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$P(A) \cup P(B) = P(A \cup B)$ iff $A \subset B$ or $B \subset A$

I just need to proof the direction where right to left direction notice here we could easily go being $B \subset A$ if we do $B \subset A \cup B$ proof:

let $x \in P(A) \cup P(B) = P(A \cup B)$

We have $P(A \cup B) \subset P(A) \cup P(B)$

Hence $A \subset x \subset A \cup B \subset (x \subset A\text{ OR }x \subset B)$

Hence $A \subset B$.

Michael Hardy
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  • never mind I noticed a mistake here we have A \subset$ x which is not true in general I can't conclude the above I will think in how to change it, then will post correct version here... –  Mar 11 '15 at 02:51
  • Related: http://math.stackexchange.com/q/246491/11994 (which is about the left-to-right direction) and http://math.stackexchange.com/q/499746/11994. – MarnixKlooster ReinstateMonica Mar 11 '15 at 06:56

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($\impliedby$) If wlog $A\subseteq B$ then $P(A)\subseteq P(B)$, so that $P(A)\cup P(B)=P(B)= P(A\cup B)$. ($\implies$) Conversely if $A\not\subseteq B$ and $B\not\subseteq A$, then $A\setminus B\neq\emptyset\neq B\setminus A$. Let $a\in A\setminus B$ and $b\in B\setminus A$. Then $\{a,b\}\in P(A\cup B)\setminus (P(A)\cup P(B))$.

EDIT: We may prove ($\implies$) directly as follows: Suppose $A\not\subseteq B$, and let $a\in A\setminus B$. Then $C:=\{a\}\cup B\not\in P(B)$, but $C\in P(A\cup B)=P(A)\cup P(B)$. Hence $C\in P(A)$. As $B\subseteq C\subseteq A$, $B\subseteq A$.

Alp Uzman
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