# I Proof check: S in C Compact implies S is closed and bounded

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1. Nov 26, 2016

### zellwwf

I am using Lang's book on complex analysis, i am trying to reprove theorem 4.1 which is a simple theorem:
Let $$Compact(S \in \mathbb{C}) \iff Closed(S) \land Bounded(S)$$
I will show my attempt on one direction of the proof only, before even trying the other direction.

• Assume S is compact
• Idea for bounded: Show by contradiction that if S is unbounded then there will be a sequence with no accumulation point in S. (same idea i think of the book, but i don't get his method)
• Idea for closed: Show that given a boundary point, you could construct a sequence that is in S that converges to that boundary point, and then show it's a point of accumulation of the sequence. Since it's compact then the boundary point is in S, then it's closed.
The first
Assume S is compact, but not bounded.
1. Let the sequence $$\{Z_n\} := n |(w)|$$, where $$w$$ is an accumulation point for the set S. We can pick any multiplier we want, since S is unbounded.
2. Let the set $$E:=\{Z_n, Z_n\in Disk(center=w,rad=\epsilon=|w|)\}$$
3. S is finite since less and less points of Z_n as n grows
4. A contradiction, so S is bounded
The second
Assume S is compact, B a boundary point of S, now i want to create sequence that converges to B, and given it's compact then whatever it converges to is in S.
1. Let the sequence $$\{z_n\}:= \frac{B}{1+\frac{1}{n}}$$
2. Let the set $$E:= \{ z_n | |z_n-B|<\epsilon\}$$
3. Need to show that the set E is infinite, but i have no idea what to do

Come to think of this, i think both ideas are on the proper track logically, but the execution is terribly shitty, can someone tell me if 1 is right? and how do i show 2.3 given that the prior steps are correct.
I mean, in the second, how do i know that the sequence, defined as it is, is actually in S.

Where am i going wrong?

2. Nov 26, 2016

### Svein

Since S is compact and it is contained in $\bigcup_{n}(z;\lvert z\rvert <n)$, a finite number of those will cover S. Assume $S\subset\bigcup_{n\leq N}(z;\lvert z\rvert <n )$, then $S\subset(z;\lvert z\rvert <N)$, hence S is bounded.

Closed: Try to show that the complement of S is open...