if $\mathbb{S}$is any non empty set , why $ \varnothing- \mathbb{S} =\varnothing $ is true? i can't understand why ?

if   $\mathbb{S}$ is any set , why   $  \varnothing- \mathbb{S} 
 =\varnothing  $  is true .

and does is hold for $  \varnothing- \varnothing   =\varnothing  $ ??
i know that for any two sets $ \mathbb{A,B}$ that $\mathbb{A-B}=  \begin{Bmatrix}
x:x\in \mathbb{A}  , x   \notin \mathbb{B}  
\end{Bmatrix}$
thanks for advance .
 A: There is no $x\in \varnothing$, so there is no $x$ such that $x\in \varnothing$ and $x\not\in S$.
A: Recall that $A-B=\{a\in A\mid a\notin B\}$.
So $\varnothing - S=\{a\in\varnothing\mid a\notin S\}$. If $a\in\varnothing-S$, then what thing happens?
A: Let's do it in a funny way: for every set $\mathrm X$, it is true that $\emptyset \subset \mathrm X$.
So for every set $\mathrm S$, we have $\emptyset \subset \mathrm X =\emptyset - \mathrm S$.
Now, we also have $\mathrm A - \mathrm B \subset A$ so, $\emptyset - \mathrm S \subset \emptyset$.
By double inclusion, $\emptyset - \mathrm S = \emptyset$.
A: You can use the fact that $A - B = A \cap B'$ And substitute $A = \varnothing, B = S$. And you're done!
A: Since $A-B \subseteq A$, we have $\varnothing- \mathbb{S} \subseteq \varnothing$, which implies $\varnothing- \mathbb{S} =\varnothing$.
A: Asaf has said it perfectly.  Another notation used here that would be the same as $A - B$ is $A \setminus B$.  This can sometimes help eliminate the idea that we are "subtracting" anything, but rather removing elements of $B$ from the set $A$.  Thus, trying to remove anything from an already empty set, $ \emptyset $, always results in the empty set again.  So, both $\emptyset \setminus A = \emptyset$ and  $\emptyset \setminus \emptyset = \emptyset$ are true, where $ A \neq \emptyset$.  
A: Hint
$  \varnothing- \mathbb{S} 
 =\varnothing\cap S^{c}=\varnothing  $
