I am working with sequences in metric spaces and I think the following may happen.

Let $(X,d)$ a complete and separable metric space, if $x \in X'$, we know that exist a sequence $(x_n)_{n \in \mathbb{N}}$ such that $x_n \to x$, I want to prove that $x_n \in X'$ for infinite elements of the sequence.

I tried to prove by reduction to the absurd assuming that there are infinite isolated points but I do not arrive at anything. Can you help me to demonstrate or refute this idea.

  • $\begingroup$ Any assumption on $X’$? $\endgroup$ – User May 28 at 2:42
  • $\begingroup$ $X$ is complete and separable. I'll edit it. $\endgroup$ – TresTresUno May 28 at 2:45
  • $\begingroup$ Ok, but I meant $X’.$ It is a subset of $X$ right? Do you have any assumption on it? Otherwise we may just take $X’=\{x\}.$ $\endgroup$ – User May 28 at 2:49
  • $\begingroup$ I have not assumption of $X'$. $\endgroup$ – TresTresUno May 28 at 2:53
  • $\begingroup$ @TresTresUno: By $X'$ you mean the derived set of $X$, i.e., the set of non-isolated points of $X$? $\endgroup$ – Brian M. Scott May 28 at 2:53

What you’re trying to prove is false. Let $X=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$ with the usual Euclidean metric; this a complete, separable metric space, $X'=\{0\}$, and the sequence $\left\langle\frac1n:n\in\Bbb Z^+\right\rangle$ converges to $0$ but is contained entirely in $X\setminus X'$.

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  • $\begingroup$ Of course, thank you very much. $\endgroup$ – TresTresUno May 28 at 3:01
  • $\begingroup$ @TresTresUno: You’re very welcome. $\endgroup$ – Brian M. Scott May 28 at 3:02

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