# Thread: Can someone verify/correct this proof?

1. ## Can someone verify/correct this proof?

Can you conclude that A=B if A,B, and C are sets such that:
A U C = B U C and A intersection C = B intersection C

I know its true but I'm not sure of the proof:

pf: Let A,B,C be sets.
Suppose x be in AUC. Then x is in BUC.
Then x is in A and x is in B or x is in.
Suppose x is in A intersect C. Then x is in B intersect C.
Then x is A, x is in B, and x is in C.
Since x is in A and x is in then x is in AUC, BUC.
Since x is in C, then x is in AUC, BUC.
Thus, from these last two lines, x in A
intersect C and x is in B intersect C.
Therefore, if x is in AUC=BUC then x is automatically in A and B or in C.
If x is in A
intersect C = B intersect C, then x is A, B, and C.
A=B since x is in the intersection of A and C and
intersection of B and C and also since x appears in the AUC and BUC regardless of C, because of the statement x is in A and x is in B or x is in C.

I really don't think this proves anything...anyone know of a more obvious or quick approach?

2. Originally Posted by mathgirl13
Can you conclude that A=B if A,B, and C are sets such that:
A U C = B U C and A intersection C = B intersection C

I know its true but I'm not sure of the proof:

pf: Let A,B,C be sets.
Suppose x be in AUC. Then x is in BUC.
Then x is in A and x is in B or x is in.
Suppose x is in A intersect C. Then x is in B intersect C.
Then x is A, x is in B, and x is in C.
Since x is in A and x is in then x is in AUC, BUC.
Since x is in C, then x is in AUC, BUC.
Thus, from these last two lines, x in A intersect C and x is in B intersect C.
Therefore, if x is in AUC=BUC then x is automatically in A and B or in C.
If x is in A intersect C = B intersect C, then x is A, B, and C.
A=B since x is in the intersection of A and C and intersection of B and C and also since x appears in the AUC and BUC regardless of C, because of the statement x is in A and x is in B or x is in C.

I really don't think this proves anything...anyone know of a more obvious or quick approach?
Let $\displaystyle x\in A$, clearly then $\displaystyle x\in A\cup C\implies x\in B\cup C$. We can see then that either $\displaystyle x\in B$ or $\displaystyle x\in C$ or it's in both. If the first or third is true we are done, so assume then that $\displaystyle x\in C,x\notin B$. Then $\displaystyle x\notin B\cap C\implies x\notin A\cap C$ but this is a contradiction since both $\displaystyle x\in A,x\in C$. Thus $\displaystyle x\in B$ and $\displaystyle A\subseteq B$. Use similar logic to show that $\displaystyle B\subseteq A$

3. uhh, thank you!! That makes much more sense!!

,

,

,

,

,

# a▲b <c→a u c=b u c

Click on a term to search for related topics.