Question about set: When will $A = (A\setminus B) \cup B$? Today , my professor has written a set formula which I don't fully understand: 
$$
A = (A\setminus B) \cup B
$$
Please help me to fill in the missing details for the above, and its proof. My notes are incomplete.
 A: Others have given examples why your posted statement is not always true.
I'll expand on a comment I left earlier. IF one of the premises we are working with is that $B \subseteq A$, then the claim does in fact hold.
In fact claim is true if and only if $A\cap B = B$ (which is equivalent to the premise $B\subseteq A$).  The important thing here is whether you see why $$B \subseteq A \implies A = (A \setminus B) \cup B\quad?$$
If we have that $B\subseteq A$, and perform $A \setminus B$, then by definition of set-minus, we take out of $A$ all the elements in $B$ (where $B\subseteq A$), then when we perform $(A\setminus B) \cup B$, by definition of set union, we are putting back all the previously removed elements back into A, leaving us with $A$ itself.
A: The claim's false. For example, take
$$A=\{1,2\}\;,\;\;B=\{1,3\}\implies A=\{1,2\}\neq \{2\}\cup\{1,3\}=\{1,2,3\}=(A\setminus B)\cup B\;.$$
A: $A\setminus B = \left\{x \mid x \in A\text{ and }x\not\in B\right\}$
$A\cup B = \left\{x \mid x \in A\text{ or }\in B\right\}$

By the way this formula false.
Take $A=\emptyset$ and $B\not= \emptyset$ (for example $\left\{0\right\}$. Then $\left(A\setminus B\right) = \emptyset$ and so $\left(A\setminus B\right)\cup B = B\not= \emptyset = A$.
A: $A\setminus B$ is the set of things that are in $A$ but not in $B$. $X\cup Y$ is the set of things that are in $X$, in $Y$, or in both $X$ and $Y$. Thus, the set $(A\setminus B)\cup B$ consists of those things that are in $A$ but not in $B$, and those things that are in $B$. (There are no things in both $A\setminus B$ and $B$: anything in $A\setminus B$ is by definition not in $B$.) The equation 
$$A=(A\setminus B)\cup B$$
therefore says that the members of $A$ are the things that are in $A$ but not in $B$, and the things that are in $B$. In particular, this says that everything in $B$ belongs to $A$. This certainly isn’t always true, and the other answers have given you some specific counterexamples.
What is true is that
$$A=(A\setminus B)\cup(A\cap B)\;.$$
$A\cap B$ is the set of things belonging to both $A$ and $B$, so this just says that $A$ consists of those things that are in $A$ but not $B$ and those things that are in $A$ and in $B$. If $a\in A$, then either $a\in B$, in which case $a\in A\cap B$, or $a\notin B$, in which case $a\in A\setminus B$. Conversely, if $a\in(A\setminus B)\cup(A\cap B)$, then either $a\in A\setminus B$, or $a\in A\cap B$, and in both cases $a\in A$.
Note, by the way, that if $B\subset A$, then $A\cap B=B$, and in this special case we do have
$$A=(A\setminus B)\cup(A\cap B)=(A\setminus B)\cup B\;.$$
A: I think it's the topic is $A=(A\cap B)\cup B$ or $A=(A\cup B)\cap B$ (note: this is inequatily modulus)
