Show that if $A$ and $B$ are sets, then $(A\cap B) \cup (A\cap \overline{B})=A$.
Show that if $A$ and $B$ are sets, then $(A\cap B) \cup (A\cap \overline{B})=A$.
So I have to show that $(A\cap B) \cup (A\cap \overline{B})\subseteq A$ and that $A \subseteq(A\cap B) \cup (A\cap \overline{B})$.
Lets begin with the first one:
If $x \in (A \cap B)$ it means $x \in A \wedge x \in B$.
If $x \in (A \cap \overline{B})$ it means $x \in A \wedge x \in \overline{B}$.
And the second one:
If $x \in A$ it means $x \in (A \cap B)$.
But here after I am confused.
$\endgroup$ 23 Answers
$\begingroup$Hint: $(A\cap B)\cup(A\cap\overline{B})=A\cap(B\cup\overline{B})=A$.
$\endgroup$ 2 $\begingroup$Following your approach of chasing elements, you can say If $x \in (A \cap \overline{B})$ it means $x \in A \wedge x \in \overline{B}$, so $x \in A \wedge x \not \in {B}$. Therefore $x \in ((A\cap B) \cup (A\cap \overline{B}))$ means $(x \in A \wedge x \in {B})\vee (x \in A \wedge x \not \in {B})$ and use the distributive principle
$\endgroup$ $\begingroup$Let $x\in (A \cap B) \cup (A \cap B^c)$. Then either $x\in (A \cap B) $ or $x\in (A \cap B^c) $. In the first case, $x\in A$ and $x\in B$, so $x\in A$. In the second case, $x\in A$ and $x\in B^c$, so $x\in A$. Thus $ (A \cap B) \cup (A \cap B^c)\subseteq A$.
Conversely, let $x\in A$. Then either $x\in B$ or $x\not\in B$. If $x\in B$, then $x\in (A \cap B)$, so $x\in (A \cap B) \cup (A \cap B^c)$. If $x\not\in B$, then $x\in B^c$. Thus $x\in (A \cap B^c)$ and $x\in (A \cap B) \cup (A \cap B^c)$. Thus $A \subseteq (A \cap B) \cup (A \cap B^c)$.
Putting these together gives $ (A \cap B) \cup (A \cap B^c)= A$.
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