Given four sets: $A, B, C, D$, suppose that $$A \triangle B \subseteq D \ \land \ B \triangle C \subseteq D $$
Prove: $A \triangle C \subseteq D$
[As $\triangle$ means symmetric difference]
My Attempt:
Notice that $A \triangle B = (A \setminus B) \cup (B \setminus A) = (A \cup B)\setminus (A\cap B)$
Therefore: $\left((A\cup B)\setminus(A\cap B)\right) \text{ and } \left((B\cup C)\setminus (B\cap C)\right) \subseteq D$
EDIT:
I'd like to prove: $(A \triangle C) \subseteq (A\triangle B)\cup(B\triangle C)$
$$\left((A\triangle B)\cup(B \triangle C)\right)=(A\cup B)\setminus(A\cap B)\cup(B\cup C)\setminus(B\cap C)$$ $$ =(A\cup B)\cup(B\cup C)\setminus(A\cap B)\cup(B\cap C)$$
$$=B\cup(A\cup C)\setminus B\cap(A\cup C)$$
$ \ $
Now — can I say that $(A\cup C)\setminus(A\cap C) \subseteq (B\cup(A\cup C)\setminus B\cap(A\cup C)$?
my previous attempt:
Notice that $A \triangle B = (A \setminus B) \cup (B \setminus A) = (A \cup B)\setminus (A\cap B)$
Therefore: $\left((A\cup B)\setminus(A\cap B)\right) \text{ and } \left((B\cup C)\setminus (B\cap C)\right) \subseteq D$
I'll divide into cases:
I. let $x \in (A\triangle B) \land x\notin C \to x\in (A\setminus C) \subseteq (A \triangle C)$
Notice that $A\triangle B \subseteq D \to x\in D$, therefore $x \in A\triangle C \land x\in D$
II. let $x'\in (B \triangle C) \land \ x'\notin A \ $, then $x'\in (C\setminus A) \subseteq (A\triangle C)$ as $(B\triangle C) \subseteq D \to \ x'\in D$
III. Let $x'' \in (B\cap C) \land x\notin A$ Then $x\in (B\setminus
A)\subseteq (A \triangle B) \subseteq D$, Hence $x'' \in D$
Notice that $x'' \in (C\setminus A) \subseteq (A \triangle C)$ , so $x'' \in D \land x''\in (A\triangle C)$.
I don't know if I covered all possible cases (and can't find a way to do so. using contradiction didn't helped).
What can be done at such case?
Best Answer
The proof is possibly good, but once you know associativity of symmetric difference, you can realize that $$ (A\mathbin{\triangle}B)\mathbin{\triangle}(B\mathbin{\triangle}C)= A\mathbin{\triangle}(B\mathbin{\triangle}B)\mathbin{\triangle}C= A\mathbin{\triangle}\emptyset\mathbin{\triangle}C=A\mathbin{\triangle}C $$ Thus you only need to show that $X\subseteq D$ and $Y\subseteq D$ implies $X\mathbin{\triangle}Y\subseteq D$, which is clear because $$ X\mathbin{\triangle}Y\subseteq X\cup Y $$
A lower level proof. Let $x\in A\mathbin{\triangle}C$. Either “$x\in A$ and $x\notin C$” or “$x\in C$ and $x\notin A$”.
First case: $x\in A$ and $x\notin C$. If $x\in B$, then $x\in B\mathbin{\triangle}C$. If $x\notin B$, then $x\in A\mathbin{\triangle}B$.
Second case: similar.