# 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$$?$$

Thanks for your attention.

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 $$B\subseteq A$$.

• From my experience on this website, it seems that most of my proofs are rather unnecessarily complicated. But proofs are very new to me. How can I improve my proofs to be more cogent? Jul 16 '20 at 19:50
• @KhashayarBaghizadeh: It’s a common problem for beginners who are trying to be very careful and justify every possible detail. Perhaps the single most helpful thing is to pay careful attention to the organization and level of detail in proofs written by more experienced mathematicians. And practice, which you’re getting, does help. Jul 16 '20 at 19:58