The common definition of open set of $\mathbb R^2$ is $$U\subset \mathbb R^2 \ \mathrm{is \ open} \ \iff \forall a\in U, \exists r>0 \ ; \ B_2(a,r)\subset U,$$ where $B_2(a,r)$ is an open ball around $a$ with radius $r.$
Thus, Euclidean topology is $\mathcal O_{\mathbb R^2}=\{ U\subset \mathbb R^2 \mid \forall a\in U, \exists r>0 \ ; \ B_2(a,r)\subset U\} \cdots (A)$
On the other hand, it is possible to define Euclidean topology of $\mathbb R^2$ as a product topology of $(\mathbb R, \mathcal O_{\mathbb R})$ and $(\mathbb R, \mathcal O_{\mathbb R})$.
Specifically, let $\mathcal B=\{U\times V \mid U\in \mathcal O_{\mathbb R}, V\in \mathcal O_{\mathbb R}\},$ then the product topology $\mathcal O_{\mathbb R \times \mathbb R}=\mathcal O_{\mathbb R^2}$ is the topology whose base is $\mathcal B$, i.e., $\mathcal O_{\mathbb R^2}=\{ U\subset \mathbb R^2 \mid \exists \ \mathrm{index\ set }\ \Lambda, \exists \{B_\lambda\}_{\lambda\in \Lambda}\subset \mathcal B \ ; \ U=\cup_{\lambda\in \Lambda}B_\lambda \}\cdots (B)$
I want to check the equivalence of $(A)$ and $(B)$.
Let $(A)=\mathcal M, (B)=\mathcal N.$
I proved $\mathcal M\supset \mathcal N,$ but I have difficulty in proving $\mathcal M\subset \mathcal N$.
Here is my proof and this doesn't seem to work.
Let $U\in \mathcal M.$
Then, $\forall a\in U, \exists r_a>0 \ ; \ B_2(a,r_a)\subset U$, thus $U=\displaystyle\bigcup_{a\in U} B_2(a,r_a)$ holds.
So if I show $\{B_2(a, r_a)\}_{a\in U}\subset \mathcal B$, I can say $U\in \mathcal N$ and the proof will finish.
For this, I have to show $\forall a\in U, \exists U, V \in \mathcal O_{\mathbb R}, B_2(a, r_a)=U\times V$,
However, from the comment in this post In Euclidean space, can open ball be written as the direct product of open sets?, such $U,V$ doesn't exist, so this proof doesn't work.
How can I show $\mathcal M\subset \mathcal N$ ? Thanks for any help.
Let $X$ be any set and $\tau, \tau'$ be two topologies on $X$ with basis $\mathcal{B}, \mathcal{B'}$ respectively.
Then $\tau\subset \tau'$ if for all $B\in\mathcal{B}$ and $\forall x\in B$ there exists a $\tau'$-basic open set $B'$ such that $x\in B'\subset B$
For the euclidean topology on $\Bbb{R}^2$ basic open sets are all open balls $\{B_2(x,r) : x\in \Bbb{R^2} ,r>0\}$
For the product topology on $\Bbb{R}^2$ basic open sets are all open rectangles, $\{(a,b)×(c,d) : a, b, c, d\in\Bbb{R} ,a<b,c<d\}$
Intuitively, if we choose a point from a open disk, we can place an open rectangle containing that point and contained in the open disc. In ther words a disc is arbitrary union of open rectangles.
And
If we choose a point from a open rectangle, we can place an open disc containing that point and contained in the open rectangle. In ther words an open rectangle is arbitrary union of open disc.
Thus two bases are equivalent and generates same topology.