Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
2  
First, what do you mean by $\cup A$? Second, what do you mean by converse? –  Ted Shifrin May 17 '13 at 16:31
1  
@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
1  
@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
1  
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

2 Answers 2

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.

share|improve this answer
    
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 $X=\bigcup A$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.