Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 2 down vote accepted

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 $A'' = A$.

share|cite|improve this answer
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.

share|cite|improve this answer

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, $A' \cap T \subset (A' \cap T)\cup(A' \cap B' \cap T)$ .

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.