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Every closed ball in X is complete, then X is complete

  1. Sep 21, 2013 #1
    The problem statement, all variables and given/known data.
    Prove that if every closed ball in a metric space X is a complete subspace, then X is complete.


    The attempt at a solution.

    Let ##\{x_n\}## be a cauchy sequence in X. Then, for ##ε=1##, ##\exists## ##n_0 \in \mathbb N##: ##\forall## ##m≥n_0##,##n≥n_0## ##d(x_m,x_n)<1##. Consider the closed ball centered at ##x_{n_0}## of radius 1. Then, the subsequence ##\{x'_n\}## such that ##x'_n=x_{n_0 + n-1}## is a subsequence of ##\{x_n\}## included in ##K(x_{n_0},1)## . By hypothesis, the ball is a complete subspace and the subsequence ##\{x'_n\}## is cauchy. But then ##\{x'_n\}## converges to some point ##x \in K(x_{n_0},1)## This implies ##\{x'_n\}## converges to the same point x in the metric space X. So the original sequence ##\{x_n\}## has a convergent subsequence that converges to the point x, this means that ##\{x_n\}## converges to x. I've proved that an arbitrary cauchy sequence in X is convergent, this implies every cauchy sequence in X is convergent, then X is a complete metric space.


    Is my proof correct?
     
  2. jcsd
  3. Sep 21, 2013 #2

    Dick

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    Why are you having doubts? Eventually the tail of every cauchy sequence is contained in a closed ball. The closed ball is complete. So the sequence is convergent.
     
  4. Sep 21, 2013 #3
    Thank you, I'm new to proofs and mathematics in general so I usually have strong doubts on whether I am doing the exercises right or not. I suppose it's a newbie's insecurity.
     
  5. Sep 21, 2013 #4

    Dick

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    Well, I thought that was a pretty fine and clear proof. I think you should start feeling less insecure.
     
  6. Sep 22, 2013 #5
    The only thing is that you don't want to say ε=1.

    Because your metric space may be "smaller" than the unit closed ball.

    You could just say for some ε. But the proof seems fine
     
  7. Sep 22, 2013 #6

    Dick

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    A 'ball' in a metric space is just the set of all points x in the space such that |d(x,x0)|<=R. It doesn't have to be spherical. The proof really doesn't need any qualification.
     
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