Does $X\setminus\emptyset=X$? Let $X$ be a non-empty set, my question is the following:

Is $X\setminus\emptyset=X$?

Instinctively, if you take off the null set from a non-empty set, it won't change anything but the null set is included in every set so if you "take it off" you ultimately change the set.
 A: By definition, $X \backslash \emptyset = \{ x | x \in X$ and $x \notin \emptyset \}$. Clearly, then, if $x \in X$ we have $x \in X \backslash \emptyset$ and vice versa. Intuitively, $X \backslash \emptyset$ is everything in $X$ except elements of $\emptyset$, of which there are none. 
A: "Instinctively, if you take off the null set from a non-empty set, it won't change anything but the null set is included in every set"
It's time to whack you on the head with a 2 by 4.  (Don't worry; we all get whacked in the head once in a while.)
"take off the null set".  What does "take off" mean?  If I said "take off 2 from the set {1,2,3,4,5}" does it mean the same thing as "take off {2} from the set of {1,2,3,4,5}" and what if instead I said "take of {2} from {$\mathbb R$, $\mathbb Q$, $\mathbb Z$, {1,2,3,4,5},  {0,2,5,6,......}, {odd numbers}}" what would that mean?
Important thing to realize is a set as an object is a different thing than then it's elements.  $\{2\} \subset \{0,1,2,3,4\}$ but $\{2\} \not \subset \{\mathbb R, \mathbb Q, \mathbb Z, \{1,2,3,4,5\},\{2\}  \{0,2,5,6,......\}, \{odd numbers\}\}$ even though $\{2\} \in  \{\mathbb R, \mathbb Q, \mathbb Z, {1,2,3,4,5},\{2\} \{0,2,5,6,.....\}, \{odd numbers\}\}$.  This is because $\{2\}$-- the set with a single element; 2-- is a different thing than $\{\{2\}\}$-- the set with a single element; the set with element 2$.
"but the null set is included in every set"
That is an ambiguous poorly formed sentence and is wrong by most accepted mathematical language.
The null set is a set with no elements.  A subset of a set is a set whose elements are all in the other set.  As the null set has no elements, they are all elements of any other set (all zero of them) so the null set is a subset of every set.
But "is included" usually does not mean "is a subset of" but instead means "is an element of". S=  {banana, $\pi$, $\sqrt{27}$} is a set with three elements.  None of them is the null set. so $\emptyset \not \in S$.  
R = {banana, {1,2,3,4,5}, $\emptyset$, $\pi$}.  Then $banana \in R$,  $\{1,2,3,4,5\} \in R$, $\pi \in R$.  $\emptyset \in R$.  $\{2,4,5\} \not \in R$.
So, no, the null set "is a subset" of every set, but the null set is not "included" in every set.
So..... What does $A\setminus B$ mean.  It means $\{a| a\in B; a \not \in B\}$ so $X\setminus \emptyset = \{x \in X; x \not \in \emptyset\}$ but as ALL $x \not \in \emptyset$,  $\{x \in X; x \not \in \emptyset\}= \{x \in X\} = X$.  So YES, $X\setminus \emptyset = X$.
BUT  if $R = \{banana, \{1,2,3,4,5\}, \emptyset, \pi\}$ then $R \setminus \{\emptyset\} = \{banana, \{1,2,3,4,5\},  \pi\}\ne R$.
A: $A\setminus B$ is the set of all elements that belong to $A$ and not belong to $B$ and, if $B=\emptyset$ there are no elements that belong to $B$, so $A\setminus B$ is the set of all elements that belong to $A$.
