If a sequence $\{x_{n}\}$ is a Cauchy sequence and the sequence has a limit point $x_{0}$ then $x_{n} \rightarrow x_{0}$

I am trying to prove that if a sequence $\{x_{n}\}$ in a set $M\in X$ ($X$ is a metric space) is a Cauchy sequence and the sequence has a limit point $x_{0}$ then $x_{n} \rightarrow x_{0}$. I wish to understand the details in such a proof whence I will try to explain my thoughts.

I know that the sequence $\{x_{n}\}_{N\in \mathbb{N}}$ is a Cauchy sequence which means that for all $\epsilon >0$ I can find an $N \in \mathbb{N}$ beyond which the distance between any to elements $x_{n}$ and $x_{m}$ of the sequence will be smaller than $\epsilon$. This means that I can make the distance between any two element of the sequence arbitrarily small.

Furthermore we know that $x_{0}$ is a limit point which means that for every $\epsilon>0$ there is a point $m \in M$ such that $m \in B(x_{0},\epsilon)$.

I wish to argue as follows:

1) I would like to argue that $x_{m}=m \in B(x_{0},\frac{\epsilon}{2})$ (I can freely choose $r=\frac{\epsilon}{2}$ since $x_{0}$ is a limit point.

2) Since $\{x_{n}\}$ is a Cauchy sequence $d(x_{n},x_{m})$ will be smaller than any $\epsilon$ I choose, so I choose $\frac{\epsilon}{2}$. Therefore $d(x_{n},x_{m})<\frac{\epsilon}{2}$.

3) Since $x_{m}=m \in B(x_{0},\frac{\epsilon}{2})$ then $d(x_{m},x_{0})<\frac{\epsilon}{2}$.

4) Using the triangle inequality we have $d(x_{n},x_{0}) \leq d(x_{n},x_{m}) + d(x_{m},x_{0}) \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

5) We can conclude that $x_{n} \rightarrow x_{0}$

Can someone please comment on the proof and the arguments presented? How can I convince myself of 1) ? How do I convince myself of the choice of epsilon (I start my choice of epsilon from the fact that $x_{0}$ is a limit point and from this I choose my epsilon for the distance between $x_{n}$ and $x_{m}$).

Many thanks

Solution 1:

We will prove that $\{x_n\}$ converges to that limit point $x_0$. Fix $\epsilon>0$ arbitrarily. Since $\{x_n\}$ is Cauchy, so there exists $N\in\mathbb{N}$ such that $d(x_n,x_m)\leq \frac{\epsilon}{2}$ for all $m,n\geq N$. Also as $x_0$ is a limit point of $\{x_n\}$, so there exists a subsequence $\{x_{n_k}\}_k$ such that $d(x_{n_k},x_0)\leq\frac{\epsilon}{2}$ for all $k\in\mathbb{N}$. Now choose some $n_k\geq N$ so that $d(x_{n_k},x_0)\leq\frac{\epsilon}{2}$ holds. Then $$d(x_n,x_0)\leq d(x_n,x_{n_k})+d(x_{n_k},x_0)\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\text{ for all }n\geq N.$$ And therefore $\{x_n\}$ converges to $x_0$.

Solution 2:

I found your usage of the set $M$ to be confusing. I had to read a few times to realize you mean $M = \{x_i| x_i \in $ the sequence $\{x_n\} \}$. In other words that $M$ is the unordered set of elements in the sequence.

You don't need to do this. A sequence is an "ordered set" and it's okay to to use regular set concepts on it.

Just say, let $S = \{x_1, x_2, ......\}$ be sequence of terms, $x_i \in X$.

You arguments are mostly good but you need to keep in mind none of your values are constant and will vary base on you choose of epsilon.

So I'd prefix all your arguments with:

Suppose we have a specific $\epsilon > 0$ in mind.

I'd rewrite 1) as.

1) As $\frac \epsilon 2>0$ we know that as $x_0$ is a limit point, that there exists an $x_i \in S$ so that $x_i \in B(x_0, \frac \epsilon 2)$.

(Note: there's no need to refer to $x_i$ by another variable $m$. And we can't refer to it by variable $m$ then we can't index it by $x_m$. $m \in X$ and $m$ is not an integer.

I'd rewrite 2) as.

2) As $S$ is a cauchy sequence we know that for $\frac \epsilon 2$ that there is an $M$ so that if $n,k > M$ then $d(x_n, x_k) < \frac \epsilon 2$.

Note: that we do not know that the $i$ in part 1: is actually larger than The $M$.

Keeping the above in mind.

I'd rewrite 3 as

3) $d(x_i,x_0)< \frac {\epsilon} 2$.

But Now we have an immediate problem with 4) in that we don't know that $i > M$.

But we can fix this.

If the $i$ in 1) is so that $i \le M$ we le $\epsilon_2 = d(x_i, x_0)$ (being a limite point we *chose $x_i \ne x_0$ so $d(x_i, c_0) > 0$) and find that the is another $x_j$ so that $d(x_j, x_0) <d(x_i, x_0)$. And if $j \le M$ we repeat. There are only finitely many integers $ \le M$ so eventually we will run out.

Or to put this more formally:

Let $\epsilon_M = $ the minimum of $d(x_j, x_0)$ so that $j \le M$ (and $x_j \ne x_0$) and $x_j \ne x_0$ and $\frac \epsilon 2$. Then

B) There is an $x_i \in S$ so that $x_i \in B(x_0, \epsilon_M)$.

Now note:

C) If $i \le M$ we would get a contradiction that $d(x_i, x_0) <\epsilon_M \le d(x_i,x_0)$ so $i > M$.

And we say

3) $d(x_i, x_0) < \epsilon_M \le \frac \epsilon 2$.

Now we haven't actually picked any $n,k$ in part 2)

We want one of them to be $x_i$ and the other can be any $n > M$.

Then we have your 4)

4) If $n \ge M$ then $d(x_n, x_0) \le d(x_n, x_i) + d(x_i, x_0) < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$.

and then your conclusion.

5) we have shown that for $\epsilon > 0$ there is an $M$ where $n > M$ will mean $d(x_n,x_0)< \epsilon$. By definition that means $\lim_{n\to \infty} x_n = x_0$ or $x_n \to x_0$.

....

To paste it all back together:

Suppose we have a specific $\epsilon > 0$ in mind

2) As $S$ is a cauchy sequence we know that for $\frac \epsilon 2$ that there is an $M$ so that if $n,k > M$ then $d(x_n, x_k) < \frac \epsilon 2$.

Let $\epsilon_M = $ the minimum of $d(x_j, x_0)$ so that $j \le M$ (and $x_j \ne x_0$) and $\frac \epsilon 2$. Then

B) There is an $x_i \in S$ so that $x_i \in B(x_0, \epsilon_M)$ (because $x_0$ is a limit point of $S$).

3) $d(x_i, x_0) < \epsilon_M \le \frac \epsilon 2$.

C) If $i \le M$ we would get a contradiction that $d(x_i, x_0) <\epsilon_M \le d(x_i,x_0)$ so $i > M$.

4) If $n > M$ (and we know $i > M$) then by the trianngle inequality, $d(x_n, x_0) \le d(x_n, x_i) + d(x_i, x_0) < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$.

5) By definition that means $\lim_{n\to \infty} x_n = x_0$ or $x_n \to x_0$