# Simple set theory question

## Homework Statement

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?

## Homework Equations

Just the axioms

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

## 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...?

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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.

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.

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...?

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

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