Why can the $n_{\epsilon}$ of the definitions of convergence and Cauchy sequence be the same in the following proposition?

Question

I have the following proposition proved in my lectures notes, but I think there are a couple of errors and there is one think I don't get:

If $p_n$ is a Cauchy sequence in a metric space $(X,d)$, the set $\{p_n| n \in \mathbb{N}\}$ is bounded. Moreover, if $p_n$ has $p_0$ as a limit point, then $p_n$ converges to $p_0$ ----->(1)

Proof

Let $\epsilon > 0$. Now for each $n \geq n_{\epsilon}$, $d(p_n, p_{n_{\epsilon}}) < \epsilon$. Then $p_n \in B_d(p_{n_{\epsilon}}, \epsilon)$ for each $n\geq n_{\epsilon}$, from which the thesis follows. Moreover, if $p_0$ is the limit of subsequence $p_{n_k}$, for $\epsilon >0$, there exists $n_{\epsilon}$ such that $d(p_{n_k},p_0)< \epsilon/2$ and $d(p_{n_k},p_n)< \epsilon/2$ for $n\geq n_{\epsilon}$ and $k\geq n_{\epsilon}$. ----->(2)

Then,

$d(p_n,p_0) \leq d(p_n,p_{n_k})+d(p_{n_k},p_0) < \epsilon$ for $n, k \geq n_{\epsilon}$ ----->(3)

from which the thesis follows

I think there are some errors in this proof. I would like some feedback/confirmation on it

(1) --> it should say " if $p_{n_k}$ has $p_0$ as a limit point" instead of " if $p_{n}$ has $p_0$ as a limit point"

(2), (3), it should be $n_k \geq n_{\epsilon}$, instead of $k \geq n_{\epsilon}$

And a small question that is really bothering me :

For all $\epsilon > 0$, If $p_n$ is a Cauchy sequence, the definition says an $n_{\epsilon}$ exist , such that for all $n_1, n_2 >n_{\epsilon}$ $d(p_{n_1},p_{n_2})<\epsilon$.

The definition of convergence says: If $p_n$ converges to $p_0$, then for all $\epsilon > 0$, there exist and $n_{\epsilon}$, such that for any $n > n_{\epsilon}$, $d(p_n, p_{n_{\epsilon}})<\epsilon$.

When they mix the definitions to prove the second part of the proposition, the one about the subsequence, they consider the $\epsilon$ and the $n_{\epsilon} $ of both definitions to be the same. I agree for the $\epsilon$, because since it must be true for all $\epsilon $ I can choose them to be equal, but then what guarantees that the $n_{\epsilon} $ of the first definition is the same as the $n_{\epsilon} $ of the second definition ? The definitions just state some $n_{\epsilon} $ exists. It is assumed they are equal without saying why and I really can't figure it out.

Answer 1

The wording ‘if $p_n$ has $p_0$ as a limit point’ is very sloppy, but not for the reason that you suggest at (1). It’s sloppy because $p_n$ is a single point, not a sequence; what the author means is ‘if $\langle p_n:n\in\Bbb N\rangle$ has $p_0$ as a limit point’ (in which any other standard notation for a sequence can be substituted for my preferred notation). What the author means here, however, is correct: if the original Cauchy sequence has $p_0$ as a limit point (I would say cluster point), then it converges to $p_0$. Not only is your suggestion not what he’s trying to say, it would make no sense: no subsequence $\langle p_{n_k}:k\in\Bbb N\rangle$ has even been defined at that point.

There is also nothing wrong with $k\ge n_\epsilon$ at (2) and (3), though it would have been helpful if the author had explained why this is the case. The point is that since $\langle p_{n_k}:k\in\Bbb N\rangle$ is a subsequence of $\langle p_n:n\in\Bbb N\rangle$, the sequence $\langle n_k:k\in\Bbb N\rangle$ is a strictly increasing sequence of natural numbers, and it’s easy to prove by induction on $k$ that this implies that $n_k\ge k$ for each $k\in\Bbb N$. Thus, $k\ge n_\epsilon$ immediately implies that $n_k\ge n_\epsilon$.

The definition of convergence in the penultimate paragraph of your question is incorrect. It should read:

$\langle p_n:n\in\Bbb N\rangle$ converges to $p_0$ if and only if for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that for each $n>n_\epsilon$, $d(p_n,\color{red}{p_0})<\epsilon$.

The last part of the proof is correct, but the author has omitted a little explanation. There is an $m_\epsilon\in\Bbb N$ such that $d(p_{n_k},p_0)<\frac{\epsilon}2$ whenever $k\ge m_\epsilon$. There is a possibly different $m_\epsilon'\in\Bbb N$ such that $d(p_{n_k},p_n)<\frac{\epsilon}2$ whenever $k,n\ge m_\epsilon'$. Now let $n_\epsilon=\max\{m_\epsilon,m_\epsilon'\}$, then $d(p_{n_k},p_0)<\frac{\epsilon}2$ and $d(p_{n_k},p_n)<\frac{\epsilon}2$ whenever $k,n\ge m_\epsilon'$, and we’re home free.

Answer 2

Why not say : suppose $(p_n)$ is Cauchy. Then apply the definition of Cauchy-ness to $\varepsilon=1$ and we find $N \in \mathbb{N}$ such that

$$\forall n,m \ge N: d(p_n, p_m) < 1$$

So we define $$M=\max\{d(p_1, p_N), \ldots, d(p_{N_1}, p_N)\}+1$$

and we have that $$\{p_n: n \in \mathbb{N}\} \subseteq B(p_N, M)$$

and so the range of the sequence is bounded.

If moreover $p_0$ is a limit point of $\{p_n: n \in \mathbb{N}\}$. Then in fact $p_n \to p_0$ as well: let $\varepsilon >0$. We find $N_1$ (from the Cauchy definition applied to $\frac{\varepsilon}{2}>0$) such that

$$\forall n,m \ge N_1: d(p_n, p_m) < \frac{\varepsilon}{2} .$$

As $p_0$ is a limit point of the range, $B(p_0, \frac{\varepsilon}{2})$ contains infinitely many $p_n$ (i.e. for infinitely many indices). So pick $p_k \in B(p_0, \frac{\varepsilon}{2})$ with $k \ge N_1$.

And then for $n \ge N_1$ we have

$$d(p_n,p_0) \le d(p_n, p_k) + d(p_k, p_0) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$

and as we found such $N_1$ for arbitrary $\varepsilon>0$, $p_n \to p_0$ as promised.