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Suppose that $A,B$ are sets. Is $$ \{ y \in B | \exists x \in A: y \in x \} $$ a set? I tried to find a formal proof (replacement? comprehension?) or to show that it doesn't exist (regularity?), but I'm not really familiar with this stuff. Of course it must have to do something with ur-elements or power sets, but I'm to confused now to figure out what is going on.

In the same spirit: If $A$ is a set, is it meaningful to ask whether there exists a set $\Omega$ such that $A=\mathbb{P}(\Omega)$?

[edit]: Hm, what about:If $A$ is a set, is it meaningful to ask whether there exists a set $\Omega$ such that $A\subseteq\mathbb{P}(\Omega)$?

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Is a subset of $B$... – Martín-Blas Pérez Pinilla Feb 13 '14 at 11:17
As for your second question: yes it is meaningful to ask but it might not exist – dani_s Feb 13 '14 at 11:18
If $A$ and $B$ are sets, rewriting the condition as $\{ y : y \in B \land \exists x (x \in A \land y \in x) \}$, you must apply Axiom schema of Separation and you will get it. – Mauro ALLEGRANZA Feb 13 '14 at 11:20
How does this give a formula $\phi$ which I can use in the Axiom of Separation? Your formula contains $A,B$? Or am I so confused? – beginner3115 Feb 13 '14 at 11:58
It is a formula $\phi(y)$ with a free variable: so it is ok. And the Axiom needs that you "separate" your new set from an existing one, in order to license you to assert that the condition gives you a set, and this is the $y \in B$ part. – Mauro ALLEGRANZA Feb 13 '14 at 12:01

About the second question. It is meaningful to ask. The answer is obviously false in general.

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