Is a rearrangement of a convergent sequence still convergent? Is a rearrangement of a convergent sequence still convergent?
I work in an arbitrary normed vector space.
I think this is the case.
Let $(x_1, x_2, \ldots)$ be a converging sequence with limit $L$.
Let $f : \mathbb{N} \rightarrow \mathbb{N}$ be a bijection.
Define $y_i = x_{f(i)}$.
Then $(y_1, y_2, \ldots)$ is convergent with limit $L$.
Proof.
Let $\epsilon > 0$ be given. Since $(x_1, x_2, \ldots)$ is convergent there exists an $N > 0$ such that for all $n > N$ we have that $||x_n - L|| < \epsilon$.
Since $N$ is finite and $f$ is a bijection we have that the set
$\{f^{-1}(i)\, | \, i = 1, 2, \ldots, N \}$ is bound by some natural number $M$. Now let $m > M$. Then $f(m) > N$. So $||y_m - L|| = ||x_{f(m)} - L || < \epsilon$.
Is this proof correct? 
 A: You've got the idea and execution just right! The only alterations I'd make involve type-setting, notation, and terminology adjustments.

Let $\epsilon > 0$ be given. Since $(x_1, x_2, \ldots)$ is convergent there exists an $N > 0$ such that for all $n > N$ we have that $||x_n - L|| < \epsilon$.

Here, rather than use ||x_n - L||, I'd use \lVert x_n-L\rVert, instead giving us $\lVert x_n-L\rVert.$ For one thing, it looks a little better (my opinion, obviously), and renders more easily. Another, more important reason comes about if we want/need to adjust sizing. For example, let's say we're in the normed vector space of continuous functions on the interval $[0,1]\to\Bbb R,$ under the norm $$\lVert f\rVert:=\max\bigl\{f(t)\mid t\in[0,1]\bigr\},$$ and we want to find the norm of the function $F(t):=\int_0^t\bigl(f(x)-g(x)\bigr)\,dx$ for some suitable functions $f$ and $g.$ Now, it's clear just by looking that the standard bars won't be tall enough to encompass the integral, so we might try using the \big function to fix it. That turns out to work just fine, at first, giving us either $\big|\big|\int_0^t\bigl(f(x)-g(x)\bigr)\,dx\big|\big|$ if we typeset it as \big|\big|\int_0^t\bigl(f(x)-g(x)\bigr)\,dx\big|\big|, or get us $\big\lVert\int_0^t\bigl(f(x)-g(x)\bigr)\,dx\big\rVert$ if we typeset it as \big\lVert\int_0^t\bigl(f(x)-g(x)\bigr)\,dx\big\rVert. However, later on, we decide for one reason or another that we'd like to center our mathematical notation, which instead gives us $$\big|\big|\int_0^t\bigl(f(x)-g(x)\bigr)\,dx\big|\big|$$ if we typeset it as \big|\big|\int_0^t\bigl(f(x)-g(x)\bigr)\,dx\big|\big|, or  $$\big\lVert\int_0^t\bigl(f(x)-g(x)\bigr)\,dx\big\rVert$$ if we typeset it as \big\lVert\int_0^t\bigl(f(x)-g(x)\bigr)\,dx\big\rVert. Clearly, we've got to adjust, so we decide to use the delimiters \left and right, instead. This yields $$\left|\left|\int_0^t\bigl(f(x)-g(x)\bigr)\,dx\right|\right|,$$ if we typeset it as \left|\left|\int_0^t\bigl(f(x)-g(x)\bigr)\,dx\right|\right|, or $$\left\lVert\int_0^t\bigl(f(x)-g(x)\bigr)\,dx\right\rVert$$ if we typeset it as \left\lVert\int_0^t\bigl(f(x)-g(x)\bigr)\,dx\right\rVert. Thus, we're saved a bit of typing and some editing, which can really pile up if we have a long array of equations to adjust, instead. Still, that's all a matter of preference.

Since $N$ is finite and $f$ is a bijection we have that the set
  $\{f^{-1}(i)\, | \, i = 1, 2, \ldots, N \}$ is bound by some natural number $M$.

Here, I'd suggest some adjustment to your terminology. It isn't particularly meaningful to say that $N$ is finite in this context, as far as I can tell. When we say "there exists an $N>0$ such that...," we mean for $N$ to be a positive real/natural number, all of which are arguably "finite" in some sense. Instead, I'd say something like: "Since $\{k\in\Bbb N\mid 1\le k\le N\}$ is a bounded subset of $\Bbb N,$ then it is finite. Thus, since $f$ is a bijection on $\Bbb N,$ then $$f^{-1}\bigl[\{k\in\Bbb N\mid 1\le k\le N\}\bigr]=\bigl\{f^{-1}(k)\mid k\in\Bbb N,1\le k\le N\bigr\}$$ is also a finite subset of $\Bbb N,$ and so is bounded above by some $M\in\Bbb N.$"
You might need to justify the equality above, that bounded subsets of $\Bbb N$ are exactly the finite subsets of $\Bbb N,$ or that pre-images of finite sets under injections are finite, if you haven't proved them/seen them proved previously.
Note here that instead of the term "bound" (which I typically see used to indicate the numbers/elements providing limits on the set in question), I use the term "bounded" (meaning possessing such limits). However, you may just be using a different terminological convention than that with which I'm familiar.

Now let $m > M$.

Here, we could be a little more careful, saying something to the effect of: "Now let $m\in\Bbb N$ such that $m>M.$" That way, we don't rely on the reader to infer that $m$ is an element of the domain of $f.$

Then $f(m) > N$.

We could be more careful, here, too. Perhaps something like: "Since $m\notin f^{-1}\bigl[\{k\in\Bbb N\mid 1\le k\le N\}\bigr],$ then $f(m)\notin\{k\in\Bbb N\mid 1\le k\le N\}$ by definition of pre-image, so $f(m)\not\le N,$ and so $f(m)>N.$"
Nice work!
