Proof that injective rearrangements of a convergent sequence preserves its limit Is this statement true? i.e. If $f(n)$ is injective and $p_n \rightarrow L$, does $p_{f(n)} \rightarrow L$?? (I realize that it's technically only a rearrangement if $f(n)$ is bijective.) 
My attempt at a proof:
Since $p_n \rightarrow L$, we have that, for all $n$ except $n \le N$, $d(p_n, L) < \epsilon$. 
Let $S = \{n \ | \ f(n)\le N\}$, let $n_0$ be the largest $n \in S$, we know there is such a largest $n$ because $f(n)$ is injective. Now we have that $\forall \ n > n_0 \ f(n) > N$ which implies that $p_{f(n)} \rightarrow L$, as required.
Is this correct? Does my proof work?
 A: In essence, your proof is correct. 
However, it should be reworded by inserting the phrases "let $\epsilon > 0$ be given," "there exists $N$ such that" and "therefore $d(p_{f(n)},L) < \epsilon$" in appropriate places. Perhaps also "$S$ is finite."
A: Let $\varepsilon >0$ be arbitrary, then as $p_n \rightarrow L$, there is some $N$ such that $$\forall n > N: d(p_n, L) < \varepsilon\text{.}$$
Now, as $f: \mathbb{N} \rightarrow \mathbb{N}$ is injective, the set 
$f^{-1}[[0,n]] = \{n: f(n) \le N \}$ is finite, as every integer below $N$ has either no pre-image (when $f$ skips it), or at most 1 by injectiveness. 
So for some $N' : f^{-1}[[0,n]] \subset [0,N']$, and then for all 
$n > N'$ we have that $f(n) \notin f^{-1}[[0,n]]$, so $f(n) > N$, and so $d(p_{f(n)}, L) < \varepsilon$, by the convergence above.
This shows that $p_{f(n)} \rightarrow L$ as well.
A: A sequence $(x_n)$ converges to a point $L$ iff 

for every $\epsilon>0$ there exists a finite set $F\subseteq\mathbb N$ such that $\{x_n:n\in\mathbb N\setminus F\}$ is containd in the ball $B_\epsilon(L)$ of radius $\epsilon$ centered at $L$. 

When you phrase convergence like this, clearly the condition does not depend on the order of $\mathbb N$ at all, and therefore a sequence converges to a point iff any bijective rearrangement does. 
You can modify this observation for the case of "injective rearrangements". 
