Solving Mathematical Proof

[angela107]

Homework Statement:

Is it TRUE that for all sets $A$ and $B$ the identity $A \setminus (A \setminus B) =A ∩ B$ holds?

Relevant Equations:

n/a

$A ∖ B$ can't include any elements that are not in $A$, so it is the same as saying $A∖(A∩B)$; it's exactly the elements of $A$ except those in $A∩B$.

$A∖(A∖(A∩B))$ is exactly the elements of $A$ except those in (exactly the elements of $A$ except those in $A∩B$). This is the same as $A∩B$.

Therefore, it is true that for all sets A and B the identity $A ∖ (A ∖B) =A ∩ B$holds.

Is this correct?

 

Well, I think you are going in the right direction. But a real formal proof (at this level) requires more details. Typically, when showing that two sets $X,Y$ are equal, you show that $X \subseteq Y$ and $Y\subseteq X$. Showing $X\subseteq Y$ can be done by fixing an arbitrary element $x\in X$ and then after some steps deducing that $x \in Y$. Similarly, you show $Y \subseteq X$. So, let us try this on your case:

Let $x \in A\setminus (A \setminus B)$. Then $x\in A$ and $x \notin A \setminus B$. The latter means that $x\notin A$ or that $x\in B$, but we already know that $x\in A$ so we must have $x\in B$. Hence, $x\in A$ and $x\in B$, which means $x\in A \cap B$.

Can you try the other direction yourself now?

Two additional remarks:

(1) Try to write a more descriptive title for your question. For example, "Prove the set equality $A\cap B = A \setminus (A \setminus B)$"
(2) Your first line in your post contains some typos.

 

[PeroK]

In this case it may help to note that both sets are subsets of $A$. You could then look at the two cases where $x$ is or is not in $B$.

The moral for these questions is to stay calm and think logically!