Null Sequences and Real Analysis I came across the following problem during the course of my study of real analysis:

Prove that $(x_n)$ is a null sequence iff $(x_{n}^{2})$ is null.

For all $\epsilon>0$,  $|x_{n}| \leq \epsilon$ for $n > N_1$. Let $N_2 = \text{ceiling}(\sqrt{N_1})$. Then $(x_{n}^{2}) \leq \epsilon$ for $n > N_2$. If $(x_{n}^{2})$ is null then $|x_{n}^{2}| \leq \epsilon$ for $n>N$. Let $N_3 = N^2$. Then $|x_n| \leq \epsilon$ for $n> N_3$. 
Is this correct? In general, we could say $(x_{n})$ is null iff $(x_{n}^{n})$ is null?

You could use the following fact:
If a function $f:X\to Y$ between two topological spaces is continuous and $x_n\to x$, then $f(x_n)\to f(x)$.
(In case you do not have learned it in this generality, you might at least know that this is true for real functions or for functions between metric spaces. In fact, in the case of real functions the above condition is equivalent to continuity.)
You can obtain your first claim by applying the fact to the continuous functions:
$f: \mathbb R\to\mathbb R$, $f(x)=x^2$ (one implication)
$f: \langle 0,\infty)\to \mathbb R$, $f(x)=\sqrt{x}$ (reverse implication)
