Suppose $A$ and $B$ are sets. Prove that $A\setminus(A\setminus B)=A\cap B$.

Question

Not a duplicate of

For sets $A,B$, prove $A \setminus (A \setminus B) = A \cap B$

Showing that $A\cap B = A\setminus (A\setminus B)$

set theory proof of $A\cap B = A \setminus(A\setminus B)$

Using disjunction to prove that $A \setminus (A \setminus B) = A \cap B$

This is exercise $3.5.3$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:

Suppose $A$ and $B$ are sets. Prove that $A\setminus(A\setminus B)=A\cap B$.

Here is my proof:

$(\rightarrow)$ Let $x$ be an arbitrary element of $A\setminus(A\setminus B)$. This means that $x\in A$ and $x\notin A\setminus B$ which in turn means $x\notin A$ or $x\in B$. Now we consider two different cases.

Case $1$. Suppose $x\notin A$ and ergo a contradiction.

Case $2$. Suppose $x\in B$. Since $x\in A$, $x\in A\cap B$.

Since we reached $x\in A\cap B$ or a contradiction, then $x\in A\cap B$.

Therefore if $x\in A\setminus(A\setminus B)$ then $x\in A\cap B$. Since $x$ is arbitrary, $\forall x\Bigr(x\in A\setminus(A\setminus B)\rightarrow x\in A\cap B\Bigr)$ and so $A\setminus(A\setminus B)\subseteq A\cap B$.

$(\leftarrow)$ Let $x$ be an arbitrary element of $A\cap B$. This means $x\in A$ and $x\in B$. Since $x\in B$, then $x\notin A\setminus B$. From $x\in A$ and $x\notin A\setminus B$, $x\in A\setminus (A\setminus B)$. Therefore if $x\in A\cap B$ then $x\in A\setminus(A\setminus B)$. Since $x$ is arbitrary, $\forall x\Bigr(x\in A\cap B\rightarrow x\in A\setminus(A\setminus B)\Bigr)$ and so $A\cap B\subseteq A\setminus (A\setminus B)$.

Since $A\setminus(A\setminus B)\subseteq A\cap B$ and $A\cap B\subseteq A\setminus (A\setminus B)$, it follows that $A\setminus(A\setminus B)= A\cap B$. $Q.E.D.$

Is my proof valid$?$

Thanks for your attention.

Answer 1

It's fine, but it's worth trying to prove both directions at once viz.$$\begin{align}x\in A\setminus(A\setminus B)&\iff x\in A\land x\notin A\setminus B\\&\iff x\in A\land x\in B\\&\iff x\in A\cap B.\end{align}$$Edit: @fleablood is right, any criterion beyond validity is aesthetic. My objective was to show how to save time with similar problems.

Answer 2

Yes, your proof is correct.

As you stated, you have proved that $A\setminus(A\setminus B)\subseteq A\cap B$ and $A\setminus(A\setminus B)\supseteq A\cap B$ (using correct logic), so from that we conclude that $$A\setminus(A\setminus B)= A\cap B.$$

Answer 3

Your proof is correct, but could be shortened by considering a sequence of 'if and only if' statements.

However, if you consider $S$ as any universal set in this context, and use $A \setminus B=A \cap B^c$, where $B^c:=S \setminus B$, then you readily get

$$ A \setminus (A \setminus B) = A \cap \big((A \cap B^c)\Big)^c = A \cap (A^c \cup B) = (A \cap A^c) \cup (A \cap B) = A \cap B. $$