I have no idea how to work on the following proof. Any Suggestions?
Prove that for any sets A and B, if P(A) ∪ P(B) = P(A ∪ B) then either A ⊆ B or B ⊆ A.
Thanks
Asked
Active
Viewed 530 times
-1
-
The union symbol makes no sense between two numbers (the two probabilities P(A) and P(B).) Also even if it were replaced with + the statement wouldn't hold. – coffeemath Nov 13 '14 at 23:47
-
It's the power set: the set of all subsets. $\mathcal P(A) = {X: X\subseteq A}$ – Graham Kemp Nov 13 '14 at 23:52
-
1Bad comment -- look at the title of the question. $P(S)$ is in this context being used as the power set of set $S$ (the set of all subsets of $S$). – Mark Fischler Nov 13 '14 at 23:53
-
@MarkFischler Agreed, read too fast! – coffeemath Nov 14 '14 at 07:15
2 Answers
2
Suppose there exists $ a \in A, a \notin B $ and $ b \notin A, b \in B $ (contradictory to the hypothesis).
Then $ \{a,b\} \in P(A\cup B) $, but $ \{a,b\} \notin P(A) \cup P(B) $ and it's a contradiction.
Jytug
- 3,169
0
We have $A \cup B \in P(A \cup B) = P(A) \cup P(B)$. Therefore $A \cup B \in P(A)$ or $A \cup B \in P(B)$.
Say, for instance, that it's the first. Then $B \subseteq A \cup B \subseteq A$.
In the second case, we likewise prove $A \subseteq B$.
Mike
- 66