For any sets $A$ and $B$, show that $(B\smallsetminus A)\cup A=B \iff A\subseteq B$.

Question

For any sets $A$ and $B$, show that $(B\smallsetminus A)\cup A=B \iff A\subseteq B$.

I know I have to do the following:

1) Assume $(B\smallsetminus A) \cup A=B$, prove $A\subseteq B$.

2) Assume $A\subseteq B$, prove $(B\setminus A)\cup A=B$.

This is what I have done so far:

1) Let $(B\smallsetminus A)\cup A=B$ and $x\in A$. Then $x\in(B\smallsetminus A)\cup A$.

Since $(B\smallsetminus A)\cup A=B$, then $x\in B\smallsetminus A$.

Therefore, $x\in B$ by definition of difference.

Thus, $A\subseteq B$.

2) Let $A\subseteq B$ and $x\in(B\smallsetminus A)\cup A$.

Then $x\in B\smallsetminus A$ or $x\in A$ by definition of union.

Then we have couple cases here:

a) If $x\in B\smallsetminus A$, then we have proved the statement.

b) Show that $x\in A$ and $x\in B$.

I have no idea if I am doing this correctly or not. Thus, any help is appreciated.

Answer 1

Hint:

Use the facts that $$ (B\setminus A)\cup A= A\cup B\quad \text{and}\quad A\cup B=B\iff A\subseteq B $$

Answer 2

For step 1), there is a point where you conclude $x \in B\setminus A $ which need not be true. The right reasoning would be $x \in A \implies x \in (B\setminus A) \cup A $ by definition of union. But $(B\setminus A) \cup A = B $ thus $x \in B $ like we wanted.

For step 2) we can think of it this way:

What is $(B\setminus A) \cup A$? The left hand side is all elements of $B $ that are not in $A $. What is the right hand side? All elements of $A $. But $A \subset B $ so we can also say all elements of $B $ that are in $A $. Hence, the union $(B\setminus A) \cup A$ is all elements that are in $B $, whether they are in $A $ or not which means $(B\setminus A) \cup A = B $ like we wanted to show