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 $B\subseteq A$.

Question

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$?$

Thanks for your attention.

Answer 1

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 $B\subseteq A$.