Prove that $(0,1)\times(0,1) \subset R^2$ is an open set I know how to prove that $(0,1) \subset R$ is an open set. But I can't apply the same logic to $R^2$.
Here's what I tried:

*

*Take $x = (x_1,x_2)$ an arbitrary element from $(0,1)\times(0,1)$


*Let $\epsilon = \min\{1 - x_1, x_1, 1 - x_2,x_2\}$


*Take $y \in B_\epsilon(x)$, arbtitrary.
Now I need to prove that $y \in (0,1)\times(0,1)$
All I can use is $\displaystyle \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2} < \epsilon$, and I need to show that $0 < y_1 < 1$ and $0 < y_2 < 1$
I saw that question: Show that $(a,b)\times (c,d)$ is an open set in $\mathbb{R}^2$ with the Euclidian metric. but it doesn't help.

Hint:  One approach would be to note that $\Bbb R^2$ has the product topology.  In particular, if you cross open sets, you get an open set.
