Set Membership

A Membership Claim Has Two Roles

In Sets and Elements, we introduced sets as collections of objects and used $x\in A$ to mean that $x$ is an element of $A$. This tutorial develops that notation as a way to make and check precise mathematical claims. The symbol $\in$ expresses a relation between two objects: the object on its left is being tested for membership in the set on its right.

For example, if $A=\{3,8,12\}$, then $8\in A$ is true because $8$ is one of the elements listed in $A$. The statement $5\in A$ is false because $5$ is not among those elements. We write $5\notin A$ to state that $5$ is not an element of $A$. In either case, the expression is a proposition: it has a definite truth value.

The positions of the symbols matter. In $8\in A$, the number $8$ is the candidate element and $A$ is the set. Reversing them to $A\in 8$ makes a different claim; it does not express the same membership relation. A symbol naming a set and a symbol naming an element may look similar, so reading the order aloud—“this object is an element of that set”—helps keep the roles clear.

Membership is not set equality. The statement $x\in A$ asks whether the object $x$ is among the elements of $A$. The statement $x=A$ compares two objects for equality. One claim can be true while the other is false; their meanings are different.

Membership in a Finite Roster

When a set is given by a finite roster, its elements are listed between braces. To decide whether an object belongs to that set, compare it with the listed elements. It belongs if it is equal to at least one of them; it does not belong if it differs from every one of them. The order of the entries and any repeated entries do not alter this test.

This test can be stated precisely. For a nonempty finite roster $A=\{a_1,a_2,\ldots,a_n\}$, where $n\geq1$, an object $x$ is an element of $A$ exactly when it equals at least one of the listed objects.

Proposition. Let $a_1,\ldots,a_n$ and $x$ be objects, where $n\geq1$, and let $A=\{a_1,\ldots,a_n\}$. Then $$ x\in A \quad\Longleftrightarrow\quad x=a_1\text{ or }x=a_2\text{ or }\cdots\text{ or }x=a_n. $$

Proof. Suppose first that $x\in A$. By the meaning of the roster $A=\{a_1,\ldots,a_n\}$, its elements are precisely the objects listed there. Thus $x$ must be equal to at least one listed object: for some index $i$ with $1\leq i\leq n$, $x=a_i$. Therefore the disjunction on the right is true.

Conversely, suppose the disjunction on the right is true. Then $x=a_i$ for some index $i$ with $1\leq i\leq n$. Since $a_i$ is one of the listed elements of $A$, and $x$ is that same object, $x$ is an element of $A$. Hence $x\in A$. Both implications hold, so the equivalence is proved.

Worked Example: Checking Several Roster Membership Claims

Let $B=\{-4,0,6,11\}$. The proposition above says that a number belongs to $B$ precisely when it equals one of $-4,0,6,11$.

For $x=6$, we have $x=6$, which matches an entry, so $6\in B$. For $x=7$, the equalities $7=-4$, $7=0$, $7=6$, and $7=11$ are all false, so $7\notin B$. For $x=-4$, one of the entries matches, so $-4\in B$. The test checks equality with the listed objects; being close to a listed number or sharing a property with it is not enough.

The proposition applies even if a roster repeats an object, because a repeated entry does not introduce a different possible element. For instance, $x\in\{2,2,9\}$ holds exactly when $x=2$ or $x=9$. The duplicated $2$ gives no additional membership condition.

Membership in a Set-Builder Description

A set can also be described by specifying a domain and a condition. For example, $\{n\in\mathbb Z:n\text{ is odd and }-3\leq n\leq5\}$ collects the integers satisfying both requirements. For membership in such a set, the candidate must first belong to the stated domain and must also satisfy the stated condition.

More generally, let $D$ be a domain and let $P(u)$ be a condition defined for each $u\in D$. The notation $$ S=\{u\in D:P(u)\} $$ describes the objects in $D$ for which $P$ is true. This gives a direct test for a membership claim.

Theorem. Let $D$ be a domain, let $P$ be a condition defined on $D$, and let $S=\{u\in D:P(u)\}$. For any object $x$, $$ x\in S \quad\Longleftrightarrow\quad x\in D\text{ and, if }x\in D,\text{ then }P(x)\text{ holds}. $$

Proof. Suppose $x\in S$. By the definition of $S$, every element of $S$ is an object in $D$ for which $P$ holds. Therefore $x\in D$ and $P(x)$.

Conversely, suppose $x\in D$ and $P(x)$. The set $S$ consists of exactly the objects in $D$ satisfying $P$. Since $x$ meets both requirements, it is one of the objects collected in $S$, and therefore $x\in S$. This proves both directions of the equivalence.

