Showing product topology on $\mathbb{R}^2$is same as the standard topology on $\mathbb{R}^2$ I have multiple ideas of how to try show this, I want to know if the approaches are correct and help correct any misunderstanding in my arguments below:
We know $B_1 = \{ \text{ all open balls in } \mathbb{R}^2 \}$ is a basis for the standard topology on $\mathbb{R}^2$ and
$B_2$ = $\{U \times V: \text{ U and V are open in } \mathbb{R} \}$ is a basis on the product topology on $\mathbb{R}^2$. If we show $B_1 = B_2$ would we be done? But clearly equality does not hold since a open ball cannot be written as a product of two sets in $\mathbb{R}$. But what if we showed $B_1$ and $B_2$ generate each other, that is a open ball can be written as union of open rectangles and vice versa, would that work?
Another approach could be to define $\mathscr T_1$ = $\{U: U \text{ open in } \mathbb{R}^2\}$ (standard topology) and 
$\mathscr T_2$ = $\{A: A \text{ is a union of sets in } B_2\}$  then try show $\mathscr T_1$ = $\mathscr T_2$. We could do this by saying any set in $\mathscr T_2$ will be open in $\mathbb{R}^2$ so will be in $\mathscr T_1$. And since the set of all open rectangles is a basis for $\mathbb{R}^2$ any open ball can be written as union of sets in $B_2$ hence is in $\mathscr T_2$. 
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A: Your second approach is too vague by far. Your first idea is much better. It amounts to showing that given any point in a disk (without boundary), there is a rectangular region (without boundary) that lies within the disk and contains the point, and that for any point in a rectangle (without boundary), there is a disk (without boundary) that contains the point and lies in the rectangular region.
A: The two bases, open rectangles and open disks, generate the same topology on R^2. This means that any set that is open with respect to one basis is also open with respect to the other basis, and vice versa.
Proof:
Open rectangles generate open disks: Let U be an open rectangle in R^2. Then, U can be written as a union of basic open rectangles, say U = ∪V_i. Each V_i is contained in an open disk with the same center and a slightly larger radius. Hence, U is contained in a union of open disks, so U is also an open disk.
Open disks generate open rectangles: Let U be an open disk in R^2. Then, U can be expressed as a union of basic open disks, say U = ∪V_i. Each V_i can be contained in an open rectangle with the same center and slightly larger sides. Hence, U is contained in a union of open rectangles, so U is also an open rectangle.
Therefore, the two bases, open rectangles and open disks, generate the same topology on R^2.
