Uniqueness of limit point of a Cauchy Sequence

Question

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!

Answer 1

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.

Answer 2

The limit point of a set :

Given a set $A$, a limit point of $A$ is a point such that there exists a sequence in $A$, converging to that point. Alternatively, every neighbourhood of a limit point contains at least one point in $A$ that isn't the limit point itself.

The limit of a sequence is given via the epsilon-delta definition.

These notions coincide when you treat a convergent sequence as a set (you can check this yourself). However, they don't coincide in general: For example, the limit points of the set $\{( -1)^n\}$ are $-1$ and $1$, although the limit of the sequence $\{ (-1)^n\}$ doesn't exist.

In the case of your question, you can do it in a few steps as follows (I'll leave the steps, but there are comments):

1) Treating the Cauchy sequence as a sequence, all it's subsequences are also Cauchy.

2) If a Cauchy sequence converges to a given point , then all it's subsequences also converge to the same point.

The crucial one:

3) If a Cauchy sequence has a convergent subsequence, then it also converges (to the same point!)

From this, we are done:

Treat the Cauchy sequence $C$ as a set. Suppose it's not convergent, then I claim it has no limit points. Suppose it had, then there would be a subsequence $D$ of $C$ converging to that point. But then, by $3$, $C$ would also have to converge to that point, and this is a contradiction.

If $C$ is convergent, then any subsequence of $C$ must also converge to that point. In short, the limit point is unique.

Hence, this completes the proof. If you do not get it yet, please inform me.