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Math Help - Metric Space

  1. #1
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    Metric Space

    Need help proving the following statements/giving counter-examples;

    (X,d) is a metric space

    1: for all a
    єX, U n=1 --> ∞, B(1/n, a) is an open set
    2:
    for all a
    єX, U n=1 --> ∞, B(1/n, a) is not an open set
    3: if xn --> x, then any subsequence of xn converges to x
    4: if any subsequence of xn converges to x, then xn converges to x
    5: if U is open then U=Int(Uconjugate)

    So I guess, 1 is true, 2 false, 3 true, 4 false, 5 no idea

    I'm generally stuck with it all, my idea for 1 is that in general you can always choose an
    ε greater than the radius of 1/n so it is open...is this right?
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  2. #2
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    Quote Originally Posted by math_help View Post
    Need help proving the following statements/giving counter-examples;

    (X,d) is a metric space

    1: for all a
    єX, U n=1 --> ∞, B(1/n, a) is an open set
    2:
    for all a
    єX, U n=1 --> ∞, B(1/n, a) is not an open set
    3: if xn --> x, then any subsequence of xn converges to x
    4: if any subsequence of xn converges to x, then xn converges to x
    5: if U is open then U=Int(Uconjugate)

    So I guess, 1 is true, 2 false, 3 true, 4 false, 5 no idea

    I'm generally stuck with it all, my idea for 1 is that in general you can always choose an
    ε greater than the radius of 1/n so it is open...is this right?
    I don't understand what you mean by "you can always choose \epsilon greater than 1/n so it is open". WHAT is open? For every n, of course B(1/n,a) is an open set because all such neighborhoods are open. But to prove that you would need to assert that you can always choose [tex]\epsilon[/itex] less than 1/n, not greater than. In any case, (1) and (2) if the question is whether they are true for all metrics, then that statement is false. \lim_{x\rightarrow a}B(1/n, a)= {a}. In the "discrete" metric (d(a,b)= 1 if a is not equal to b, 0 otherwise) all singleton sets are both open and closed so (2) is not true but if in, say, R, d(a,b)= |a- b|, then all singleton sets are closed but not open so (2) in that case is true.

    As far as (4) is concerned the English word "any" is ambiguous. I could see interpreting this as "If every subsequence... " in which case it is true. I could also see interpreting it as "if some subsequenc ..." in which case it is false.

    For (5), I don't know what "Uconjugate" means. Did you mean U complement? In that case it is certainly not true! Or did you mean "Uclosure"? In that case it is true.
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  3. #3
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    Sorry, never stated some things clearly.

    It simply says "everywhere (X,d) is a metric space" so i'd guess the question is implying it can be any metric space.

    For 1 and 2 I mean Union, but it looks like you knew that one. 5 isn't union and yes U compliment sorry, stupid mistake there.

    Yeah 4 is worded badly, my interpretation is that the question is "if there exists a of subsequence xn that converges to x, then xn also converges to x.

    So 3 is true, how can I go about proving this? and for number 4 i'm confused to the concept of a sequence in a metric space to be honest, naturally I want to say sin(k*pi/2) is subsequence of sin which converges for 1, where K is an integer, however sin doesn't converge, but can't get my head around it. Thanks in advance for any tips, and the original ones!
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  4. #4
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    For #3, do you fully understand the definition of subsequence?
    There must be a increasing sequence of positive integers, \alpha (n), such that \left\{ {x_{\alpha (n)} } \right\} \subseteq \left\{ {x_n } \right\}.
    If p is the limit point of x_n then for any open set O if p \in O then O contains almost all the terms of x_n.
    So is that not also true of any subsequence?
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  5. #5
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    Quote Originally Posted by math_help View Post
    Need help proving the following statements/giving counter-examples;

    (X,d) is a metric space

    1: for all a єX, U n=1 --> ∞, B(1/n, a) is an open set
    2: for all a
    єX, U n=1 --> ∞, B(1/n, a) is not an open set
    3: if xn --> x, then any subsequence of xn converges to x
    4: if any subsequence of xn converges to x, then xn converges to x
    5: if U is open then U=Int(Uconjugate)

    So I guess, 1 is true, 2 false, 3 true, 4 false, 5 no idea

    I'm generally stuck with it all, my idea for 1 is that in general you can always choose an
    ε greater than the radius of 1/n so it is open...is this right?

    Statement 4 is false ,as the following counter example shows:


    Take the sequence : 1,-1,1,-1,1,-1............................the sequence does not converge, it only has 1,-1 as accumulation points.

    But every subsequence of the sequence has the limit 1 or -1
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  6. #6
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    Quote Originally Posted by benes View Post
    Statement 4 is false ,as the following counter example shows: Take the sequence : 1,-1,1,-1,1,-1............................the sequence does not converge, it only has 1,-1 as accumulation points.
    But every subsequence of the sequence has the limit 1 or -1
    Sorry, but that is not a counterexample to the statement.
    Did you carefully read the statement if the statement?
    ďany subsequence of x_n converges to \color{blue}x.
    \color{blue}x is one number not two as in your so called example.

    I also disagree with an earlier post. Surely x_n is a subsequence of itself.
    So part 4 is true!
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  7. #7
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    Quote Originally Posted by Plato View Post
    Sorry, but that is not a counterexample to the statement.
    Did you carefully read the statement if the statement?
    ďany subsequence of x_n converges to \color{blue}x.
    \color{blue}x is one number not two as in your so called example.

    I also disagree with an earlier post. Surely x_n is a subsequence of itself.
    So part 4 is true!
    YES you right i misread the question,thank you.

    But which earlier post??
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