A set without the empty set By definition, the empty set is a subset of every set, right? Then how would you interpret this set: $A\setminus\{\}$? On one hand it looks like a set without the empty set, on the other hand, the empty set is in every set... Can you explain?
 A: Just write it
$$
A\setminus\{\}=\Big\{a~\mid~ a\in A \text{ and } a \not \in \{\} \Big\} =A
$$
A: 
By definition, the empty set is a subset of every set, right?

Yes.

Then how would you interpret this set: $A\setminus\{\}$?

The set $A\setminus\{\}$ is the set of members of $A$ which are not members of $\{\}$. However, $\{\}$ has no members, so $A\setminus\{\}=A$.

On one hand it looks like a set without the empty set, on the other hand, the empty set is in every set...

If you wish to remove the empty set from $A$, you should do $A\setminus\{\{\}\}$.

On one hand it looks like a set without the empty set, on the other hand, the empty set is in every set...

The empty set is not a member of every set, it is a subset of every set. $A\subseteq B$ means that for all $x\in A$: $x\in B$. If $A=\{\}$, regardless of what kind of set $B$ is, this statement is always true. This is because there are no $x\in\{\}$.
A: The notation $A-\{\}$ roughly translates to "the set $A$ without the elements of $\{\}$."  The difference is that the empty set is not an element of $A$ and this notation just means you're not removing any elements from your original set.
A: The construction $A\setminus B$ can be axiomatized as follows:


*

*$(A\setminus B)\subseteq A$

*$(A\setminus B)\cap B = \{\}$

*$A\setminus B$ is the largest set satifisfying (1) and (2).


Condition (2) captures axiomatically the idea that "$B$ isn't in $A\setminus B$".  Consider replacing condition (2) with the axiom


*

*$B\not\subseteq(A\setminus B)$


Let $A = \{1, 2, 3\}$ and $B = \{2, 3\}$.  Then $\{1, 3\}$ and $\{1, 3\}$ are both subsets of $A$ satisfying the incorrect condition *, but we want the set difference to be smaller than either of them.  So we use condition (2), which says none of the elements of $B$ is in $A\setminus B$.
But if $B = \{\}$, condition (2) is just $(A\setminus \{\})\cap\{\} = \{\}$, which is true regardless of what $A\setminus\{\}$ is.  So we have conditions (1) and (3) left to fulfill: $(A\setminus\{\})\subseteq A$, and $A\setminus\{\}$ is the largest set satisfying (1).  But $A$ is the largest subset of $A$, so $A\setminus\{\} = A$.
