# Given 4 sets: $A, B, C, D$, Prove that if $A \triangle B \subseteq D, B \triangle C \subseteq D$, then $A \triangle C \subseteq D$

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?

• Verify that $(A\Delta C) \subset (A\Delta B) \cup (B\Delta C)$. – Kavi Rama Murthy Jul 4 '19 at 12:27
• When showing that $X\subseteq Y$, you usually start with "Let $x\in X$", and end with "and therefore, $x\in Y$.", with some reasoning in-between. In this case, you therefore ought to start with "let $x\in A\Delta C$". – Arthur Jul 4 '19 at 12:29
• @KaviRamaMurthy but the problem is I can't prove that as I'm not sure how to cover all possibilities – Jneven Jul 4 '19 at 13:04

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.

This is actually a very easy proof. I think you are getting lost among all those fancy math symbols

Let $$x \in A\triangle C$$. Therefore $$x \in A$$ and $$x \notin C$$ are both true.

If $$x \in B, x \in B \triangle C \subseteq D$$

Otherwise, $$x \notin B$$. This implies $$x \in A \triangle B \subseteq D$$

• From $x\in A\triangle C$ it does not follow that $x\in A$ and $x\notin C$. – Christian Blatter Jul 4 '19 at 15:03
• yes, regarding @ChristianBlatter comment -- $x \in A$ and $x \notin C \to x\in A\triangle C$, but the other direction isn't necessarily true – Jneven Jul 5 '19 at 16:49