Show that $[-1,1] \times [-1,1]$ is a closed set. Show that $A = [-1,1] \times [-1,1]$ is a closed set.
I know that I have to show $A^c$ is open. So I have to find $\epsilon > 0$ 
sufficiently small such that for all $x \in A^c$, $B(x,\epsilon) \subset A^c$. I am a bit blocked at this point­. I think I have to use the triangle inequality and Cauchy-Schwarz.
Is anyone can give me a hint?
 A: If $(x,y) \in A^c$ then $|x| > 1$ or $|y| > 1$. If the former, consider a ball around $(x,y)$ with radius $|x| - 1$... (draw a picture).
A: You have set yourself an impossible task.  Instead you could aim to show that for any $x \in A^c$ you can find $\epsilon > 0$ 
sufficiently small such that $B(x,\epsilon) \subset A^c$ (note the change in order: $\epsilon$ depends on $x$).
Any point in $A^c$ has the coordinates $(1+c,d)$, $(-1-c,d)$, $(d, 1+c)$, or $(d, -1-c)$ for real $d$ and positive $c$. So given one of these perhaps choose $\epsilon = \frac{c}{2}$ and show that the result is of one of those forms.
A: It is an easy exercise to show that the following are open subsets of $\mathbb{R}^2$ for any $r\in\mathbb{R}$:
\begin{align*}
(r,\infty)\times\mathbb{R} &&
(-\infty,r)\times\mathbb{R} &&
\mathbb{R}\times(r,\infty) &&
\mathbb{R}\times(-\infty,r) &&
\end{align*}
Now notice that $([-1,1]\times[-1,1])^C$ can be described as a union of sets of this type for certain appropriate choices of $r$.  Thus the complement is open as desired.
