Proof that $A^C\cap A = \varnothing$

Question

Proof. Consider $x\in A \overset{\text{def}}{\Longrightarrow} x \notin A^C$. Therefore $A\nsubseteq A^C$.

Consider also $x\in A^C \overset{\text{def}}{\Longrightarrow} x \notin A$. Therefore $A^C\nsubseteq A$.

Since $A^C\nsubseteq A$ and $A\nsubseteq A^C$ we prove that $A^C \cap A = \varnothing$.

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I have a simple question: does my proof contain errors?

EDIT:

I think I have got it now.

Proof. Let $A \subseteq \Omega$.

Assume there is an element such that $x \in A\cap A^C$. This is equivalent to $x\in A$ and $x\in A^C. $

Since $x \in A \overset{\text{def}}{\Longrightarrow}x\notin A^C$, there exists no $x$ such that $x\in A$ simultaneously as $x\in A^C$.

Therefore, $\nexists x: x\in A \wedge x\in A^C$ and $A\cap A^C=\varnothing$.

Answer 1

You cannot conclude $X\cap Y = \varnothing $ from $X\nsubseteq Y$ and $Y\nsubseteq X$ (take, for instance, $X=\{a,b\}$ and $Y=\{b,c\}$). That being said, your first two lines are true, and any one of your two implications can be used to deduce that no $x$ can be in both $A$ and its complement.

Answer 2

You state that $A^C \not\subseteq A$ and $A \not\subseteq A^C$ implies $A^C \cap A = \emptyset$. This is not true. For example, $A = \{1,2,3\}$ and $B = \{3,4,5\}$ have the property $A \not\subseteq B$ and $B \not\subseteq A$, but their intersection is $\{3\}$, which is nonempty.
For a correct proof of this fact, we need to show that there exist no elements in the intersection, not just that they are not subsets of each other.
To this end, suppose there is an element $x$, in $A^C \cap A$. Since $x$ in $A^C \cap A$, then $x \in A^C$, by definition of $\cap$. But since $x \in A^C$, then $x \not\in A$, by the definition of the complement. So $x \in A^C \cap A$ implies that $x \not\in A$. If $x \not\in A$, then $x \not\in A^C \cap A$. But this contradicts the assumption that $x \in A \cap A^C$. Thus, there cannot exist any elements in $A^C \cap A$, which means $A^C \cap A = \emptyset$.