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

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 $$A^C \cap A = \varnothing$$.

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 $$A\cap A^C=\varnothing$$.

• In your title you mention $A'$ but nowhere do you mention it in your body. I assume you mean $A'=A^c$. Your proof contains errors assuming this is what you mean because $A \not\subset B$ and $B \not\subset A$ doesn't imply $A \cap B = \emptyset$. For example, the nonnegative numbers and the nonpositive numbers do not contain each other, but 0 is in both sets. – Connor Malin Feb 19 at 19:23
• Um.. why does $X \not \subset Y$ and $Y\not \subset X$ imply $X\cap Y =\emptyset$? Is that an axiom? A definition? I think for a proof about basic concepts one has to be extra precise about definitions. Frankly I have to wonder about anyone being asked to "prove" this. It is essentially definition. $A^c$ is defined to be the set of precisely the elements that are not contained in $A$ so there can't be any elements in both. – fleablood Feb 19 at 20:08
• I've made an edit. I keep the previous mistakes as reference. – user615771 Feb 19 at 20:10
• It might help your proof to begin it with a clear, self-contained definition of $A^C$, e.g., "If $A$ is a set, its complement is the set $A^C=\{x: x\not\in A\}$." – Barry Cipra Feb 19 at 20:53

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.
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 $$A^C \cap A = \emptyset$$.
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.