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Let $(X,\tau)$ be a topological space and let $$\Omega(X)=\{A\subseteq X: \text{there is surjective continuous }f:X\to A\}.$$ Can we express $\Omega(X\times X)$ in terms of $\Omega(X)$? (We assume $X\times X$ to carry the product topology.)

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  • $\begingroup$ You don't assume anything regarding connectivity of $X$? $\endgroup$ – H. H. Rugh Aug 15 '16 at 10:42
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    $\begingroup$ How do you formalize the verb "to express"? One possible way is that you are given $X$ and $\Omega(X)$ but not the topology of $X$ and the question is whether you can find $\Omega(X\times X)$ but I'm not sure if that's what you had in mind. $\endgroup$ – fedja Aug 16 '16 at 15:47
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    $\begingroup$ What's the motivation for this question? $\endgroup$ – Todd Trimble Oct 14 '16 at 11:43
  • $\begingroup$ @fedja We don't need both $X$ and $\Omega(X)$ as input, since $X$ is determined by $\Omega(X)$, as the largest element of it. I interpret the question as: can we calculate the continuous images of the product $X\times X$ from the set of continuous images of $X$ in $X$? In other words, is $\Omega(X\times X)$ a function of $\Omega(X)$? For a negative answer, you need to find a set $X$ with two topologies on it, such that they have the same set of continuous images $X\to A\subset X$, but different continuous images of the product $X\times X$ in itself. $\endgroup$ – Joel David Hamkins Oct 14 '16 at 13:10
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First it is clear that $$\{ A \times B: A, B\subset \Omega (X) \} \subset \Omega(X \times X) \}$$ However, just thinking about $(0,1)$, it is easy to see that this is not an equality, and that, in fact, it is much larger. For example $$\Omega( (0,1) ) = \{ I : I \textrm{ interval} \} $$ whereas $$ \Omega( (0,1)^2 ) \supset \{ U: U \textrm{ homeomorphic to a square } \}$$ which includes not only squares, but ellipses and whichever other shape you can think of.

Cheers

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    $\begingroup$ Both displayed equations are wrong: no reason for $U$ to be open. $\endgroup$ – Adam Przeździecki Sep 14 '16 at 10:51
  • $\begingroup$ Correct. Edited $\endgroup$ – D G Sep 14 '16 at 11:06
  • $\begingroup$ Do you mean $A,B\in\Omega(X)$ rather than $\subset$ in the first line? $\endgroup$ – Joel David Hamkins Oct 14 '16 at 13:02

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