Show that if $A\subseteq \mathbb R^n $ is closed and bounded(compact) Show that if $A\subseteq \mathbb R^n $  is  closed and bounded(compact), then $A\times A \subseteq \mathbb R^n\times \mathbb R^n $ is closed and bounded (compact).
Someone could give me a hint, 
I do not know how to start.
Sorry for my english.
 A: It helps to write the definitions of A being closed and bounded. 
Let's say $(x, y)$ is an element of $A\times A$, what is the relationship between $|(x,y)|$ and $|x|$, $|y|$?
If $\{(x_n, y_n)\}$ is a sequence in $A\times A$ converging to $(x, y)$,
what do $\{x_n\}$ and $\{y_n\}$ converge to? And what does the closedness of $A$ tell you about the limits of these two sequences?
A: Let $\pi_1, \pi_2$ be projections onto the first and second coordinate, i.e. $\pi_X(x,y) = x$ and $\pi_Y(x,y) = y$. We know that  $\pi_1, \pi_2$ are continuous.
Then 
$$A \times A = \underbrace{\pi_X^{-1}(A)}_{\text{closed}} \cap \underbrace{\pi_Y^{-1}(A)}_{\text{closed}}$$
is closed as an intersection of two closed sets.
Equip $\mathbb{R}^n \times \mathbb{R}^n$ with the norm $\|(x,y)\| = \max\{\|x\|_\infty, \|y\|_\infty\}$. $A$ is bounded so $\exists M > 0$ such that $\|x\|_\infty \le M,\forall x\in A$.
For $(x,y) \in A\times A$ we have
$$\|(x,y)\| =\max\{\|x\|_\infty, \|y\|_\infty\} \le M$$
so $A \times A$ is bounded.
Therefore $A \times A$ is closed and bounded and hence compact.
