Find whether the set $S$ in $\mathbb{R}^3$ is open or closed. My set $S=\{(x, y, z) :-1 \le x \le 1,-1 < y <1, -1<z<1\}$.i am trying to prove that the set $S$ is open .
attempt  a:
So I pick up a point $(x_0,y_0,z_0)$ then I would try to show that it is an interior point. I pick up the space $(x_0-r,y_0-r,z_0-r) \times (x_0+r,y_0+r,z_0+r)$
.From here I should get that $x_0-r \ge -1$ and $x_0+r \le 1$ from here it is easy to conclude that $r \le x_0+1$ and $r \le 1-x_0$ from here we can conclude that $r \le 1$ and from the other equationsi am also able to prove that $r$ should be $0 <r < 1$.
attempt b:
I am trying to show that  closure($S$) $\ne S$.Then I am done. Now I want to show that the closure of $S$ is essentially $[-1,-1,-1] \times [1,1,1]$.We pick up a sequence $(-1,-1-\frac{1}{n},-1-\frac{1}{n})$then it is a sequence in $S$ converging to $(-1,-1,-1)$.Sothe point $(-1,-1,-1)$ must be in the set $S$. Since cl($S$) $\ne S$ we are done.
attempt c:
Now I wanted to find a continuous map $f$ from $S$ such that the map $f(S)$ is open. How do I find such an $f$?

The set $S$ is not open. Any ball centered at $(1,0,0)$ contains points $(x,y,z)$ with $x>1$, and no such point belongs to $S$.
But the set $S$ is not closed either, since, say $(1,1,1)\in\overline S$, but $(1,1,1)\notin S$.
