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!