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Not a duplicate of

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

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 $

Set theory: Prove that $C \subseteq A \Delta B \iff C \subseteq A \cup B \wedge A \cap B \cap C = \emptyset$

This is exercise $3.5.21$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:

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

Here is my proof:

$(\rightarrow)$ Suppose $C\subseteq A\Delta B$.

$(1)$ Let $x$ be an arbitrary element of $C$. From $C\subseteq A\Delta B$ and $x\in C$, $x\in A\Delta B$. Now we consider two cases.

Case $1.$ Suppose $x\in A\setminus B$. Ergo $x\in A\cup B$.

Case $2.$ Suppose $x\in B\setminus A$. Ergo $x\in A\cup B$.

Since the above cases are exhaustive, $x\in A\cup B$. Thus if $x\in C$ then $x\in A\cup B$. Since $x$ is arbitrary, $\forall x(x\in C\rightarrow x\in A\cup B)$ and so $C\subseteq A\cup B$. Therefore if $C\subseteq A\Delta B$ then $C\subseteq A\cup B$.

$(2)$ Suppose $A\cap B\cap C\neq\emptyset$. So we can choose some $x_0$ such that $x_0\in A$, $x_0\in B$, and $x_0\in C$. From $C\subseteq A\Delta B$ and $x_0\in C$, $x_0\in A\Delta B$. Now we consider two cases.

Case $1.$ Suppose $x_0\in A\setminus B$. Ergo $x_0\notin B$ which contradicts $x_0\in B$ and so it must be the case that $A\cap B\cap C=\emptyset$.

Case $2.$ Suppose $x_0\in B\setminus A$. Ergo $x_0\notin A$ which contradicts $x_0\in A$ and so it must be the case that $A\cap B\cap C=\emptyset$.

Since the above cases are exhaustive, $A\cap B\cap C=\emptyset$. Therefore if $C\subseteq A\Delta B$ then $A\cap B\cap C=\emptyset$.

From parts $(1)$ and $(2)$ we can conclude that if $C\subseteq A\Delta B$ then $C\subseteq A\cup B$ and $A\cap B\cap C=\emptyset$.

$(\leftarrow)$ Suppose $C\subseteq A\cup B$ and $A\cap B\cap C=\emptyset$. Let $x$ be an arbitrary element of $C$. From $C\subseteq A\cup B$ and $x\in C$, $x\in A\cup B$. Now we consider two cases.

Case $1.$ Suppose $x\in A$. Now we consider two cases.

Case $1.1.$ Suppose $x\in A\setminus B$. Ergo $x\in A\Delta B$.

Case $1.2.$ Suppose $x\notin A\setminus B$ and so $x\notin A$ or $x\in B$. Now we consider two cases.

Case $1.2.1.$ Suppose $x\notin A$ which is a contradiction.

Case $1.2.2.$ Suppose $x\in B$ which is a contradiction since $A\cap B\cap C=\emptyset$.

Since cases $1.2.1$ and $1.2.2$ lead to a contradiction then case $1.2$ leads to a contradiction. From case $1.1$ or case $1.2$ we can conclude $x\in A\Delta B$.

Case $2.$ Suppose $x\in B$ and a similar argument shows $x\in A\Delta B$.

Since case $1$ and case $2$ are exhaustive, $x\in A\Delta B$. Thus if $x\in C$ then $x\in A\Delta B$. Since $x$ is arbitrary, $\forall x(x\in C\rightarrow x\in A\Delta B)$ and so $C\subseteq A\Delta B$. Therefore if $C\subseteq A\cup B$ and $A\cap B\cap C=\emptyset$ then $C\subseteq A\Delta B$.

From $(\rightarrow)$ and $(\leftarrow)$ we can conclude $C\subseteq A\Delta B$ iff $C\subseteq A\cup B$ and $A\cap B\cap C=\emptyset$. $Q.E.D.$

Is my proof valid$?$ Is my proof unnecessarily redundant or every step is needed$?$

Thanks for your attention.

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  • $\begingroup$ You know. Saying it is not a duplicate, doesn't mean it actually isnt a duplicate. $\endgroup$
    – fleablood
    Jul 26, 2020 at 16:59
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    $\begingroup$ Of course. But I checked those posts and I am definitely sure that my proof is different. $\endgroup$ Jul 26, 2020 at 17:04

5 Answers 5

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Your proof is correct. Here is a proof that avoids any mention of specific elements (following the theme of my answer to one of your previous questions). The key statements we use are the following:

(a) If $X$ and $Y$ are sets then $X \subseteq Y$ iff $X \setminus Y = \emptyset$.

