# Checking simple complex analysis proofs, S compact iff S closed and bounded

I need help checking and going over my attempted proofs, I list two here, simply because they are related to each other. Please tell me if i did a mistake, or show me a nicer way to do it, still learning, and thanks for reading! I am using Serge Lang's Complex Analysis book

Let's start

Serge Lang's definition of accumulation point of an infinite set S:

Definition: a complex number v, such that given an open set U containing v, there exists infinitely many elements of S lying in in U

Definition of Compact Set:

Definition: a set of complex numbers S is compact if every sequence of elements of S has a point of accumulation in S.

I want to prove theorem 4.1 (which he does in the book, but i don't understand some of his ideas). The theorem states:

Theorem 4.1: A set of complex numbers is compact if and only if it is closed and bounded.

Before attempting this proof I proved something else that i used here to make my proof shorter.

I attempted a proof on this

Theorem A: A set S is closed if and only if it contains all of its accumulation points.

Proof A:

forward direction (Assume S is closed)

$S$ is closed, let $w$ be an acc. point of $S$, then either $w$ is an interior point, exterior point, or boundary point of $S$. If $w$ is either an interior point, or boundary point, and $S$ is closed, then $w$ is in $S$.

Assume $w$ is an exterior point, then $w$ has an open neighborhood that contains no elements of $S$, this contradicts the definition of acc. point. End of proof

backward direction (Assume S contains all its acc. points)

Let $B$ be a boundary point of $S$ then create this sequence $\{B\}:(B,B,..B)$, clearly the acc point of this random boundary point is $B$, and $S$ contains all of its acc. points, end of proof.

Now I can attempt to prove 4.1

Proof 4.1:

Forward (S is compact)

By the theorem we proved earlier $S$ is closed. Now assume $S$ is unbounded, then $\exists z\in S$ s.t. $\forall C>0 |z|>C$. Pick that one, call it z, then create this sequence $\{z_n\}:(|z|, 2|z|, 3|z|, ...,n|z|)$, and since $S$ is unbounded, we guarantee that each $z_n \in S$. Thus this sequence has no accumulation point, contradicting S being compact, thus S is bounded.

Backwards (S is closed and bounded)

Assume B is the bound $\forall z\in S$, given $z = x+iy$ then $|x+iy|<B \implies |x|<B \land |y|<B$ then by the Bolzano-Weierstrass theorem of real analysis the set $\{x : \forall z \in S\land z=x+iy\}$ and its $y$ equivalent have points of accumulation, and since S is closed, these points are in S. Hence S is compact.

This is it? are my steps right? reasoning? ... ? are there any redundancies?

• I think you have to use the definition of compactness: A set B is compact if for any open covers for B, there exist a finite sub cover. en.m.wikipedia.org/wiki/Compact_space. In fact, a complex space is a metric space which the proof of such property (Euclidean 2 space) in metric space is similar. We can view complex numbers as R^2 space. Hope that helps! – Novice Nov 27 '16 at 6:23
• While i get this topological approach, it's actually stated as theorem number 4.9 so i am guessing technically, i am not allowed to use the finite subcover technique yet, and in analysis i find these topological proofs harder since i am bound to use sequences and stuff. Thanks – zellwwf Nov 27 '16 at 6:34
• I cannot verify your reasoning because it is not clear what def'ns you are using for closed and for compact/ – DanielWainfleet Nov 27 '16 at 7:19
• The toplogical def'n of "$S$ is compact " is that whenever $F$ is a family of open sets with $S\subset \cup F$ there is a finite $G\subset F$ with $S\subset \cup G$.... If $S$ is unbounded let $F=\{B_n: n\in \mathbb N\}$where $B_n$ is the open disc of radius $n$, centered at $0$. Then no finite $G\subset F$ can satisfy $S\subset \cup G$ because for some $n\in \mathbb N$ we will have $\cup G\subset B_n$ and $S$ is not a subset of any $B_n$. – DanielWainfleet Nov 27 '16 at 7:31
• I added the books definition of compact, verbatim. – zellwwf Nov 27 '16 at 13:30