# Is a set with the form $\bigcup_{i,j\in I}U_i\times V_j$ an element of the base of the product topology?

Given $$(X,\tau)$$ a top. manifold, consider an open set $$O\in\tau^2$$(in the product topology of $$X\times X$$) with the form

$$O=\bigcup_{i,j\in I}U_i\times V_j$$

(note that not every open set has this form).

Is it true that $$O=\text{Dom }O\times\text{Im } O$$?

Here is my proof:

\begin{align} O&=\bigcup_{i,j\in I}U_i\times V_j \\ &=\bigcup_{i\in I}(\bigcup_{j\in I}U_i\times V_j) \\ &=\bigcup_{i\in I}(U_i\times(\bigcup_{j\in I}V_j)) \\ &=\bigcup_{i\in I}(U_i\times\text{Im }O) \\ &=(\bigcup_{i\in I}U_i)\times\text{Im }O \\ &=\text{Dom }O\times\text{Im }O \end{align}

Is this correct?

Thanks.

Instead of $$O=\text{Dom }O\times\text{Im }O$$ one could say equivalently that $$O=A\times B$$ for some $$A,B\in\tau$$.

Edit:

$$\text{Im}O=\{y\mid \exists x:(x,y)\in O\}$$ and $$\text{Dom}O=\{x\mid \exists y:(x,y)\in O\}=\text{Im}O^T$$.

• What is $\text{Dom}(O)$ and $\text{Im}(O)$? – G. Chiusole Aug 29 at 10:43
• So what you're saying is that all open sets are rectangles? The unit disc will be very very sorry to hear that. – Asaf Karagila Aug 29 at 10:48
• How is $\text{Dom}(O)$ defined? – G. Chiusole Aug 29 at 10:58
• That is not a general open set. It's the union of every possible product of two families of sets.. and so yes, i think this holds. The unit ball is not a set of this kind. – astrobarrel Aug 29 at 11:00
• Clearly a general open set is of the form $\bigcup A_j \times B_j$ and this is not, in general, a rectangle. Different is the situation where we take $\bigcup _{i,j} A_j \times B_i$ – astrobarrel Aug 29 at 11:02