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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?

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

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  • $\begingroup$ 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? $\endgroup$ – Y.X. Apr 20 '17 at 9:52
  • $\begingroup$ You didn't answer the question in my hint. $\endgroup$ – Robert Israel Apr 20 '17 at 17:16

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