# Cancellation of sets

• Mar 11th 2011, 02:11 PM
topsquark
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
$\displaystyle B = C \Leftrightarrow A \cup B = A \cup C$

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

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

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

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

Thanks in advance for the help.

-Dan
• Mar 11th 2011, 02:38 PM
Plato
Quote:

Originally Posted by topsquark
I can't figure out how to start the other half: $A \cup B = A \cup C \implies B = C$. If $x \in A$

Do you really think that it is true?
Let $A=\{1,2,3\},~B=\{1,2\},~\&~C=\{2,3\}$.
• Mar 11th 2011, 03:15 PM
topsquark
Quote:

Originally Posted by Plato
Do you really think that it is true?
Let $A=\{1,2,3\},~B=\{1,2\},~\&~C=\{2,3\}$.

(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