# Given a set $A$, how do I prove that there exists a set of all sets $x$ such that $\bigcup x=A$?

I am working with Zermelo-Fraenkel axioms. Specifically, I am allowed to assume the Axiom of Pair, Axiom Schema of Comprehension, Axiom of Union, and Axiom of Power Set, etc. (not yet allowed to use Axiom of Choice but that really should not be relevant to this question)

I'm given that there's a set $A$.

I know that by the Axiom of Union, the following collection is a set:

$$\bigcup A = \left\{x \mid \exists\, y \in A \text{ such that } x \in y\right\}$$

I also know how to prove that any set is a subset of the power set of the union of itself. For $A$, this would be $$A \subseteq \mathcal{P}\left(\bigcup A\right)$$

I need to show that the set of all sets whose unions are $A$ exists. This set should be a subset of the power set of $A$ but I'm not sure how to actually prove that it exists.

I want to contruct the power set of $A$ and use the Axiom Schema of Comprehension to show that this set $B$ exists, with

$$B = \left\{x \mid \bigcup x = A\right\}$$

Any help would be appreciated ... again, I feel like this is going to be a one-liner :/

• Are you using $\mathcal U$ instead of $\cup$? – Git Gud Dec 17 '14 at 2:13
• @Ian Is $B$ a subset of $\mathcal{P}\mathcal{P}A$? I thought it was just a subset of $\mathcal{P}A$. Anyway, how exactly do I show that $B$ can be written as a subset of $\mathcal{P}A$, or an element of $\mathcal{P}\mathcal{P}A$? This is my problem ... – Vincent Luo Dec 17 '14 at 2:19
• @GitGud I'm using $\mathcal{U}$ as this thing. I think Brian M. Scott fixed my notation (thanks for that). – Vincent Luo Dec 17 '14 at 2:21
• My eyes nearly got cancer from that notation $\ddot \smile$ – Git Gud Dec 17 '14 at 2:21
• @Vincent: \bigcup gives you $\bigcup$, and \cup gives you $\cup$. Similarly with \bigcap and \cap for intersection symbols. – Brian M. Scott Dec 17 '14 at 2:22

You want a set $B$ that you can define from $A$ that is big enough so that if $\bigcup x=A$, then $x\in B$. Suppose that $\bigcup x\subseteq A$. If $y\in x$, then every element of $y$ is an element of $A$, so $y\subseteq A$, and $y\in\wp(A)$. This means that $x\subseteq\wp(A)$, and therefore $x\in\wp(\wp(A))$. Thus, you can take $B=\wp(\wp(A))$: it contains as an element every $x$ that could possibly satisfy $\bigcup x=A$.
• @GitGud: I’ve always thought that it would make sense to let even a first-timer vote on answers to his or her own question, but the $15$ rep limit is across the board. – Brian M. Scott Dec 17 '14 at 2:52