A set without the empty set

4,449

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 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 $A\setminus\{\} = A$.

Solution 4

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.

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4,449

Arthur

Updated on October 17, 2020

Comments

Arthur over 2 years

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?
- Ben West over 8 years
  
  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 $A$.
- Loki Clock over 8 years
  
  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.
- Martín-Blas Pérez Pinilla over 8 years
  
  Subset $\ne$ element.
- aes over 8 years
  
  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?
Mathemagician1234 over 8 years

This is one of those ultra trivial constructions that's a quirk of our strange foundations.
Alec over 8 years

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?
Regret over 8 years

@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)
Alec over 8 years

Sorry, I was thinking of the rule that the empty set is a subset of every set.
Regret over 8 years

@Alec: Do you have any other questions or clarifications?
Arthur over 8 years

@Regret: In $A \backslash${{}}, does it mean {} is an element of {{}} or a subset?
Regret over 8 years

@infinitesimal: Nothing is a member of the empty set, hence the name. The statement $x\in\{\}$ is always false, so the statement "for all $x\in\{\}$:..." is always true.
Regret over 8 years

@Arthur: Sorry for the late response. $\{\{\}\}$ is the set with only the empty set as its member.
Regret over 8 years

@Arthur: If you have any other questions, please let me know.
Bill Dubuque over 8 years

i.e. $\,\ C\subseteq A\setminus B\iff C\subseteq A,\, C\cap B =\{\}\ \ $