# $A' \cap (A' \cup B') \cap T= A' \cap T$

Let $A'$ denotes the complement of A with respect to $\mathbb{R}$ and $A,B,T$ are subsets of $\mathbb{R}$. I am trying to prove $A' \cap (A' \cup B') \cap T= A' \cap T$, but I got some problems along the way.

$A' \cap (A' \cup B') \cap T= (A' \cap A') \cup (A' \cap B') \cap T= A' \cup (A \cup B) \cap T =(A' \cup A)\cup B \cap T= \mathbb{R} \cup B \cap T = B \cap T$ Something wrong?

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How did you get $A^\prime \cap B^\prime$ to equal $A \cup B$? – Dilip Sarwate Jul 30 '12 at 1:17
@Dilip de morgan – Daniel Jul 30 '12 at 1:18
Try again. What do DeMorgan's Laws really say as opposed to what you think they say? – Dilip Sarwate Jul 30 '12 at 1:22
@DilipSarwate shoot that is not De Morgan's law. I should be able to remedy it – Daniel Jul 30 '12 at 1:24
There really is no reason to think of $A'$ as a complement. – Ink Jul 30 '12 at 1:25

Well, given arbitrary sets $X,Y$ we always have $X\cap(X\cup Y) = X$ since the intersection of a set with a union of that same set and anything else is just the first set itself (I hope that's not too wordy). Try drawing all possibilites with sets to see this. So $A'\cap (A' \cup B') = A'$ and that gives you your answer.

Your fault is that in the second equality does not hold: for arbitrary sets $X,Y$ in general $(X \cap X) \cup (X\cap Y) \neq X \cup (X'\cup Y')$. Again, drawing all possibilites really helps to make this clear. Here I'm generalizing slightly from what you have, so for you $X$ is $A'$ and $Y$ is $B'$. Recall $A'' = A$.

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you are right on this one. I am too hasty to think that I am using the De Morgan's law correctly. – Daniel Jul 30 '12 at 1:45

Let $x \in A' \cap (A' \cup B) \cap T$. Then $x \in A'$ and $x \in T$, so $x \in A' \cap T$. This proves that $A' \cap (A' \cup B) \cap T \subset A' \cap T$. The reverse inclusion is similar.

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Dilip's comments point out the flaw re: DeMorgan's Laws.

To avoid that mess altogether, a both-directions-subset proof works concisely. On the one hand, it is clear that $A' \cap (A' \cup B') \cap T \subset A' \cap T$, since the LHS intersects all of the RHS's elements with another set.

On the other hand, expanding the LHS yields $A' \cap ((A' \cap T) \cup(B' \cap T)) = (A' \cap T)\cup(A' \cap B' \cap T)$ . Certainly, $A' \cap T \subset (A' \cap T)\cup(A' \cap B' \cap T)$ .

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$A' \cap (A' \cup B') \cap T= ((A' \cap A') \cup (A' \cap B')) \cap T= (A' \cup (A' \cap B')) \cap T$. But, $A' \cap B' \subset A'$ and so $A' \cup (A' \cap B') = A'$ giving $(A' \cup (A' \cap B')) \cap T = A' \cap T$ without dragging DeMorgan's Laws into it.

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