# Prove that for any sets $A$ and $B$, if $\mathscr P(A)\cup\mathscr P(B)=\mathscr P(A\cup B)$ then either $A\subseteq B$ or $B\subseteq A$.

Not a duplicate of

Suppose $\mathcal{P} (A) \cup \mathcal{P} (B) = \mathcal{P} (A \cup B)$. Then either $A \subseteq B$ or $B \subseteq A$.

Prove that if $\mathcal P(A) \cup \mathcal P(B)= \mathcal P(A\cup B)$ then either $A \subseteq B$ or $B \subseteq A$.

Prove that if $\mathcal{P}(A)\cup\mathcal{P}(B)$=$\mathcal{P}(A\cup B)$ then $A\subseteq B$ or $B\subseteq A$

Proof verification: $P(A\cup B)=P(A)\cup P(B)\rightarrow A\subseteq B\vee A\supseteq B$

How do you prove $P(A) \cup P(B) = P(A \cup B) \Rightarrow (A \subseteq B) \lor (B \subseteq A)$

This is exercise $$3.5.8$$ from the book How to Prove it by Velleman $$(2^{nd}$$ edition$$)$$:

Prove that for any sets $$A$$ and $$B$$, if $$\mathscr P(A)\cup\mathscr P(B)=\mathscr P(A\cup B)$$ then either $$A\subseteq B$$ or $$B\subseteq A$$.

Here is my proof:

Let $$A$$ and $$B$$ be arbitrary sets. Suppose $$\mathscr P(A)\cup\mathscr P(B)=\mathscr P(A\cup B)$$. Now we consider two different cases.

Case $$1.$$ Suppose $$A\subseteq B$$. Ergo $$A\subseteq B$$ or $$B\subseteq A$$.

Case $$2.$$ Suppose $$A\nsubseteq B$$. So we can choose some $$x_0$$ such that $$x_0\in A$$ and $$x_0\notin B$$. Let $$y$$ be an arbitrary element of $$B$$. Since $$A\cup B\in\mathscr P(A\cup B)$$, then $$A\cup B\in\mathscr P(A)\cup\mathscr P(B)$$. So either $$A\cup B\subseteq A$$ or $$A\cup B\subseteq B$$. Again we consider two different cases.

Case $$2.1.$$ Suppose $$A\cup B\subseteq A$$. Since $$y\in B$$, $$y\in A\cup B$$. Ergo $$y\in A$$.

Case $$2.2.$$ Suppose $$A\cup B\subseteq B$$. Since $$x_0\in A$$, $$x_0\in A\cup B$$. Ergo $$x_0\in B$$ which is a contradiction.

From $$y\in A$$ or a contradiction we obtain $$y\in A$$. Thus if $$y\in B$$ then $$y\in A$$. Since $$y$$ is arbitrary, $$\forall y(y\in B\rightarrow y\in A)$$ and so $$B\subseteq A$$. Ergo $$A\subseteq B$$ or $$B\subseteq A$$.

Since case $$1$$ and case $$2$$ are exhaustive, $$A\subseteq B$$ or $$B\subseteq A$$. Therefore if $$\mathscr P(A)\cup\mathscr P(B)=\mathscr P(A\cup B)$$ then either $$A\subseteq B$$ or $$B\subseteq A$$. Since $$A$$ and $$B$$ are arbitrary, $$\forall A\forall B\Bigr(\mathscr P(A)\cup\mathscr P(B)=\mathscr P(A\cup B)\rightarrow(A\subseteq B\lor B\subseteq A)\Bigr)$$. $$Q.E.D.$$

Is my proof valid$$?$$

It’s correct, but it’s much too wordy and far more complicated than necessary. To prove a theorem of the form $$X\implies Y\text{ or }Z$$, it suffices to show that if $$X$$ holds and $$Y$$ does not, then $$Z$$ must hold. Here that means that we need only show that if $$\wp(A\cup B)=\wp(A)\cup\wp(B)$$ and $$A\nsubseteq B$$, then $$B\subseteq A$$. This can be done in five lines, even writing it up in fairly wordy fashion:
Suppose that $$\wp(A\cup B)=\wp(A)\cup\wp(B)$$, but $$A\nsubseteq B$$. $$A\cup B\in\wp(A\cup B)$$, so $$A\cup B\in\wp(A)\cup\wp(B)$$, and therefore $$A\cup B\in\wp(A)$$, or $$A\cup B\in\wp(B)$$. $$A\subseteq A\cup B$$, so if $$A\cup B\in\wp(B)$$, then $$A\in\wp(B)$$, and therefore $$A\subseteq B$$, contradicting our assumption that $$A\nsubseteq B$$; thus, we must instead have $$A\cup B\in\wp(A)$$. And $$B\subseteq A\cup B$$, so this implies that $$B\in\wp(A)$$ and hence that $$B\subseteq A$$.