(b) If $X$ and $Y$ are sets then $X \cup Y = \emptyset$ iff $X = \emptyset$ and $Y = \emptyset$.

(We discussed both of these before, so let's not reprove them!)

Now, in this problem we care about when $C \subseteq A \Delta B$. So, guided by property (a), we should examine $C \setminus (A\Delta B)$. Use axioms of set operations (e.g., De Morgan etc) to prove: $$ C \setminus (A\Delta B) = \big(C \setminus (A\cup B)\big) \cup \big(A \cap B \cap C\big)\tag{1} $$

I have hidden the proof of $(1)$ at the bottom of this answer; but try it yourself first. It's also a sensible thing to say out loud: $A \Delta B$ is the set of elements that are in either $A$ or $B$, but not both. So being in $C \setminus (A \Delta B)$ is the same as either being in $C$ and not in $A$ or $B$, or being in $C$ and in both $A$ and $B$.

Once you have $(1)$, the rest is very straightforward.

\begin{align} C \subseteq A \Delta B &\iff C \setminus (A \Delta B) = \emptyset \tag{using (a)} \\ &\iff \big(C \setminus (A\cup B)\big) \cup \big(A \cap B \cap C\big) = \emptyset \tag{using (1)}\\ &\iff C \setminus (A \cup B) = \emptyset \text{ and } A \cap B \cap C = \emptyset \tag{using (b)}\\ &\iff C \subseteq A\cup B \text{ and } A\cap B\cap C = \emptyset \tag{using (a)} \end{align}

Proof of $(1)$:

Recall that $$A \Delta B = (A \cup B) \setminus (A \cap B) = (A \cup B) \cap \neg(A \cap B)\tag{2}$$ So \begin{align}C \setminus (A \Delta B) &= C\cap \neg\big((A \cup B)\cap \neg (A \cap B)\big) \tag{by (2)} \\ &= C \cap \big(\neg (A \cup B) \cup (A \cap B)\big) \tag{De Morgan} \\ &= \big(C \cap \neg (A \cup B)\big) \cup \big(C \cap (A \cap B)\big) \tag{distributivity} \\ &= \big(C \setminus (A \cup B)\big) \cup \big(A \cap B \cap C\big)\end{align} In the last line we used the definition of set difference on the left side, and associativity/commutativity of intersection on the right side.

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    $\begingroup$ I cannot thank you enough. $\endgroup$ Jul 26, 2020 at 17:13
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The first inclusion follows from the fact that the symmetric difference is inside the union. The second condition foolows from the fact that symmetric difference is disjoint from the intersection.

As for your proof, it is correct but too long.

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In the first part, in both the second cases(where it says Case $2$) you can simply refer to the similar arguments as in the first cases but with $B \setminus A$ instead of $A\setminus B$.

Since you assume in Case 1 (in the converse part) that $x\in A$, the cases including and after Case 1.2 can be shortened to: "If $x\notin A\setminus B$ then $x\in B$ contradicting $A\cap B \cap C = \emptyset$".

The rest seems good!

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You can shorten you proof by writing $A \cup B=(A \bigtriangleup B) \cup (A \cap B)$. First assume that $C \subseteq A \bigtriangleup B$. Since, $A \bigtriangleup B \subseteq A\cup B$, then $C \subseteq A \bigtriangleup B \implies C \subseteq A \cup B$. Also, $A \bigtriangleup B $ is disjoint from $A \cap B$. And so, $A \cap B \cap C= \phi$. The reverse implication follows by observing that $C \subseteq A \cup B= (A \bigtriangleup B) \cup (A \cap B)$ but $A\cap B \cap C = \phi$ and so, $C \subseteq A \bigtriangleup B$.

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  • $\begingroup$ @JCAA Why? I don't see anything wrong there. $\endgroup$
    – Poorwelsh
    Jul 27, 2020 at 22:10
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I think the "($\to$)" direction of your proof is fine. The "($\leftarrow$)" direction is correct but could be shortened. There was no need to break case 1 into cases 1.1, 1.2, 1.2.1, and 1.2.2. You could have completed case 1 like this:

Case 1. Suppose $x \in A$. If $x \in B$ then $x \in A \cap B \cap C$, which contradicts the fact that $A \cap B \cap C = \emptyset$. Therefore $x \notin B$. Since $x \in A$ and $x \notin B$, $x \in A \bigtriangleup B$.

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