I'm trying to show that the closure of a Cauchy sequence is countable, and the hint for the question is to show that if $(x_n)$ is Cauchy, then it has at most one limit point.
I'm a little bit confused between the idea between the idea of a limit point of a set and a limit of a sequence. I know that if $\{x_1,x_2,…\}$ is a set, then $x$ is a limit point if $\forall r >0, \exists x_k \neq x $ such that $x_k \in B_r(x)$.
Now I also know what a limit of a sequence is, but since this isn't necessarily a complete metric space we don't know if $(x_n)$ converges. My first thought was to use the fact that Cauchy sequences are bounded, but I don't know where to go with that.
Would appreciate any hints!
Best Answer
Note that from your definition of the limit point of a sequence it follows that each neighborhood of a limit point contains infinitely many terms of the sequence. You can use this statement for your proof as follows:
Suppose for the sake of contradiction that your cauchy sequence $(x_n)$ has two limit points $x \neq y$. Then $\varepsilon := d(x,y) > 0$, by the definiteness of the metric. Since $(x_n)$ is Cauchy, there exists $N \in \mathbb{N}$ such that for all $n, m \geq N$ you have $$d(x_n, x_m) < \frac{\varepsilon}{ 4}.$$ Furthermore, since $x$ is a limit point you find $x_n \in B_{\frac{\varepsilon}{4}}(x)$ with $n \geq N$. But then, for all $m \geq N$ you have $$ d(x_m, x) \leq d(x_m, x_n) + d(x_n, x) < \frac{\varepsilon}{2}. $$ So, for all $m \geq N$ this gives you $x_m \in B_{\frac{\varepsilon}{2}}(x)$ which contradicts the assumption that $y$ is another limit point.