Problem in understanding solution to problem 4a, chapter 22 of Spivak Calculus In chapter 22 of Spivak calculus, problem 4a is

Prove that if a subsequence of a Cauchy sequence converges to $l$, then the sequence itself converges to $l$.

This is the solution in Combined answer book

Since $a_{n_k}$ converges to $l$. There exits $J\in\mathbb{N}$ such that
  $$
k>J\Rightarrow |a_{n_k}-l|<\frac{\epsilon}{2}
$$
  Since $a_n$ is Cauchy. There exists $N_1\in\mathbb{N}$ such that
  $$
n,m>N_1\Rightarrow|a_n-a_m|<\frac{\epsilon}{2}
$$
  Let $N=\max(N_1,n_J)$. If $n>N$, then
  $$
|a_n-a_{n_{j+1}}|<\frac{\epsilon}{2}
$$
  and
  $$
|a_{n_{j+1}}-l|<\frac{\epsilon}{2}
$$
  Consequently $|a_n-l|<\epsilon$.

I do not understand, how when $n>N$, then
$$
|a_n-a_{n_{j+1}}|<\frac{\epsilon}{2}
$$
and
$$
|a_{n_{j+1}}-l|<\frac{\epsilon}{2}
$$
Could someone please explain it to me?
 A: The proof seems to be incorrect.

In order to deduce 
  \begin{align*}
|a_n-a_{n_{J+1}}|<\frac{\epsilon}{2}\tag{1}
\end{align*}
  we have to assure that $n>N_1$ and $n_{J+1}>N_1$.
Since $n>N=\max(N_1,n_J)\geq N_1$ we have $n>N_1$. But we don't know if $n_{J+1}>N_1$.

We know $n_{J+1}>n_J$ which is sufficient to show
\begin{align*}
|a_{n_{J+1}}-l|<\frac{\epsilon}{2}
\end{align*}
but not sufficient to show (1).
A: Proof goes as follows:
Let $a_{n}$ be a Cauchy sequence. 
Suppose that $a_{j_{n}}$ is a convergent subsequence of $a_{n}$. 
First, note that the sequence $j_{n} \geq n$ for all $n \in \mathbb{N}$ (You can prove this as an exercise). (*)
We want to proof that $\forall \epsilon > 0$ there exist a $N \in \mathbb{N}$ such that for all $n \geq N$, $|a_{n}-L|<\epsilon$ (for some $L$ where $L$ is the limit we want).
Let $\epsilon > 0$. 
Since $a_{n}$ is a Cauchy sequence, then exist an $N_{1} \in \mathbb{N}$ such that for all $m,n \geq N_{1}$, $|a_{n}-a_{m}|<\dfrac{\epsilon}{2}$.
In the other hand, we have that $a_{j_{n}}$ is convergent (say, to $M$). Then, exist $N_{2}$ such that for all $n \geq N_{2}$, $|a_{j_{n}}-M|<\dfrac{\epsilon}{2}$
If we choose $N_{0}=max(N_{1},N_{2})$, then let $n \in \mathbb{N}$ such that $n \geq N_{0}$.
By (*) we have $j_{n} \geq n \geq N_{0}$. In particular, $j_{n}, n \geq N_{1}$ then, since $a_{n}$ is a Cauchy sequence by hypotesis, $|a_{j_{n}}-a_{n}|< \dfrac{\epsilon}{2}$
But also $n \geq N_{2}$, then $|a_{j_{n}}-M|< \dfrac{\epsilon}{2}$.
By triangle inequality, we have $$|a_{n}-M|=|(a_{n}-a_{j_{n}})+(a_{j_{n}}-M)|\leq |a_{n}-a_{j_{n}}|+|a_{j_{n}}-M|=|a_{j_{n}}-a_{n}|+|a_{j_{n}}-M|<\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon$$
In other words: $$|a_{n}-M|<\epsilon$$
And $M$ is the $L$ we wanted at the begin of the proof.
