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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!

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  • $\begingroup$ Using boundedness is clearly not sufficient. But Cauchy sequence means that the members get closer and closer with each other for sufficiently large N. $\endgroup$ Apr 6, 2017 at 6:25
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    $\begingroup$ So if you have two limit points, the two limit points must be separated at some distance…so why is it impossible to get all points as close as you want no matter how large you make the n to be? $\endgroup$ Apr 6, 2017 at 6:27
  • $\begingroup$ Off the top of my head, let $x$, $y$ be limit points with distance $r$ between them. Since they are limit points, there are infinitely many points from $(x_n)$ in both $B_{r/2}(y)$ and $B_{r/2}(x)$...then take a point in one of the balls, use the Cauchy property and maybe the triangle inequality to get a contradiction between the distance of the limit points? $\endgroup$ Apr 6, 2017 at 6:55

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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.

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    $\begingroup$ That's a contradiction because the rest of the sequence ($x_m$ with $m > N$) being in that ball around $x$ necessarily means they can't be in $B_{\epsilon /2}(y)$ right? $\endgroup$ Apr 6, 2017 at 7:05
  • $\begingroup$ Correct! Per constructionem, we have $B_{\varepsilon/2}(x) \cap B_{\varepsilon/2}(y) = \emptyset$. $\endgroup$
    – el_tenedor
    Apr 6, 2017 at 7:06
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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.

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  • $\begingroup$ wait… does your proof require the Cauchy sequence to have a convergent subsequence? That will be a problem…since we can take the rational as a metric space and we can make a continued fraction that converges to sqrt(2)…but it converges to no where in R? $\endgroup$ Apr 6, 2017 at 6:36
  • $\begingroup$ If you take continued fractions converging to $\sqrt 2$, then either all the subsequences will converge to $\sqrt 2$, or none of these sequences are convergent (since we are restricting ourselves to rationals). $\endgroup$ Apr 6, 2017 at 6:42
  • $\begingroup$ Oh I see… you are using the fact that if it is a limit point then there is a subsequence converge to that. $\endgroup$ Apr 6, 2017 at 6:45
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    $\begingroup$ Why is it the case that if there were a limit point, we could find a subsequence converging to it? $\endgroup$ Apr 6, 2017 at 7:01
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    $\begingroup$ @Excalibur42: This follows directly from the definition of the limit point: For each $k \in \mathbb{N}$ you can choose $x_{n_k} \in B_{1/k}(x)$ and per constructionem this is a convergent subsequence. $\endgroup$
    – el_tenedor
    Apr 6, 2017 at 7:10

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