Empty Sets

What Does “Empty” Mean?

In Set Builder Notation, a set was described by specifying a domain and a condition that determines which domain elements are included. Sometimes no element of the domain satisfies the condition. The resulting set has no elements; it is called the empty set and is written $\varnothing$. Thus, for every object $x$, $$ x\notin\varnothing. $$ The symbol $\varnothing$ names a set, not an element that is hidden inside the set.

A set is empty because of its membership, not because of how it is written. A set-builder expression may look complicated and still describe $\varnothing$; a roster with no entries can also be written $\varnothing$. Conversely, a short expression can describe a nonempty set if at least one object meets its defining condition. To determine which case holds, test whether there is a candidate in the stated domain that satisfies the condition.

The empty set is not the same as $\{\varnothing\}$. The braces in $\{\varnothing\}$ form a set whose one element is the empty set itself. In symbols, $\varnothing\notin\varnothing$, while $\varnothing\in\{\varnothing\}$. The second set is therefore not empty.

Keep elements and sets distinct. The empty set has no elements. The set $\{\varnothing\}$ has one element: the empty set. Braces indicate membership, so placing $\varnothing$ between braces changes the set.

Recognizing an Empty Set

For a condition-defined set $S=\{x\in D:P(x)\}$, the membership characterization from Set Membership says that an object $y$ belongs to $S$ exactly when $y\in D$ and $P(y)$. Therefore, $S$ is empty precisely when there is no $y\in D$ for which $P(y)$ holds. Showing that a set is empty is often an impossibility argument: assume that an element exists, use the membership requirements, and derive a contradiction.

Assume a member exists.
Choose an arbitrary candidate $y$ and suppose $y\in S$.

Unpack membership.
For a condition-defined set, conclude that the candidate is in the domain and satisfies the defining condition.

Reach an impossibility.
Show that the membership requirements cannot hold together. This establishes that no candidate belongs to $S$.

Worked Example: An Impossible Equation in the Reals

Consider $$ A=\{x\in\mathbb R:x^2+1=0\}. $$ Suppose that $x\in A$. By the membership characterization, $x\in\mathbb R$ and $x^2+1=0$. For a real number $x$, its square is nonnegative, so $x^2\geq0$. It follows that $x^2+1\geq1>0$, contradicting $x^2+1=0$. There is no real $x$ satisfying the condition, and hence $A=\varnothing$.

The domain matters here. The conclusion concerns real numbers because $\mathbb R$ is the stated domain. It is the contradiction between the equation and a property of real squares that rules out every possible member.

The following characterization makes the proof goal precise. It also lets us identify a set as empty by comparing its elements with those of $\varnothing$, using the Equality by Double Inclusion theorem from Set Equality.

Theorem (Characterization of the Empty Set). A set $A$ is empty if and only if no object belongs to $A$. Equivalently, $$ A=\varnothing \quad\Longleftrightarrow\quad \text{there is no object }x\text{ such that }x\in A. $$

Proof. First suppose $A=\varnothing$. Since no object belongs to $\varnothing$, no object belongs to $A$.

Now suppose no object belongs to $A$. We show that $A$ and $\varnothing$ have the same elements. There is no element of $A$, so every element of $A$ belongs to $\varnothing$: there are no elements of $A$ that could fail this requirement. Thus $A\subseteq\varnothing$. Also, $\varnothing\subseteq A$, by the Empty Set Subset theorem from Subset and Proper Subset. The two inclusions imply $A=\varnothing$ by the Equality by Double Inclusion theorem. Therefore $A=\varnothing$, as required.

This proof illustrates a useful feature of the empty set: a statement about every element of an empty set has no counterexample. It does not mean that we have checked a list of elements and found each one to work. There are no elements to check. The precise logical consequence is developed below.

Worked Example: A Condition with No Integer Solutions

Let $$ B=\{n\in\mathbb Z:n^2=2\}. $$ Suppose $n\in B$. Then $n$ is an integer and $n^2=2$. Every integer is either even or odd. If $n$ is even, write $n=2k$ for some integer $k$. Then $$ n^2=4k^2, $$ which is divisible by $4$, whereas $2$ is not divisible by $4$. This contradicts $n^2=2$.

If $n$ is odd, write $n=2k+1$ for some integer $k$. Then $$ n^2=(2k+1)^2=4k^2+4k+1=4k(k+1)+1, $$ so $n^2$ is odd. But $2$ is even, again contradicting $n^2=2$. Both possible cases lead to a contradiction. Thus no integer $n$ belongs to $B$, and the Characterization of the Empty Set gives $B=\varnothing$.

Empty Domains and Quantified Statements

A domain may itself be empty. Quantifiers over that domain still have definite truth values. A universal statement $\forall x\in D,\ P(x)$ says that every element of $D$ satisfies $P$. An existential statement $\exists x\in D,\ P(x)$ says that at least one element of $D$ satisfies $P$. When $D=\varnothing$, the existential statement is false because there is no possible witness. The universal statement is true because there is no element of $D$ that could violate $P$.

This is sometimes called vacuous truth. It is not a special exception to the meanings of “every” and “there exists”; it follows from those meanings. A universal claim is false only if a counterexample in its domain can be produced. An empty domain supplies no such counterexample. An existential claim, in contrast, requires an actual element satisfying the condition.

Theorem (Quantifiers over the Empty Domain). Let $P(x)$ be any condition, and take its domain to be $\varnothing$. Then $$ \forall x\in\varnothing,\ P(x)\text{ is true}, \qquad \exists x\in\varnothing,\ P(x)\text{ is false}. $$

Proof. First consider the universal statement. By the definition of a universal statement, it would be false only if there were an $x\in\varnothing$ for which $P(x)$ is false. But there is no $x\in\varnothing$, since the empty set has no elements. Therefore no counterexample exists, and the universal statement is true.

