Limit of sequence of real numbers Let $\{x_n\}$ be sequence of real numbers such that $\lim_{n\to\infty} x_n=p$. 
Also if $m\to \infty$ then $n\to \infty$ and converse.
How to prove strictly $\lim_{m\to\infty} x_n=\lim_{n\to\infty} x_m=p$?
I met this feint in many books and also in my last problem that I solved but I can't understand this.
 A: Let $m(n)$ be a monotone function $\mathbb N\to\mathbb N$, then $(x_m):=(x_{m(n)})$ is a subsequence of $(x_n)$. Because each subsequence of a convergent sequence converges to the same limit as the original sequence, you have $\lim_{n\to\infty} x_{m(n)} = p$.
A: Well let's go back to the definition of limits as other answers seems to call the result which you willing to demonstrate :
Def of $(x_{n})$ converges to $p$ L : $\forall \varepsilon > 0, \exists N \in \mathbb N, \forall n>N,  x_{n}\in B(p,\varepsilon) $
For $\varepsilon > 0$, as $(x_{n})$ converges to $p$ let N be such an integer, as $m(n) \rightarrow \infty $ , let $M$ be an integer such that $  \forall n>M, m(n)>N $, and we have $\forall n>M,  x_{m(n)}\in B(p,\varepsilon) $
A: Hint. Let $m\colon\mathbf N\to\mathbf N$ be a function of $n$, i.e., $m(n)$. With $m\to\infty$ as $n\to\infty$, we are stating $\lim_{n\to\infty}m_n=\infty$, i.e., for any $M>0$, there exists a $N\ge0$ such that $m_n>M$ for every $n\ge N$.
By defnition of convergence of $(x_n)^\infty_{n=0}$: for any $\epsilon>0$, there exists a $N\ge0$ such that $|x_j-p|<\epsilon$ for every $j\ge N$.
With these facts: ¿we can pick a integer $N':=m(j)$ for some $j\in\mathbf N$ such that $N'\ge N$, and this $N'$ satisfies the definition of convergence?

To see the relation, you can prove
Let $(a_n)^\infty_{n=m}$ be a sequence of real numbers, and let $L\in\mathbf R$. Then the following two statementes are logically equivalent:


*

*The sequence $(a_n)^\infty_{n=m}$ converges to $L$.

*Every subsequence of  $(a_n)^\infty_{n=m}$ converge to $L$.

