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

In summary, the conversation discusses using Lang's book on complex analysis to reprove theorem 4.1, which states that Compact(S) is equivalent to Closed(S) and Bounded(S). One direction of the proof involves showing that if S is unbounded, then there will be a sequence with no accumulation point in S, and if S is a boundary point, then a sequence can be constructed that converges to that boundary point. The conversation also discusses potential issues with the execution of these ideas and how to show that the sequence is actually in S. Ultimately, it is determined that the first idea is correct and the second idea needs further clarification.
  • #1
zellwwf
34
0
[tex] [/tex]I am using Lang's book on complex analysis, i am trying to reprove theorem 4.1 which is a simple theorem:
Let [tex] Compact(S \in \mathbb{C}) \iff Closed(S) \land Bounded(S)[/tex]
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 [tex] \{Z_n\} := n |(w)|[/tex], where [tex]w[/tex] is an accumulation point for the set S. We can pick any multiplier we want, since S is unbounded.
  2. Let the set [tex] E:=\{Z_n, Z_n\in Disk(center=w,rad=\epsilon=|w|)\} [/tex]
  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 [tex] \{z_n\}:= \frac{B}{1+\frac{1}{n}} [/tex]
  2. Let the set [tex]E:= \{ z_n | |z_n-B|<\epsilon\}[/tex]
  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?
 
Mathematics news on Phys.org
  • #2
Since S is compact and it is contained in [itex]\bigcup_{n}(z;\lvert z\rvert <n) [/itex], a finite number of those will cover S. Assume [itex]S\subset\bigcup_{n\leq N}(z;\lvert z\rvert <n )[/itex], then [itex] S\subset(z;\lvert z\rvert <N)[/itex], hence S is bounded.

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

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

1. What is the definition of a compact set?

A compact set is a set of points in a metric space that is closed and bounded. This means that every sequence in the set has a convergent subsequence whose limit is also in the set.

2. Why does a compact set imply closed and bounded?

A compact set is defined as a set that has every sequence in it converging to a point in the set. This means that the set must contain all of its limit points, making it closed. Additionally, since all sequences in the set have a limit point in the set, the set must be bounded.

3. Can you provide an example of a compact set that is not closed or bounded?

No, a compact set must be both closed and bounded. If a set is not closed, then it cannot contain all of its limit points, and if it is not bounded, then there exist sequences that do not have a limit point in the set. Therefore, a compact set must be closed and bounded.

4. Is every closed and bounded set a compact set?

No, a set can be closed and bounded without being compact. For example, the set of rational numbers between 0 and 1 is closed and bounded, but it is not compact because it does not contain all of its limit points.

5. How is the compactness of a set related to its topology?

The compactness of a set is a topological property, meaning it is dependent on the topology of the space the set is in. In general, a set is compact in a topological space if and only if it is closed and bounded in that space.

Back
Top