Convergence of sequence with one to one function We are working in some metric space $(X,d)$. I am given that $\{x_n\}_{n=1}^\infty$ converges to $x$ and we suppose that $f: \mathbb{N} \rightarrow \mathbb{N}$ is a one to one function. We want to show that $\{x_{f(n)}\}_{n=1}^\infty$ converges to x.
What I did was the following: Since we know $\{x_n\}_{n=1}^\infty$ converges, then that implies that for every $\epsilon >0 \ \exists M$ such that $ n \geq M$ implies that $d(x_n,x)<\epsilon$. Then, since $f(x)$ is one to one, then there exists an $x$ such that $f(x)=M$ and there exists a $y$ such that $f(y)=n$. Thus, we can write that for every $\epsilon >0 \ \exists M=f(x)$ such that $ n=f(y) \geq M$ implies that $d(x_{f(y)},x)<\epsilon$. Hence we can say that $\{x_{f(n)}\}_{n=1}^\infty$ converges to x, as desired. 
Is this prove right, or am I missing something? Thanks!
 A: You have only proven that there is a single value of $n \ge M$ such that $|f(x_{f(n)} - f(x)| < \epsilon$.  But showing that for a single value of $f(n)$ doesn't mean anything.  You must show it for all values for $f(n); n \ge M$.
Which, believe it or not, you can.  As {$f(i)$} are infinite, all but finite of them are larger than any Real $M$.
In other words, for any real $M$, we can let $A_M$ = {$i \in \mathbb N| f(i) < M$}. $A_M$ is finite and has a max element.
So let's try again.
$x_n \rightarrow x$.  So for any $\epsilon > 0$ there exists an $M$ s.t. $n \ge M \implies |f(x_n) - f(x)| < \epsilon$.
Well let $\mathbb M$ = $\max $ $A_M$.  Then $n > \mathbb M \implies f(n) \ge M \implies |f(x_{f(n)}) - f(x)| < \epsilon$. 
So $x_{f(n)} \rightarrow x$.
NOW we have proven it. 
A: You have to prove that for all $\epsilon > 0$, there exists $L \in \mathbb N$ such that for all $n \ge L$ you have $d(x_{f(n)},x) < \epsilon$.
There are several issue with what you did. First you use $x$ with as the limit of the sequence $(x_n)$ and as an index of the sequence. Second, if I understand well is that you took $L$ such that $f(L)=M$. but then there is no reason that for $n \ge L$ you have $d(x_n,x) < \epsilon$.
You need to explain why you can find $L$ (large enough) such that for $n \ge L$ you have $f(n) \ge M$. Try to analyze why $$L > \max f^{-1}(\{1, \dots , M\})$$ would work.
A: So we have that $x_n$ tends to $x$. Let's fix some $\varepsilon > 0 $ and find number $N$ such that $\forall n \geq N ~~~~ d(x_n, x) < \varepsilon$.
Each of numbers $1,...,N$ has preimage under $f$ since the function is surjective. Let's condier the set $\Phi = \{ f^{-1}(1), ...,f^{-1}(N)\}$. 
Let $M$ be the largest number in $\Phi$. Then $\forall n \geq M+1$ the number $f(n)$ doesn't belong to $\{1, ..., N \}$ (we use injectivity of function $f$ here). So $f(n) > N$  and thus $d(x_{f(n)},x) < \varepsilon$.
We have proved that for arbitrary $\varepsilon > 0$ there is number $M+1$ such that $\forall n \geq M+1 ~~d(x_{f(n)},x)<\varepsilon$ which means that $x_{f(n)}$ tends to $x$. 
A: In a metric space, the sequence $(x_n)_n$ converges to $x$ iff, for every nbhd $U$ of $x$, the set $V_U=\{n :x_n\not \in U\}$ is finite. Let $y_n=x_{f(n)}$ for each $n.$ For a nbhd $U$ of $x$ we have $\{n :y_n\not \in U\}=\{n: f(n)\in V_U\}$ which is finite because $V_U$ is finite and $f$ is 1-to-1.
