Prove that for any two sets A;B , we have (A n B)^c = A^c u B^c

I think im supposed to start expansion with the de morgan laws my teacher said I was doing it wrong I'm pretty sure this is the correct approach but i dont know how to continue

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- Apr 17th 2010, 03:35 PMtreethetaProve Set theroy
Prove that for any two sets A;B , we have (A n B)^c = A^c u B^c

I think im supposed to start expansion with the de morgan laws my teacher said I was doing it wrong I'm pretty sure this is the correct approach but i dont know how to continue - Apr 17th 2010, 04:19 PMDeadstar
Lol usually if your lecturer tells you it's wrong it's wrong...

Consider an element $\displaystyle x \in (A \cap B)^c$.

Then x is NOT contained in the intersection of A and B.

Hence x is NOT contained in A or B.

Hence x IS contained in $\displaystyle A^c$ and also $\displaystyle B^c$.

So $\displaystyle x \in A^c \cup B^c$.

Then to prove the reverse...

Well, you try it. - Apr 17th 2010, 07:22 PMtreetheta
so then

x is an element of A^c u b^c

hence, X is an element of A^c and B^c

so x is not contained in A or B

so x is an element of (AnB)^c since AnB is everything thats in both A and B

hence

(A n B)^c = A^c u B^c

is that the correct approach? - Apr 19th 2010, 08:24 AMDefunkt
It is the correct approach.

Now, after stating $\displaystyle x \in A^c \cup B^c \Rightarrow x \in A^c$ OR $\displaystyle x \in B^c$, separate to 2 cases:

1) If $\displaystyle x \in A^c$, is x in $\displaystyle (A \cap B)^c$?

2) If $\displaystyle x \in B^c$, is x in $\displaystyle (A \cap B)^c$?

Try to formally write this, it usually helps..