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Note: $P$ is power set. It's easy to prove that this inclusion holds. But when is other inclusion true? I can't even think of one example...

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First, what do you mean by $\cup A$? Second, what do you mean by converse? – Ted Shifrin May 17 '13 at 16:31
@TedShifrin $\cup A = \!\bigcup\limits_{a \in A}\! a$. AFAIK it's a standard notation. – kahen May 17 '13 at 16:39
Wow, @kahen, maybe I've avoided set theory my whole career for a good reason:) I guess this only makes sense when $A$ is a collection and not a set? But I only think of the power set of a set, not of a collection! Ugh :) – Ted Shifrin May 17 '13 at 16:44
@TedShifrin in ZF(C) everything is a set, so it makes sense to write even truly bizarre things like $\cup \pi$, the union of the elements of the set $\pi$. Of course it's probably not going to give you anything useful... – kahen May 17 '13 at 16:46
Regardless, you can imagine $\cup\cdot$ as a little boy who runs around with a hammer smashing all the "set bubbles" of the elements of the set it's operating on. It helps drawing a picture of what happens to sets like $\{\{0\},\{1\},\{2\}\}$ and $\{\{0,1\},\{0,2\},\{1,2\}\}$. – kahen May 17 '13 at 16:51
up vote 7 down vote accepted

Here is one example: $A=\{\varnothing\}$.

Also is $A=V_{\alpha+1}$ for any ordinal $\alpha$, then $A=P(V_\alpha)$, therefore $\bigcup A=V_\alpha$, and the equality holds.

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Isn't $P(A)$ a set with two elements then? – N. S. May 17 '13 at 16:32
But $P(\bigcup A)$ is not. – Asaf Karagila May 17 '13 at 16:32

If equality holds, then $A=\mathcal P(X)$ for some set $X$, namely $X=\bigcup A$. Conversely, note that if $A=\mathcal P(X)$, then $\bigcup A=X$. That is: $A=\mathcal P(\bigcup A)$ iff $A$ is the power set of some set $X$, in which case $X=\bigcup A$.

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