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 $f$?