Let $(x_n)$ be a Cauchy sequence in a metric space $(X, d)$. Prove that the closure of $\lbrace x_1, x_2, x_3, . . . \rbrace$ in $X$ is countable.

Let $(x_n)$ be a Cauchy sequence in a metric space $(X, d)$. I need to show that the closure of $\lbrace x_1, x_2, x_3, . . . \rbrace$ in $X$ is countable. To show this, I need to prove that $\lbrace x_1, x_2, x_3, . . . \rbrace$ has at most one limit point.

Assume that $x\in X$ and $\mathscr X \in X$ are any two limit points of the set $S$. We need to show that $x = \mathscr X$. By definition of limit points, we have that $(\forall \varepsilon_1 >0)(\exists x_n \in S) ((x_n \neq x) \land (d(x_n,x)<\varepsilon_1))$ and $(\forall \varepsilon_2 >0)(\exists x_m \in S) ((x_m \neq \mathscr X) \land (d(x_m,\mathscr X)<\varepsilon_2))$. I don't really know how to proceed. May I ask for some help?

I am not sure how to proceed. Can someone help me,please?

• Hint: A Cauchy sequence need not have a limit point in $X$, but it can't have more than one. – quasi Apr 20 '17 at 6:05

Hint: if $a$ and $b$ are two different limit points, what can you say about $d(x_n, x_m)$ when $x_n$ is close to $a$ and $x_m$ is close to $b$?
• Assume that $x\in X$ and $\mathscr X \in X$ are any two limit points of the set $S$. We need to show that $x = \mathscr X$. By definition of limit points, we have that $(\forall \varepsilon_1 >0)(\exists x_n \in S) ((x_n \neq x) \land (d(x_n,x)<\varepsilon_1))$ and $(\forall \varepsilon_2 >0)(\exists x_m \in S) ((x_m \neq \mathscr X) \land (d(x_m,\mathscr X)<\varepsilon_2))$. I don't really know how to proceed. May I ask for some help? – Y.X. Apr 20 '17 at 9:52