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

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  • $\begingroup$ What is $\text{Dom}(O)$ and $\text{Im}(O)$? $\endgroup$ – G. Chiusole Aug 29 at 10:43
  • $\begingroup$ So what you're saying is that all open sets are rectangles? The unit disc will be very very sorry to hear that. $\endgroup$ – Asaf Karagila Aug 29 at 10:48
  • $\begingroup$ How is $\text{Dom}(O)$ defined? $\endgroup$ – G. Chiusole Aug 29 at 10:58
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    $\begingroup$ 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. $\endgroup$ – astrobarrel Aug 29 at 11:00
  • $\begingroup$ 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 $\bigcup _{i,j} A_j \times B_i$ $\endgroup$ – astrobarrel Aug 29 at 11:02
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Yes, that's correct. Your equalities all work perfectly.

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