# Homework Help: Simple set theory question

1. Feb 19, 2010

### dreyvas

1. The problem statement, all variables and given/known data

We're working more or less with the standard ZF axioms.

Prove that $$A \subseteq \mathcal{P}(\bigcup A)$$ for any set A, whose elements are all sets. When are they equal?

2. Relevant equations
Just the axioms

I) Extensionality
II) Emptyset and Pairset
III) Separation
IV) Powerset
V) Unionset
VI) Infinity

3. The attempt at a solution

I can prove the first part pretty easily. We have that the union and power set of the union are sets directly from V and IV.

Briefly, suppose $$X \in A$$. Then $$X \subseteq \bigcup A$$ by def. of union. So $$X \in \mathcal{P}(\bigcup A)$$.

The part I'm stuck on is when they're equal. I attempted to prove that $$\mathcal{P}(\bigcup A) \subseteq A$$ in order to get an idea of what condition A would need to meet.

My best guess is that they're only equal when A (at the very least) contains the emptyset and the singletons of every element in the union of all of A. Seems like a pretty circular definition, though, since I need to know what A is to know what the union of all of A is. Maybe I could say that A needs to be composed of the emptyset and the remaining elements must be singletons or else be the union of singletons already contained in A? I have thought about this a fair bit, and I'm pretty sure my condition ensures equality, so this isn't a random guess. Help would be appreciated.

Also, we have not really gotten into cardinality yet, but I do know that $$|S| \leq |\bigcup S|$$ if S is countable. But I also know that $$|\bigcup S| < |\mathcal{P}(\bigcup S)|$$. So for our equality to hold, S must be uncountable...?

Last edited: Feb 19, 2010
2. Feb 19, 2010

### ystael

Because you do not have any a priori structure on $$A$$ ($$A$$ is not assumed to be closed under unions or intersections or anything like that), there is no "simpler" condition that $$A$$ be all of $$\mathcal{P}\left(\textstyle\bigcup A\right)$$. To say, as you have, that $$A$$ must contain the empty set, singletons, and all unions of singletons contained in $$A$$, is more or less exactly to say that $$A = \mathcal{P}\left(\textstyle\bigcup A\right)$$: that is, it's not really a useful sufficient condition.

Your first statement here is false: take $$S = \{\emptyset, \{0\}, \{1\}, \{0, 1\}\}$$; then $$|S| = 4$$ while $$\left|\textstyle\bigcup S\right| = |\{0, 1\}| = 2$$.

Last edited: Feb 19, 2010