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

[mahler1]

1. Homework Statement .

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?

 

[Adrmay]

A:Let $x_1,x_2,\ldots$ be a countable infinite subset of $P$. Then $$X\setminus\bigcup_{n=1}^{\infty}B\left(x_n,\frac{1}{n}\right)$$ is a closed subset of $X$, and it must contain an element of $P$ (why?). This contradicts the fact that $x_n\in P$ for all $n$.