For the existential statement to be true, there would have to be an $x\in\varnothing$ such that $P(x)$ holds. Again, the empty set has no elements, so no such $x$ exists. The existential statement is therefore false. These conclusions do not depend on the content of $P$; they follow from the domain being empty.

Worked Example: A Universal Claim with No Cases

Take the domain to be $D=\varnothing$, and consider the condition $P(x): x>0$. The statement $$ \forall x\in\varnothing,\ x>0 $$ is true. To disprove it, one would need a member of $\varnothing$ that is not greater than $0$. No member of $\varnothing$ exists, so such a counterexample cannot be given. This does not assert that some number in the empty set is positive; it asserts that every member meets the condition, and there are no members that fail it.

By contrast, $$ \exists x\in\varnothing,\ x>0 $$ is false. It requires a witness $x$ in the domain, and the domain has none. The universal and existential statements have different truth values because their requirements differ.

Empty Condition-Defined Sets

The same reasoning applies when the domain is nonempty but the condition selects no elements. The domain tells us which candidates are allowed; the condition may still rule out all of them. In logical form, the defining expression $\{x\in D:P(x)\}$ is empty exactly when there is no $x\in D$ satisfying $P(x)$. This is the connection between an empty set and a false existential statement over its domain.

Proposition (Empty Condition-Defined Set). Let $D$ be a domain and let $P(x)$ be a condition defined on $D$. Then $$ \{x\in D:P(x)\}=\varnothing \quad\Longleftrightarrow\quad \neg\exists x\in D,\ P(x). $$

Proof. Suppose first that $\{x\in D:P(x)\}=\varnothing$. If there were an $x\in D$ satisfying $P(x)$, the set-builder membership characterization would give $x\in\{x\in D:P(x)\}$. That contradicts the set being empty. Thus no such $x$ exists.

Conversely, suppose there is no $x\in D$ satisfying $P(x)$. If some object $y$ belonged to $\{x\in D:P(x)\}$, the membership characterization would imply $y\in D$ and $P(y)$. This would provide an element of $D$ satisfying $P$, contrary to the assumption. Hence no object belongs to the condition-defined set. By the Characterization of the Empty Set, that set equals $\varnothing$.

Worked Example: A Condition That Selects No Real Numbers

Consider $$ C=\{x\in\mathbb R:|x|<0\}. $$ For every real number $x$, absolute value is nonnegative: $|x|\geq0$. Therefore $|x|<0$ cannot hold for any real $x$. There is no witness to the existential statement $\exists x\in\mathbb R,\ |x|<0$. By the Empty Condition-Defined Set proposition, $C=\varnothing$.

The strict inequality is essential to the conclusion. If the condition were $|x|\leq0$, then $x=0$ would satisfy it, because $|0|=0$. The resulting set would contain $0$, and so it would not be empty. A small change in the condition can change whether any elements are selected.

Common Pitfalls

One mistake is to treat “no elements” as if it were a description of a particular element. The empty set is not a number, and it is not an unspecified missing object. It is a set whose membership test fails for every object. Another mistake is to confuse $\varnothing$ with $\{\varnothing\}$. The first has zero elements; the second has one.

A further mistake is to infer that a universal statement over an empty domain is false because there are no examples verifying it. A universal statement is refuted by a counterexample, not by the absence of examples. With an empty domain, counterexamples cannot occur, so the statement is true. An existential statement has the opposite requirement: it needs an example, and the empty domain cannot provide one.

Expression	Meaning
$\varnothing$	The set with no elements.
$\{\varnothing\}$	A set with one element, namely $\varnothing$.
$\forall x\in\varnothing,\ P(x)$	True: there is no counterexample in the domain.
$\exists x\in\varnothing,\ P(x)$	False: there is no possible witness in the domain.
$\{x\in D:P(x)\}=\varnothing$	No element of $D$ satisfies $P$.

Proof strategy. To prove a set is empty, assume a candidate belongs to it, unpack its membership condition, and derive a contradiction. To prove it is nonempty, give one valid member and verify both its domain membership and the defining condition.

For each question, distinguish the set itself from its elements, and use the stated domain when deciding whether a condition has a witness.

Check Your Understanding

Explain why $\varnothing$ and $\{\varnothing\}$ are different sets. Which one has an element?
Show that $\{x\in\mathbb R:x^2+4=0\}=\varnothing$. Which property of real squares gives the contradiction?
Let $D=\varnothing$. State whether each claim is true or false: $\forall x\in D,\ x=0$, and $\exists x\in D,\ x=0$. Explain the difference in their requirements.
Determine whether $\{n\in\mathbb Z:n=2k+1\text{ for some }k\in\mathbb Z\}$ is empty. If it is not, give one member and verify it.
What would need to be shown to prove $\{x\in D:P(x)\}=\varnothing$ using the Empty Condition-Defined Set proposition?
Give an example of a condition-defined set over a nonempty domain that is nevertheless empty, and explain why the condition has no witness.

Expression	Meaning
\(\varnothing\)	The set with no elements.
\(\{\varnothing\}\)	A set with one element, namely \(\varnothing\).
\(\forall x\in\varnothing,\ P(x)\)	True: there is no counterexample in the domain.
\(\exists x\in\varnothing,\ P(x)\)	False: there is no possible witness in the domain.
\(\{x\in D:P(x)\}=\varnothing\)	No element of \(D\) satisfies \(P\).

What You'll Learn