# Let $A,B,C$ be sets such that $\mathcal{P}(A) \cup \mathcal{P}(B) = \mathcal{P}(C)$. Then either $A=C$ or $B=C$.

Let $A,B,C$ be sets such that $\mathcal{P}(A) \cup \mathcal{P}(B) = \mathcal{P}(C)$. Then either $A=C$ or $B=C$.

I just need a little direction with this. I know that since $\mathcal{P}(A) \cup \mathcal{P}(B)$, then $(x\in \mathcal{P}(A)) \lor (x\in \mathcal{P}(B))$ and also $[\mathcal{P}(A) \cup \mathcal{P}(B) = \mathcal{P}(C)] \Rightarrow [(\mathcal{P}(A) \cup \mathcal{P}(B))\subseteq \mathcal{P}(C)] \land[\mathcal{P}(C)\subseteq (\mathcal{P}(A) \cup \mathcal{P}(B))]$, but I honestly have no idea what to do with this information. Should I break the union down into truth statements like so: $\mathcal{P}(A) \cup \mathcal{P}(B)\Rightarrow (x\in \mathcal{P}(A)) \lor (x\in \mathcal{P}(B))$?

Any direction would be greatly appreciated!

• math.stackexchange.com/questions/345978/… and its duplicate and related threads. – Asaf Karagila Mar 22 '16 at 19:54
• HINT: $C \in \mathcal{P}(C)$, $A \in \mathcal{P}(A)$, $B \in \mathcal{P}(B)$. – Crostul Mar 22 '16 at 19:56
• @AsafKaragila So, in the case of the question you linked, which I duplicated, I can just use the fact that $A \subseteq A \cup B \subseteq B$ or $B \subseteq A \cup B \subseteq A$ to show that $A=C$ or $B=C$? – CreasyBear Mar 22 '16 at 20:05
• @Crostul Thanks, I think that helped. Here's what I've got, maybe you can let me know if I'm looking at it right! We have: $A\in \mathcal{P}(A)$, $B\in \mathcal{P}(B)$, and $C\in \mathcal{P}(C)$ $\Rightarrow A,B\in \mathcal{P}(A) \cup \mathcal{P}(B) = \mathcal{P}(C)$ $\Rightarrow A,B,C \in \mathcal{P}(C)$ At this point, I can only assume that there is something I should be able to infer, but I'm not seeing it. – CreasyBear Mar 22 '16 at 21:28
• You have concluded that $A,B,C$ are all subsets of $C$. But now, $C \in \mathcal{P}(A) \cup \mathcal{P}(B)$, so that $C$ is a subset of $A$ or $C$ is a subset of $B$. – Crostul Mar 22 '16 at 21:44