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Math Help - complex analysis

  1. #1
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    complex analysis

    Suppose A and B are nonempty closed bounded subsets of \mathbb{C}.
    1. Show that there exist a_0 \in A, b_0 \in B such that |a_0-b_0|=\inf\{|a-b|:a\in A, b\in B\}.
    2. Show that this is not true if the boundedness condition on A and B is dropped.
    3. What if only one of A or B is bounded but both are still closed?

    Question about #1, if A, B are closed and bounded, does this mean every sequence in A, B is bounded?

    Also, I'm clueless about #2 and #3. Some help please.
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  2. #2
    Member mabruka's Avatar
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    About your question of #1 the answer is yes,


    think about it:

    You have a bounded set A, u take a sequence there, say \{ x_n \}_n \subset A.


    A is bounded means that there is a bound M such that:

    |y| \leq M \forall y \in A

    even for those elements in your sequence so


    |x_n| \leq M \forall n \in \mathbb N

    You dont even need the sets to be closed for this.
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  3. #3
    Member mabruka's Avatar
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    About #2 think of this pair of sets:


    <br />
A = \{x+iy \in \mathbb C : y\geq \frac{1}{x} for x>0 \}

    and take B=\{x+iy :  y\leq 0 \}

    Both sets are closed but unbounded. And this
    |a_0-b_0|=\inf\{|a-b|:a\in A, b\in B\}<br />
does not hold.

    If by any chance why it does not hold remains unclear, draw both sets and you will see.
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  4. #4
    Member mabruka's Avatar
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    #3 Must be true. I think #1 proof should work with a slight modification.
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  5. #5
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    I am stuck in #3. For #1, I used the fact that both A,B are bounded so that I have two bounded sequences, and then apply the Bolzano-Weierstrass Theorem.
    How can I prove that it is true with only A or  B bounded?
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by mabruka View Post
    About your question of #1 the answer is yes,


    think about it:

    You have a bounded set A, u take a sequence there, say \{ x_n \}_n \subset A.


    A is bounded means that there is a bound M such that:

    |y| \leq M \forall y \in A

    even for those elements in your sequence so


    |x_n| \leq M \forall n \in \mathbb N

    You dont even need the sets to be closed for this.
    You do! What would a_0, b_0 be if the two sets were the open disc |z|<1 and the open disc |z-2|<1?
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  7. #7
    Member mabruka's Avatar
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    Quote Originally Posted by Bruno J. View Post
    You do! What would a_0, b_0 be if the two sets were the open disc |z|<1 and the open disc |z-2|<1?
    Every sequence in any of those sets is bounded of course!

    I was answering his question about problem #1:

    Quote Originally Posted by dori1123
    Question about #1, if are closed and bounded, does this mean every sequence in is bounded?
    It is enough the set be bounded so that every sequence in there is bounded.



    Of course you need them to be closed for the existence of a_0, b_0.

    Sorry about the confusion.
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