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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?

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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."

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  • $\begingroup$ This makes sense, thank you $\endgroup$ – jackson5 Feb 5 '17 at 5:33
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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.

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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".

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  • $\begingroup$ I know that this is true but could you please you point me to a proof of the iff statement? The $\Rightarrow$ is obvious but could you please point me to a proof of the $\Leftarrow$? Thanks $\endgroup$ – jackson5 Feb 5 '17 at 6:07
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    $\begingroup$ suppose the condition holds, and let $\epsilon$ be positive. The condition gives us an $F$: let $N$ be any integer larger than the maximum element of $F$. Every element in the sequenc after the $N$th one is at less than $\epsilon$ distance from $L$. $\endgroup$ – Mariano Suárez-Álvarez Feb 5 '17 at 10:20
  • $\begingroup$ makes perfect sense, thank you $\endgroup$ – jackson5 Feb 5 '17 at 10:21

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