# Prove Set theroy

• April 17th 2010, 03:35 PM
treetheta
Prove 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
• April 17th 2010, 04:19 PM
Quote:

Originally Posted by treetheta
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

Lol usually if your lecturer tells you it's wrong it's wrong...

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

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 $A^c$ and also $B^c$.

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

Then to prove the reverse...

Well, you try it.
• April 17th 2010, 07:22 PM
treetheta
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?
• April 19th 2010, 08:24 AM
Defunkt
Quote:

Originally Posted by treetheta
so then

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

hence, X is an element of A^c and B^c No: x is an element of $A^c$ OR $B^c$. This renders the rest of the proof invalid..

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?

It is the correct approach.
Now, after stating $x \in A^c \cup B^c \Rightarrow x \in A^c$ OR $x \in B^c$, separate to 2 cases:

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

Try to formally write this, it usually helps..