# Suppose $A$, $B$, and $C$ are sets. Prove that $C ⊆ A △ B$ iff $C ⊆ A ∪ B$ and $A ∩ B ∩ C = ∅$.

This is Velleman's exercise 3.5.21 (And NO, not a duplicate of "Suppose $A, B$, and C are sets. Prove that $C\subset A\Delta B \Leftrightarrow C \subset A \cup B$ and $A \cap B \cap C = \emptyset$", my question is different):

Suppose $A$, $B$, and $C$ are sets. Prove that $C \subseteq A\,\triangle\,B$ iff $C \subseteq A \cup B$ and $A \cap B \cap C = \emptyset$.

And here's my proof of it:

Proof.

($\rightarrow$) Suppose $C ⊆ A △ B$ and let $x$ be an arbitrary element of $C$, then we have $x ∈ A △ B$. We now consider two cases:

Case 1. $x ∈ A\setminus B$, which means $x ∈ A$ but $x ∉ B$. Thus $x ∈ A ∪ B$.

Case 2. $x ∈ B\setminus A$, which means $x ∈ B$ but $x ∉ A$. Thus $x ∈ A ∪ B$.

Now suppose $A ∩ B ∩ C \neq ∅$. From $x ∈ A △ B$ we have that either $x ∉ A$ or $x ∉ B$ which in either case is a contradiction and hence $A ∩ B ∩ C = ∅$.

We have $x ∈ A ∪ B$ and $A ∩ B ∩ C = ∅$ and therefore, if $C ⊆ A △ B$, then $C ⊆ A ∪ B$ and $A ∩ B ∩ C = ∅$.

($\leftarrow$) Suppose $C ⊆ A ∪ B$ and $A ∩ B ∩ C = ∅$ and let $x$ be an arbitrary element of $C$. We now have two different cases to consider:

Case 1. $x ∈ B\setminus A$, then clearly $x ∈ A △ B$.

Case 2. $x ∉ B\setminus A$, which means $x ∈ A$ but $x ∉ B$. Since $A ∩ B ∩ C = ∅$ is equivalent to $∀x(x ∈ A \Rightarrow (x ∈ C \Rightarrow x ∉ B))$, then by $x ∈ A$ we have $(x ∈ C \Rightarrow x ∉ B)$. Since we had $x ∈ C$, then $x ∉ B$ and then $x ∈ A\setminus B$. Ergo $x ∈ A △ B$.

From both case we have $x ∈ A △ B$. Since $x$ was arbitrary, $C ⊆ A △ B$ and therefore, if $C ⊆ A ∪ B$ and $A ∩ B ∩ C = ∅$, then $C ⊆ A △ B$.

By ($\rightarrow$) and ($\leftarrow$) we have $C ⊆ A △ B$ iff $C ⊆ A ∪ B$ and $A ∩ B ∩ C = ∅$.

Now here are my questions:

1. Is my proof valid?

2. In part one (i.e. ($\rightarrow$)), is there anything wrong with the proof of $A ∩ B ∩ C = ∅$?

3. In part two (i.e. ($\leftarrow$)), the second case seems to be a little redundant to me! Is that correct (i.e. is my proof of it correct)?

In the forward direction, I don't see how you obtain a contradiction after "From $x\in A\,\triangle\,B$ we have that either $x\not\in A$ or $x\not\in B$..." Just because we have $A\cap B\cap C\ne\emptyset$ doesn't mean that everything in $A$ must also be in $B$, or visa versa. The fix is to just assume you have something in the intersection instead of working with the same $x$ as from before:

Since $A\cap B\cap C\ne\emptyset$, then there is $y\in A\cap B\cap C$. Since $y\in C$, then $y\in A\,\triangle\,B$. Thus $y\not\in A$ or $y\not\in B$, which contradicts $y\in A\cap B$.

In case 2 of the reverse direction, you wrote "...which means $x\in A$ but $x\not\in B$". This is true, but you are skipping a few steps. Since $x\in C$ and $x\not\in B\setminus A$, then we must have $x\in A\setminus B$ or $x\in A\cap B$. However the later is impossible because $A\cap B\cap C=\emptyset$. From here there is no need for the next line, because $x\in A$ and $x\not\in B$ means $x\in A\setminus B$.

After these fixes you have a perfectly valid proof. However I would consider different cases in the reverse direction. You know that $x\in C\subseteq A\cup B$. Thus it would seem more natural to consider the cases:

Case 1: $x\in A$. Since $x\in A\cap C$ and $A\cap B\cap C=\emptyset$, we deduce $x\not\in B$. Thus $x\in A\setminus B$ and consequently $x\in A\,\triangle\,B$.

Case 2: $x\in B$. Similar to case $1$, we get $x\in B\setminus A$ and thus $x\in A\,\triangle\,B$.

• Thanks a lot and great as always. One more thing. I don't want to be rude but is it possible for you to answer this another unanswered question of mine: "math.stackexchange.com/questions/2401172/…" as well? – Heptapod Aug 21 '17 at 16:14
• @Heptapod Sure I'll take a look! – John Griffin Aug 21 '17 at 16:16