Vrftjkry

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 $Ninmathbb{N}$ such that for $nge N$, we have $left| x_n right|<epsilon$. Now, define the set $A:={nin mathbb{N} : sigma(n) < N }$. 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 $nge sigma(M)$, we have $|y_n|< epsilon$.

Is this proof correct? What are some alternative proofs?

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 $n geq M = max A + 1$.

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

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

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

And in the last line you want $n>M, $ because $n>Mimplies sigma(n)geq Nimplies |x_{sigma(n)}|<epsilon$, 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.)