Assume that $x_n \to 0$. Let $σ: N → N$ be a bijection. Define a new sequence $y_n := x_{σ(n)}$, Show that $y_n → 0$. Let $(x_n)$ be a sequence. Assume that $x_n \to 0$. Let $σ: N → N$ be a bijection.
Define a new sequence $y_n := x_{σ(n)}$, i.e. $\sigma$ is a permutation of the set of natural numbers. Show that $y_n → 0$.
I have moved in the following direction:
Since the sequence $x_n \to 0$, then there exists a certain $N \in \Bbb N$ such that for all $n > N$ the tail of the sequence comes within a preferred nbd. 
Since $N$ is finite we located all the numbers from $1,2,...,N$ in the image set of $\sigma$ and find out among the numbers $\sigma(1),\sigma(2), \cdots ,\sigma(N)$ which is located last among them in the rearrangement. Then choosing any number $M$ greater than that number will make the tail of $y_n \to 0$ and hence $y_n \to 0$.
Is the solution correct...Thank You!
 A: The basic idea is correct. Let me propose a more rigorous operationalization.
Fix $\varepsilon>0$. Since $x_n\to 0$, there must exist some $N\in\mathbb N$ such that for any integer $n> N$, one has $|x_n|<\varepsilon$. For each $i\in\{1,\ldots,N\}$, one can find a unique integer $m_i$ such that $\sigma(m_i)=i$. Let $M\equiv\max_{i\in\{1,\ldots,N\}}m_i$.
Suppose that $m$ is an integer greater than $M$. I claim that $\sigma(m)>N$. Indeed, if it were the case that $\sigma(m)\leq N$, then $\sigma(m)=i$ for some $i\in\{1,\ldots,N\}$. Then, it would follow that $\sigma(m)=i=\sigma(m_i)$, and since $\sigma$ is a bijection, $m=m_i$. But $m_i\leq M<m$, a contradiction.
Therefore, $m\in\mathbb N$ and $m>M$ imply that $\sigma(m)>N$. As a consequence, $$|y_m|=|x_{\sigma(m)}|<\varepsilon.$$ It follows that $y_m\to0$ as desired.
A: Fix $\epsilon > 0$. Consider the subset of the image of $\sigma$ such that for all $c$ in this subset, $x_c$ is not within $\epsilon$ of $0$. This is finite. In particular, the preimage of this subset has a maximal element, say $M$ since $\sigma$ is a bijection. Hence for $n > M$, $|y_n|<\epsilon$, QED. 
A: I don't think I understand your proof, but here is one
For $\epsilon > 0$, there is $N$ such that $n > N$ gives $|x_n| < \epsilon$. Let $A = \{n \in \Bbb N: \sigma(n) \le N\}$. Let $n_0 = \max A$. Then, for all $n > n_0$, $n \notin A$, so $\sigma(n) > N$, hence $|x_{\sigma(n)}| < \epsilon$, i.e. $|y_n| < \epsilon$.
