How to prove that $A = \{(x,y):0<x<1, \; 2<y<3\}$ is open

Question

$A = \{(x,y):0<x<1,\; 2<y<3\}$ is intuitively open.

Let's choose $x_{0},y_{0}$. We know that

$$x_{0}>0,\; 1-x_{0}>0$$ $$y_0-3>0,\;4-y_{0}>0$$

So,

$$\delta_{1}=\min\{x_{0},1-x_{0}\},\;\; \delta_{2}=\min\{y_{0}-3,4-y_{0}\}$$

But I can't go further. Thanks in advance!

Answer 1

Choose $(x_0,y_0)\in A$. Since $\{x: 0 < x< 1\}$ is open in $\mathbb R$, there exists $r_1 > 0$ such that $(x_0 - r_1, x_0 + r_1)\subset \{x: 0 < x< 1\}$. Similarly, there exists $r_2 > 0$ such that $(y_0 - r_2, y_0 + r_2) \subset \{y: 2 < y< 3\}$. Take $r := \min\{r_1,r_2\}$. Can you see that $(x_0 - r, x_0 + r) \times (y_0 - r, y_0 + r) \subset A$? If you want a ball contained inside $A$, just take $B((x_0,y_0), r)$. Certainly, $B((x_0,y_0),r) \subset (x_0 - r, x_0 + r) \times (y_0 - r, y_0 + r)$, so $B((x_0,y_0),r) \subset A$ as required.

Hence, $A$ is open.