# If for every $\epsilon>0$ there exist infinitely many $n$ such that $d(x_n,c)<\epsilon$ then $(x_n)$ has a subsequence converging to $c$.

Suppose $(X,d)$ is a metric space. I am trying to show that:

If for every $\epsilon>0$ there exist infinitely many $n$ such that $d(x_n,c)<\epsilon$ then $(x_n)$ has a subsequence converging to $c$ where $c\in X$.

Is this proof fine?

For each $k\in \mathbb{N}$ we can find $x_{n_k}$ such that $d(x_{n_k},c)<1/k$. Letting $k \rightarrow \infty$ we get that $d(x_{n_k},c)\rightarrow 0$ meaning that $(x_{n_k})$ is a sub-sequence of $(x_n)$ that converges to $c$.

• Looks fine to me. – DonAntonio May 20 '16 at 13:20
• That's basically correct, although to be more fully correct you should arrange that $n_k$ is strictly increasing as a function of $k$, that being a usual requirement for a subsequence. – Lee Mosher May 20 '16 at 13:21

You are right. Since $\{B(c,\frac1n): n \in Z^+\}$ is a neighbourhood base of $c$, then $x_{n_k}$ converges to $c$.