Annoying set theory question

**apologies if this is the wrong part of the forum... it was the closest I could find :) **

Whilst doing questions in preparation for uni this year, I've been working on some abstract algebra. I was doing pretty well until I came across this:

$\displaystyle P=(A \cap B) \cup (C \cap D)$

$\displaystyle Q=(A \cup B) \cap (C \cup D)$

Prove that $\displaystyle P \subseteq Q$

It looks like it should be easy but it's been resisting all attempts that I've tried. I've tried a direct proof that $\displaystyle x \in P \Rightarrow x \in Q$ but that didn't go anywhere, nor did a contradiction.

Can anyone please make me look a fool and pull out a solution that is from THE BOOK? (Nerd)

Re: Annoying set theory question

Quote:

Originally Posted by

**cac2008** $\displaystyle P=(A \cap B) \cup (C \cap D)$

$\displaystyle Q=(A \cup B) \cap (C \cup D)$

Prove that $\displaystyle P \subseteq Q$

Let $\displaystyle A=\{1,2\},~B=\{2,3\},~\&~C=\{3,4\},~D=\{4,5,\},~$

Is it true?

2 Attachment(s)

Re: Annoying set theory question

Quote:

Originally Posted by

**Plato** Let $\displaystyle A=\{1,2\},~B=\{2,3\},~\&~C=\{3,4\},~D=\{4,5,\},~$

Is it true?

With those, P={2,4} and Q={3} so no, it isn't.

Just done a Venn diagram (why did I not do that earlier!?) and it looks like the question is written wrong as:

Attachment 28009

Attachment 28010

Please tell me that this is not a common thing with Uni textbooks!

Re: Annoying set theory question

Plato has given a counter example to the theorem, which evidently disproves the theorem. If the question doesn't place constraints stating that all the sets have to be different,

we could try making simpler counter examples revolving around the empty set.

A = B = {1}

C = D = ∅

Then, {1} is a subset of ∅ will obviously be false.