Q:

Closed or open + proof

Proof if $H = \{ (x,y)\mid 0<x<1, 0<y<1\}$ is closed, open or neither. 
So I would strongly suggest that it is open because if you draw it you recognize that the complementary set contains all boundary points that intersect $H$ and $H^c$ and is therefore closed. But now I am struggling with proofing it formally since we have never done that before in class. I would choose some $\epsilon$ and try to show that a ball around it is contained fully in $H$ but I can only show $x>0$ and not that it  $x<1$.
I would appreciate some help or suggestions and thank you in advance.

A:

Note that $H= (0,1) \times (0,1)$, i.e it is the product of $2$ open sets. Thus it is open in the product topology.

A:

Let $X_0=(x_0,y_0) \in H$, let $r = \frac{1}{2}\cdot \text{min}(x_0,y_0,1-x_0,1-y_0)> 0 \to B(X_0,r) \subset H\to H$ is open.

