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Complete metric space can't have a countably infinite perfect space

  1. Sep 25, 2013 #1
    1. The problem statement, all variables and given/known data.

    Let ##(X,d)## be a complete metric space. Prove that if ##P \subset X## is perfect, then P is not countably infinite.

    3. The attempt at a solution.

    Well, I couldn't think of a direct proof, I thought that in this case it may be easier to assume is countably infinite and arrive at an absurd conclusion but I have "technical"problems with my proof.

    So, suppose P is countably infinite. Let's take an arbitrary ##x \in P## and denote it by ##x_1##. Let ##ε_1>0## and consider the ball ##B_1(x_1, ε_1)##. Now, P is perfect, so there must be at least one (in fact, infinite) ##x_2≠x_1## such that ##x_2## is in ##B_1##. Let r be the distance between ##x_1## and ##x_2## and consider the ball centered at ##x_2## of radius ##ε_2=r/2## and denoted by ##B_2##. Again, I can pick another element ##x_3## from ##B_2## because none of the elements of P are isolated. I take ##ε_3## to be r'/2 where r' is the distance between ##x_2## and ##x_3## and take the ball centered at ##x_3## of radius ##ε_3##. Keeping in this way, I construct the ball ##B_n## centered at ##x_n##, of radius ##ε_n=(d(x_{n-1}, x_n))/2##.
    Observe that P is a closed metric space in X, so P is also complete. The balls ##B_n## satisfy that ##B_{n+1}## ##\subset B_n##; to see this, take x in ##B_{n+1}## , then, ##d(x,x_n)≤d(x,x_{n+1}+d(x_{n+1},x_n)##... Well, in this part, I got totally stuck, I draw the picture of the balls and it is obvious that the ball B_n is contained in all the previous balls, but I don't know how to prove it formally. So, suppose I could prove that part, then we also have that the diameters of the balls ##δ(B_n)=2ε_n<2ε_1/2^{n-1}##, which tends to 0 when n→∞. Then I am under the hypothesis of Cantor's theorem, which says that given this conditions, there is an ##x \in## X such that for all ##ε>0## ##B_n(x_n,ε_n) \cap B(x,ε)≠0## ##\forall n##. Moreover, the exists ##n_0(ε)## such that ##B_n \subset B(x,ε)##. We can see that this means x must be in an accumulation point of P. Since P is perfect, x has to be in P.

    I would love to say that x can't be in P by the way I constructed the balls but I'm not at all sure this is true. I think that I'm not going in the right way doing all these, but were the only ideas I got. Another idea I have is trying to construct a Cauchy sequence in P such that the limit of the sequence is not in P, but again, how can I construct the wanted sequence? Another thing I noticed is that in my previous attempt of a proof I really didn't use the fact that P is countably infinite, I just used the fact that P is infinite and perfect.

    Can someone explain me another way of proving the statement or how could I improve this attempt of a proof?
     
  2. jcsd
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