I’m stuck on this one problem in my textbook regarding proofs in set theory. I’ve done the following so far:
Let $ x \in B $. As $ B \subseteq A \cup B $, we have $ x \in A \cup B $, so $ x \in A \cup C $ because $ A \cup B = A \cup C $. Hence, $ x \in (A \cup B) \cap (A \cup C) $, which yields $ x \in A \cup (B \cap C) $.
Now, either $ x \in A $ or $ x \in B \cap C $. If $ x \in B \cap C $, then $ x \in C $ because $ B \cap C \subseteq C $, so $ B \subseteq C $.
This is only half of the proof, though, as I have to prove that $ C \subseteq B $ as well; and I also haven’t even considered the case $ x \in A $ for the last step in what I have so far.
I’ve drawn many different Venn diagrams, and although I see why the statement is true, I just can’t formalize it. Any pointers, help or guidance is very much appreciated.
Thanks!
Best Answer
Using the symmetric difference property that $A \Delta B = (A \cup B) \setminus (A \cap B)$, it follows from the given equalities that:
$$A \Delta B = (A \cup B) \setminus (A \cap B) = (A \cup C) \setminus (A \cap C) = A \Delta C$$
Using the associativity of the symmetric difference, and given that $A \Delta A = \emptyset$, and $\emptyset \Delta X = X$:
$$B = (A \Delta A) \Delta B = A \Delta (A \Delta B) = A \Delta (A \Delta C) = (A \Delta A) \Delta C = C$$