^{𝑛}\left({𝑧}_0,\frac{1}{{𝑘}_{𝑗}}\right)}|𝑗\in {\mathbb{N}}_{\le 𝑚}\right\}$. Then,

and that $𝑋\cap \stackrel{‾}{{𝐵}^{𝑛}\left({𝑧}_0,\frac{1}{max\left\{{𝑘}_1,\dots ,{𝑘}_{𝑚}\right\}}\right)}=⌀$. However, by the definition of the accumulation point, every open neighborhood of the accumulation point must intersect $𝑋$. Therefore, by contradiction, $𝑋$ is closed.

We then prove the converse. By the assumption that $𝑋$ is bounded, $\exists 𝑅>0$ such that $𝑋$ is contained within the closed cube

Assume that there exists an infinite open cover $\mathrm{𝒰︀}$ of $𝑋$ without finite subcovering.

Bisect ${𝑄}_0=𝑄$ into $2^{2𝑛}$ sub-cubes (for real and complex parts), choose ${𝑄}_1$ such that ${𝑄}_1\cap 𝑋$ has no finite subcover of $\mathrm{𝒰︀}$.

Under the previous assumptions, this is possible since if every $\text{sub-cube}\cap 𝑋$ had finite subcovering, then ${𝑄}_0\cap 𝑋=𝑋$ would have finite subcovering. Similarly, choose ${𝑄}_2$ by bisecting ${𝑄}_1$ similarly, and recursively obtain a sequence of cubes:

Since the side length of each cube tends to $0$, ${\bigcap }_{𝑗=0}^{\infty }{𝑄}_{𝑗}$ consists of a single point ${𝑧}_{\infty }\in {ℂ}^{𝑛}$. Since $𝑋$ is closed, ${𝑧}_{\infty }\in 𝑋$, and since $\mathrm{𝒰︀}$ covers $𝑋$, $\exists 𝑈\in \mathrm{𝒰︀}$ such that ${𝑧}_{\infty }\in 𝑈$. Since $𝑈$ is open, $\exists 𝜀>0$ such that ${𝐵}^{𝑛}\left({𝑧}_{\infty },𝜀\right)\subseteq 𝑈$. $\exists 𝑁\in \mathbb{N}$ such that $\forall 𝑘>𝑁$, ${𝑄}_{𝑘}\subset {𝐵}^{𝑛}\left({𝑧}_{\infty },𝜀\right)$. Then taking the intersection with $𝑋$ on both sides,

This contradicts the assumption that for every $𝑘$, ${𝑄}_{𝑘}\cap 𝑋$ has no finite subcovering, since $\left\{𝑈\right\}\subset \mathrm{𝒰︀}$ clearly covers ${𝑄}_{𝑘}\cap 𝑋$, as it is a single open set that covers a nonempty subset. Therefore by contradiction, every open cover has finite subcovering. ⁠$\square$

Since $𝑋$ is compact, it is bounded by Theorem 1.1.9$\text{<a href="../../pdf/notes.pdf\#loc-11">Theorem 1.1.9</a>}\text{<a href="\#thm-theorem-1-1-9">Theorem 1.1.9</a>}$, so there exists a closed cube $𝑄\subset {ℂ}^{𝑛}$ such that $𝐴\subseteq 𝑋\subset 𝑄$.

Bisect ${𝑄}_0=𝑄$ into $2^{2𝑛}$ congruent sub-cubes. Since $𝐴$ is infinite and the sub-cubes are finite in number, at least one of the sub-cubes contains infinitely many points of $𝐴$, and choose one to be ${𝑄}_1$.

Bisect ${𝑄}_1$ into $2^{2𝑛}$ sub-cubes, and choose a sub-cube ${𝑄}_2\subset {𝑄}_1$ that contains infinitely many points of $𝐴$. We then obtain the recursive sequence

Because the side lengths shrink to zero and the cubes are nested, the intersection

consists of exactly one point. Call this point ${𝑧}_{\infty }\in {ℂ}^{𝑛}$.

For each $𝑘$, ${𝑄}_{𝑘}$ contains infinitely many points of $𝐴$. Because the side length of ${𝑄}_{𝑘}$ tends to zero, for any $𝜀>0$, $\exists 𝑁\in \mathbb{N}$ such that $\forall 𝑘\ge 𝑁$, ${𝑄}_{𝑘}\subset {𝐵}^{𝑛}\left({𝑧}_{\infty },𝜀\right)$.

Then, ${𝐵}^{𝑛}\left({𝑧}_{\infty },𝜀\right)$ also contains infinitely many points of $𝐴$. Therefore, ${𝑧}_{\infty }$ is an accumulation point of $𝐴$.

We now show that ${𝑧}_{\infty }\in 𝑋$. Suppose for contradiction that ${𝑧}_{\infty }\notin 𝑋$. Since $𝑋$ is compact, it is closed by Theorem 1.1.9$\text{<a href="../../pdf/notes.pdf\#loc-11">Theorem 1.1.9</a>}\text{<a href="\#thm-theorem-1-1-9">Theorem 1.1.9</a>}$, thus ${ℂ}^{𝑛}\setminus 𝑋$ is open, and $\exists 𝛿>0$ such that

But then, for sufficiently large $𝑘$, we have ${𝑄}_{𝑘}\subset {𝐵}^{𝑛}\left({𝑧}_{\infty },𝛿\right)$, and hence ${𝑄}_{𝑘}\cap 𝑋=⌀$. This contradicts the construction of ${𝑄}_{𝑘}$, which ensures that ${𝑄}_{𝑘}$ contains infinitely many points of $𝐴\subset 𝑋$. ⁠$\square$