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I am having difficulties with this proof--Hints will be appreciated.

Prove that for any sets $A$ and $B$, if $\mathcal{P}(A) \cup \mathcal{P}(B) = \mathcal{P}(A\cup B)$ then either $A$ is a subset of $B$ or $B$ is a subset of $A$.

Andrés E. Caicedo
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Pascal
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    I assume that P is being used as power set notation, not as probability, yes? Consider the contrapositive statement then, if $A\not\subseteq B$ and $B\not\subseteq A$, then that says that there is some $a\in A$ and $b\in B$ such that $a\notin B$ and $b\notin A$. What can you say about the subset ${a,b}$ then? – JMoravitz Feb 22 '15 at 15:34
  • Ah, thank you so much!(that was subtle) – Pascal Feb 22 '15 at 15:42

4 Answers4

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Hint: We have $A\cup B\in\mathcal P(A\cup B)$. By hypothesis $\mathcal P(A)\cup\mathcal P(B)=\mathcal P(A\cup B)$, we obtain $A\cup B\in\mathcal P(A)\cup\mathcal P(B)$. Now, by definition of union, we have $A\cup B\in\mathcal P(A)$ or $A\cup B\in\mathcal P(B)$, and so $A\cup B\subseteq A$ or $A\cup B\subseteq B$ by definition of power set. In particular, we know that $A\subseteq A\cup B$ and $B\subseteq A\cup B$; thus the claim follows.

Cristhian Gz
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    +1 although I don't see how this is a "hint" :-) – k_g Feb 22 '15 at 19:49
  • Missing the obvious statements: why $A\cup B\in\mathcal P(A\cup B)$ (start) and why $A\subseteq B$ or $B\subseteq A$ (end). Indeed, now I think it is not a hint. :-) – Cristhian Gz Feb 22 '15 at 20:01
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Suppose that we have the assumption but neither of the inclusions holds. Then $A \nsubseteq B$ so there is some $a \in A, a \notin B$ and also $B \nsubseteq A$ and so there is some $b \in B, b \notin A$. Now consider $\{a,b\} \subseteq A \cup B$, so $\{a,b\} \in \mathcal{P}(A \cup B)$.

user26486
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Henno Brandsma
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As $A \cup B \in \mathcal{P} (A \cup B) = \mathcal{P} (A ) \cup \mathcal{P} (B)$ it follows that $A \cup B \in \mathcal{P} (A)$ or $A \cup B \in \mathcal{P} (B)$.

Thus $A \cup B \subset A$ or $A \cup B \subset B$. Therefore $B \subset A \cup B \subset A$ or $A \subset A \cup B \subset B$.

N. S.
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I have to admit I was confused by the question until I noticed the following. When you think of $\mathcal{P}(A) \cup \mathcal{P}(B)$ watch out. It isn't true that:

$$\mathcal{P}(A) \cup \mathcal{P}(B) = \{\alpha \cup \beta: \alpha \subset A, \beta \subset B\}$$

but rather

$$\mathcal{P}(A) \cup \mathcal{P}(B) = \{\alpha: \alpha \subset A \ or \ \alpha \subset B \}.$$

equin
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