# Accumulation points of a sequence in a metric space

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.

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