A set without the empty set
Solution 1:
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 $x\in\{\}$.
Solution 2:
Just write it $$ A\setminus\{\}=\Big\{a~\mid~ a\in A \text{ and } a \not \in \{\} \Big\} =A $$
Solution 3:
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.
Solution 4:
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 $A\setminus\{\} = A$.