If $x_n \to 0$ and $\sigma : \mathbb{N} \to \mathbb{N}$ is a bijection, then show that $x_{\sigma (n)} \to 0$

Question

Here's the question:

Let $(x_n)$ be a sequence. Assume that $x_n \to 0$. Let $\sigma : \mathbb{N} \to \mathbb{N} $ be a bijection. Define a new sequence $y_n:= x_{\sigma(n)}$. Show that $y_n \to 0$.

Here's my attempted proof:

Let $\epsilon >0$ be arbitrary. Then there exists $N\in\mathbb{N}$ such that for $n\ge N$, we have $\left| x_n \right|<\epsilon$. Now, define the set $A:=\{n\in \mathbb{N} : \sigma(n) < N \}$. Clearly, $A$ is finite set (since $\sigma$ is a bijection and there are precisely $N-1$ elements in the set). Let $M=\max A +1$. It is evident by the definition of $(y_n)$ that for $n\ge \sigma(M)$, we have $|y_n|< \epsilon$.

Is this proof correct? What are some alternative proofs?

Answer 1

Something slightly strange has happened. I think where you say $n \geq \sigma(M)$ you just want $n \geq M$. Let's see it your way first:

Your argument

You say $M = \max A + 1$.

Then you take $n \geq \sigma(M)$.

Now $y_n = x_{\sigma(n)}$ and you want $|y_n| < \epsilon$, i.e. you want $|x_{\sigma(n)}| < \epsilon$.

This would be true if $\sigma(n) \geq N$, which is equivalent to saying $n \notin A$. But why is this true?

Proposed correction

Say $M = \max A + 1$.

Then take $n \geq M$.

Now $y_n = x_{\sigma(n)}$ and we want $|y_n| < \epsilon$, i.e. we want $|x_{\sigma(n)}| < \epsilon$.

This would be true if $\sigma(n) \geq N$, which is equivalent to saying $n \notin A$. Clearly $n \notin A$ becasue $n \geq M = \max A + 1$.

Answer 2

Nitpicking. If $N=1$ then $A=\emptyset$ and $\max A$ does not exist.

You can fix this by saying "Then there exists $N\in \Bbb N$ such that $N>1$ and such that $\forall n\geq N\,(|x_n|<\epsilon).$"

Or you can fix this by saying "Let $M\in \Bbb N$ such that $\forall n\in A\,(M>n).$" (which is logical whether or not $A$ is empty, because $A$ is finite).

And in the last line you want $n>M, $ because $n>M\implies \sigma(n)\geq N\implies |x_{\sigma(n)}|<\epsilon$, as explained in the Answer from T_M.

Other than that, it's fine.

BTW. If we weaken the condition on $\sigma$ to having finite fibers, that is, if $\{m:f(m)=n\}$ is finite for each $n$, and without $\sigma$ necessarily being surjective, then we still have $x_{\sigma(n)}\to 0 $ because $A$ is still finite. (Although $A$ may have more than $N-1$ members.)