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

  1. Nov 26, 2016 #1
    [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?
  2. jcsd
  3. Nov 26, 2016 #2


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    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...
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