Prove $P(A \cup B)= P(A)\cup P(B) \leftrightarrow (A \subseteq B) \vee (B \subseteq A)$

Question

I want to prove this claim: $$P(A \cup B)= P(A)\cup P(B) \leftrightarrow (A \subseteq B) \vee (B \subseteq A)$$ I need to prove the both sides? Or which side I need to start with?
I would like to get some hints.
thanks!

It's generally best to give some sense of context when asking questions. I know very well what a power set is... — dfeuer, Jul 18 '13 at 07:28
Yes, but I initially figured it was a probability question, because no information was given about what $P$, $A$, and $B$ might represent. — dfeuer, Jul 18 '13 at 07:29
@dfeuer Sorry, thought its obvious. next time I will write it. — Ofir Attia, Jul 18 '13 at 07:30
@OfirAttia for the powetset symbol use $\mathcal P$ (tex: \mathcal P). — Ittay Weiss, Jul 18 '13 at 07:31
See also: http://math.stackexchange.com/questions/246491/stuck-with-proof-for-forall-a-forall-b-mathcalpa-cup-mathcalpb-mathc — Martin Sleziak, Apr 01 '16 at 05:33

score 0 · Answer 1 · answered Jul 18 '13 at 07:31

The reverse implication should be entirely clear to you.

Suppose now that $\mathcal P(A \cup B) = \mathcal P(A) \cup \mathcal P(B)$.

Then in particular $\mathcal P(A\cup B) \subseteq \mathcal P(A) \cup \mathcal P(B)$.

Can you see how to proceed from there? Think about what sets are in $\mathcal P(A) \cup \mathcal P(B)$.

score 0 · Answer 2 · answered Jul 18 '13 at 07:33

You need to prove both sides of the implication. So, assume first that either $A\subseteq B$ or $B\subseteq A$. Without loss of generality, suppose $A\subseteq B$. Now porve that $\mathcal P(A\cup B)=\mathcal P(A)\cup \mathcal P(B)$. This is most easily done by noticing that in this case $A\cup B=B$.

Next, assume that $\mathcal P(A\cup B)=\mathcal P(A)\cup \mathcal (B)$. You now wish to show that either $A\subseteq B$ or $B\subseteq A$. Suppose neither holds. Can you find a contradiction?

score 0 · Answer 3 · answered Jul 18 '13 at 08:10

Suppose $\mathcal P(A \cup B)= \mathcal P(A)\cup \mathcal P(B)$. Since $A\cup B\subseteq A\cup B$, $A\cup B \in \mathcal P(A\cup B)$ and thus $A\cup B\in \mathcal P(A) \cup \mathcal P(B)$. So either, $A\cup B\subseteq A$ or $A\cup B\subseteq B$ and it is easy to show that the two statements imply $B \subseteq A$ and $A \subseteq B$ respectively.

Suppose $A \subseteq B$ or $B \subseteq A$. Without loss of generality assume $A \subseteq B$. Then it is easy to show that $\mathcal P(A)\subseteq \mathcal P(B)$. Now prove that $X\subseteq Y \implies X\cup Y=Y$. So $A\cup B = B$ and $\mathcal P(A)\cup\mathcal P(B)= \mathcal P(B)$. Thus $\mathcal P(A \cup B)= \mathcal P(B)= \mathcal P(A)\cup \mathcal P(B)$.

score 0 · Accepted Answer · answered Jul 18 '13 at 08:18

($\Longleftarrow$) If $A \subseteq B$, then $A \cup B = B$, also $P(A) \subseteq P(B)$ and finally $P(A) \cup P(B) = P(B)$, hence $$P(A \cup B) = P(B) = P(A) \cup P(B).$$ Analogously for $B \subseteq A$.

($\Longrightarrow$) Suppose that $A \not\subseteq B$ and $B \not\subseteq A$, then there exist elements $a \in A$ and $b \in B$ such that $a \notin B$ and $b \notin A$. We have that $$\{a,b\} \subseteq A \cup B,\quad \text{ so }\quad \{a,b\} \in P(A \cup B),$$ but $$\{a,b\} \not\subseteq A \quad \text{ implies }\quad \{a,b\} \notin P(A),$$ $$\{a,b\} \not\subseteq B\quad\text{ implies }\quad\{a,b\} \notin P(B),$$ so $$P(A \cup B)\ \ \ni\ \ \{a,b\}\ \ \notin \ \ P(A) \cup P(B),$$ hence, $P(A \cup B) \neq P(A) \cup P(B)$.

I hope this helps ;-)

Prove $P(A \cup B)= P(A)\cup P(B) \leftrightarrow (A \subseteq B) \vee (B \subseteq A)$

4 Answers4