# 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?

• The set difference $A\setminus B$ means you're removing the elements in $B$ from $A$. You're not removing the set $B$ itself from $A$. So $A\setminus\{\}$ means we remove everything in the empty set from $A$. So we're removing nothing, so it's still just $A$. Feb 20, 2015 at 7:04
• You can have all of the cake I have left once I've eaten it all. If I've eaten all the cake $A$, then the portion that's left is the empty subset of the cake. Removing all crumbs in that subset from the cake, what's left is the crumbs I have eaten - all of them. Every last one. The empty set is not a crumb in the cake, but it is a portion of the cake, or a subset of the cake. Feb 20, 2015 at 7:25
• Subset $\ne$ element. Feb 20, 2015 at 8:18
• A man walks down the street with a pocket full of change. He comes to a fountain and he throws every penny he has in his pocket into the fountain. As he walks away, he mumbles "I still have every penny I started with." How?
– aes
Feb 20, 2015 at 23:55

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 $x\in\{\}$.

• In $A \setminus \{\{\}\}$, if you have a $B\subset A$, and given the rule that every set contains the empty set, wouldn't the empty set be in $B$ and thus in $A$ still?
– Alec
Feb 20, 2015 at 7:18
• @Alec: given the rule that every set contains the empty set What do you mean? Not every set contains the empty set. (for example, the empty set does not contain the empty set) Feb 20, 2015 at 7:20
• Sorry, I was thinking of the rule that the empty set is a subset of every set.
– Alec
Feb 20, 2015 at 7:21
• @Alec: Do you have any other questions or clarifications? Feb 20, 2015 at 7:28
• @Regret: In $A \backslash${{}}, does it mean {} is an element of {{}} or a subset? Feb 20, 2015 at 7:31

Just write it $$A\setminus\{\}=\Big\{a~\mid~ a\in A \text{ and } a \not \in \{\} \Big\} =A$$

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.

• This is one of those ultra trivial constructions that's a quirk of our strange foundations. Feb 20, 2015 at 7:00

The construction $A\setminus B$ can be axiomatized as follows:

1. $(A\setminus B)\subseteq A$
2. $(A\setminus B)\cap B = \{\}$
3. $A\setminus B$ 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 $A\setminus\{\} = A$.

• i.e. $\,\ C\subseteq A\setminus B\iff C\subseteq A,\, C\cap B =\{\}\ \$ Feb 20, 2015 at 23:05