# Proof of Set Identity

• Oct 12th 2010, 11:40 PM
mremwo
Proof of Set Identity
Show that if (A∩C)=(B∩C) then A=B, where A, B, C are sets.

*by using a 2- way containment method without using set identities.
*2-way method meaning showing that two sets are equal by showing they contain each other

for example, D=E (where D=(A∩C) and E=(B∩C)) iff D⊆E ^ E⊆D.
so i would split these two up (proof by cases) and prove:
firstly i) D⊆E by proving (x∈D→x∈E)
secondly ii) E⊆D by proving (x∈E→x∈D)

*Continuing with this example, I would show how this implication (of parts i and ii) as a whole implies that A=B

But every time I attempt to do this, I feel like I go around in circles. Or at least that I'm doing too much and making it too complicated so that I forget what I'm trying to prove. please help!
• Oct 13th 2010, 12:38 AM
Swlabr
The result isn't true. For example, just take $C = A \cap B$.

(An explicit counterexample would be: take A= $\{1, 2, 3\}$, $B=\{2, 3, 4\}$ and $C=\{2, 3\}$. Then $A\capC = \{2, 3\} = B\cap C$...)
• Oct 13th 2010, 02:30 AM
Ackbeet
The result might be true if you assume that for all $C,\;A\cap C=B\cap C.$ Might that be what the OP is getting at?
• Oct 13th 2010, 06:16 AM
mremwo
**** WOW, sorry guys! BIG mistake
Asking this question might make more sense if you knew that I screwed up while asking it, of course.
What I mean to ask about is the union of these sets NOT the intersection. As in, everywhere on this problem where I put the intersection is supposed to be the union.

I actually came up with a counterexample very similar to that of Swlabr

• Oct 13th 2010, 06:21 AM
Swlabr
Quote:

Originally Posted by mremwo
**** WOW, sorry guys! BIG mistake
Asking this question might make more sense if you knew that I screwed up while asking it, of course.
What I mean to ask about is the union of these sets NOT the intersection. As in, everywhere on this problem where I put the intersection is supposed to be the union.

I actually came up with a counterexample very similar to that of Swlabr

Yeah, this isn't true either, and your right, in that counter-examples are very similar to those for the intersection. Just take $A=C$, $B \subsetneq C$ ( $B$ a proper subset of $C$)...