Worked Example: Checking a Domain and a Condition

Define $$ C=\{m\in\mathbb Z:m^2<20\}. $$ To test whether $3\in C$, check both parts of the membership criterion. First, $3\in\mathbb Z$. Second, $3^2=9<20$. Both requirements hold, so $3\in C$.

For $5$, we have $5\in\mathbb Z$, but $5^2=25\not<20$. The condition fails, so $5\notin C$. For $\sqrt{5}$, the numerical inequality $(\sqrt{5})^2=5<20$ holds, but $\sqrt{5}\notin\mathbb Z$. It is outside the stated domain, so $\sqrt{5}\notin C$. Passing the condition alone is not sufficient when the domain requirement fails.

Check the domain.
Determine whether the candidate object belongs to the domain stated after the set-builder bar.

Evaluate the condition.
If the object is in the domain, substitute it into the condition and decide whether the resulting statement is true.

Combine the two checks.
The object belongs to the described set exactly when it is in the domain and satisfies the condition.

The word “and” is important. A set-builder description with a domain and a condition imposes both requirements. If either fails, the membership claim is false. When the candidate is not in the domain, the condition may not even be defined for it; the domain restriction is not optional.

When the Element Is Itself a Set

An element of a set need not be a number. It can itself be a set, and the outer braces indicate which objects are elements of the collection. Consequently, membership must be read from the outside inward: identify the elements immediately enclosed by the outermost braces before considering any inner braces.

Worked Example: Membership with Nested Sets

Let $D=\{1,\{1\},\{1,2\}\}$. The elements of $D$ are the number $1$, the singleton set $\{1\}$, and the set $\{1,2\}$. Thus $1\in D$, $\{1\}\in D$, and $\{1,2\}\in D$.

But $\{2\}\notin D$, since the set $\{2\}$ is not any of the three listed elements. Also, $2\in\{1,2\}$ is true, but that does not make $2\in D$: the number $2$ is an element of one of the elements of $D$, not one of the elements listed directly in $D$. The braces distinguish these levels.

This distinction prevents a frequent error: a property of an element inside a nested set does not automatically become a membership claim about the outer set. From $2\in\{1,2\}$ and $\{1,2\}\in D$, one may conclude that $\{1,2\}$ is in $D$; these two facts alone do not say that $2$ is in $D$.

Membership in the Empty Set

The empty set $\varnothing$ has no elements. Therefore, for every object $x$, the statement $x\in\varnothing$ is false and $x\notin\varnothing$ is true. This conclusion follows directly from what it means for the empty set to contain no objects: there is no candidate that could appear among its elements.

Be careful to distinguish $\varnothing$ from $\{\varnothing\}$. The first has no elements, whereas the second has exactly one element, namely the empty set itself. In particular, $\varnothing\notin\varnothing$, but $\varnothing\in\{\varnothing\}$. A set may be empty, or it may contain the empty set as an element; those are different situations.

Worked Example: The Empty Set as a Candidate Element

Consider $E=\{\varnothing,4\}$. The outer roster lists two elements: $\varnothing$ and $4$. Hence $\varnothing\in E$ and $4\in E$. The empty set has not become a number or ceased to be empty; it is simply one of the objects collected in $E$.

By contrast, neither $4\in\varnothing$ nor $\varnothing\in\varnothing$ is true, because the empty set has no elements at all. The membership question always concerns the set on the right side of $\in$, so changing that set can change the truth value.

Read membership at the correct level. For $x\in A$, check whether the object $x$ itself is an element of $A$. Do not replace that question with whether $x$ is related to an element of $A$, appears inside an element of $A$, or satisfies a property that the set does not require.

For each question, identify the candidate object and the set being tested. For set-builder descriptions, check the domain as well as the condition.

Check Your Understanding

Let $A=\{-2,1,5,10\}$. Decide whether $1\in A$, $3\in A$, and $10\in A$. State how you checked each claim.
Let $B=\{k\in\mathbb Z:k+4=1\}$. Is $-3\in B$? Is $1\in B$? Verify the domain and condition for both candidates.
Let $C=\{0,\{0\},\{0,1\}\}$. Which of $0$, $\{0\}$, and $\{1\}$ are elements of $C$?
For $S=\{x\in\mathbb R:x^2=9\}$, state the membership test for an arbitrary real number $a$. Then decide whether $-3\in S$ and $3.5\in S$.
Is $\varnothing\in\varnothing$ true or false? Is $\varnothing\in\{\varnothing\}$ true or false? Explain the difference.
In your own words, state the two requirements an object must meet to belong to a set described as $\{u\in D:P(u)\}$.

What You'll Learn