show that $\mathcal{K}=\{(a,a+r) \times (b,b+r):a,b\in \mathbb R, \, r>0\}$ is equal to Borel

Question

In my homework I want to show that $\mathcal{K}=\{(a,a+r) \times (b,b+r):a,b\in \mathbb R, \, r>0\}$ is equal to $\mathcal{B}(\mathbb{R}^2)$

So I want to show that $\sigma(\mathcal{K})=\mathcal{B}(\mathbb{R}^2)$

$\subseteq:$ Is straight forward as open squares are a subset of open rectangles

$\supseteq:$ This one is causing me problems. I am thinking to show that any (open) rectangle can be divided into squares. However this proof does not seem trivial

EDIT:

Let R be a rectangle with points $x_1,y_1,x_2,y_2$ hvor $x_1<x_2$ og $y_1 < y_2$

Assume $w=x_2-x_1 < y_2-y_1$

Then $R=\bigcup_{i=y_1}^{y_2-w} (x_1+w,i+w)$

Any help would be appreciated

Answer 1

Let $\tau$ be the standard topology on $\mathbb{R}^2$. Then $\mathcal{K}$ is a basis of $\tau$. Therefore, $\mathcal{K}$ generates the Borel $\sigma$-field on $\mathbb{R}^2$ , i.e. $\sigma(\mathcal{K})=\mathcal{B}(\mathbb{R}^2)$ (see, e.g., this question).