What Does It Mean for a Set Not to Contain the Empty Set?

“A set without the empty set” most commonly means a set that does not have the empty set as one of its elements. In symbols, the condition is $\varnothing \notin A$. There is no special standard name for such a set; it is simply a set that does not contain $\varnothing$ as an element.

The wording can be confusing because “is an element of” and “is a subset of” describe different relationships. The empty set is a subset of every set, but it is not necessarily an element of every set.

Element versus subset

The notation $x \in A$ says that $x$ is an element, or member, of $A$. The notation $B \subseteq A$ says that every element of $B$ is also an element of $A$. These are not interchangeable.

For example, let

$$A = \{1,2\}.$$

Neither element of $A$ is the empty set, so $\varnothing \notin A$. But $\varnothing \subseteq A$: every element of $\varnothing$ is in $A$, because the empty set has no elements that could violate this condition.

By contrast, if

$$B = \{\varnothing,1,2\},$$

then $\varnothing \in B$. The empty set is an element of $B$, and it is also a subset of $B$.

Not the same as a nonempty set

A set is nonempty if it has at least one element. In symbols, $A \ne \varnothing$. That condition does not say whether $\varnothing$ itself is an element of $A$.

Set	Is it nonempty?	Is $\varnothing$ an element?
$\varnothing$	No	No
$\{1,2\}$	Yes	No
$\{\varnothing\}$	Yes	Yes
$\{\varnothing,1\}$	Yes	Yes

The set $\{\varnothing\}$ is a particularly useful example. It contains one element—the empty set—so its cardinality is $1$. The empty set itself contains no elements, so its cardinality is $0$. Thus $\{\varnothing\} \ne \varnothing$.

Removing the empty set from a set

If a set $A$ may contain $\varnothing$ and you want a set that excludes it, use set difference:

$$A \setminus \{\varnothing\}.$$

This removes the element $\varnothing$ from $A$, if it is present, and leaves all other elements unchanged.

For a concrete example, take $A=\{\varnothing, 3, 5\}$. Then

$$A \setminus \{\varnothing\} = \{3,5\}.$$

If instead $A=\{3,5\}$, then $\varnothing$ was not in $A$ to begin with, so

$$A \setminus \{\varnothing\} = \{3,5\}.$$

The braces matter: $\{\varnothing\}$ is a set whose only element is $\varnothing$. Set difference takes away that element if it occurs in $A$.

A proof that the empty set is a subset of every set

To show $\varnothing \subseteq A$, use the definition of subset: every element of $\varnothing$ must also be an element of $A$. Since $\varnothing$ has no elements, there is no element that could fail this condition. Therefore $\varnothing \subseteq A$ for every set $A$.

This includes $A=\varnothing$, so $\varnothing \subseteq \varnothing$ as well. It does not follow that $\varnothing \in A$; that is a separate claim about membership.

Common mistakes

• Confusing $\in$ and $\subseteq$. The statement $\varnothing \notin A$ does not contradict $\varnothing \subseteq A$.
• Treating $\{\varnothing\}$ as the empty set. The former has one element; the latter has none.
• Assuming “nonempty” means “does not contain $\varnothing$.” A set such as $\{\varnothing\}$ is nonempty and does contain the empty set.
• Reading $A \setminus \{\varnothing\}$ as removing an empty set of elements. It removes the element $\varnothing$ from $A$; the set being subtracted has one element.

When you see the phrase “a set without the empty set,” first check whether it refers to membership, $\varnothing \notin A$, or to the operation of removing that element, $A \setminus \{\varnothing\}$. Neither statement changes the general fact that $\varnothing$ is a subset of every set.