# Thread: Cancellation of sets

1. ## Cancellation of sets

I'd appreciate it if someone would look this over for me. This is probably easier than I am making it.

Let A, B, and C be subsets of a set X. I wish to show
$\displaystyle \displaystyle B = C \Leftrightarrow A \cup B = A \cup C$

First: $\displaystyle \displaystyle B = C \implies A \cup B = A \cup C$

If $\displaystyle x \in B$ then $\displaystyle x \in A \cup B$, but $\displaystyle x \in B \implies x \in C$, thus $\displaystyle \forall x \in B \text{, } x \in A \cup C$

Thus $\displaystyle A \cup B \subset A \cup C$. Since the roles of B and C are interchangable in the above by symmetry we also have that $\displaystyle A \cup C \subset A \cup B$. Thus $\displaystyle A \cup B = A \cup C$.

I can't figure out how to start the other half: $\displaystyle A \cup B = A \cup C \implies B = C$. If $\displaystyle x \in A$ I can't seem to go anywhere and if $\displaystyle x \in B - A$ I get down to a statement that says $\displaystyle B - A = C - A$, which seems to be a rather circular result. (I think this means $\displaystyle B = C$ since $\displaystyle x \not \in A$, but I'm a little sketchy on the logic here.)

Thanks in advance for the help.

-Dan

2. Originally Posted by topsquark
I can't figure out how to start the other half: $\displaystyle A \cup B = A \cup C \implies B = C$. If $\displaystyle x \in A$
Do you really think that it is true?
Let $\displaystyle A=\{1,2,3\},~B=\{1,2\},~\&~C=\{2,3\}$.

3. Originally Posted by Plato
Do you really think that it is true?
Let $\displaystyle A=\{1,2,3\},~B=\{1,2\},~\&~C=\{2,3\}$.
(sighs) I got bogged down in the proof and didn't even look at the obvious stuff. Thanks for the catch.

(I'm definately going to hit my Set Theory text.)

-Dan