Write the set \(1)/(2), (2)/(3), (3)/(4), (4)/(5), (5)/(6), (6)/(7)\ in the set-builder form.

Worked Examples · Example 4

\left\{\dfrac{1}{2}, \dfrac{2}{3}, \dfrac{3}{4}, \dfrac{4}{5}, \dfrac{5}{6}, \dfrac{6}{7}\right\}

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Concept understanding — Set Membership

Set Membership

The idea in plain words

Every set is defined by exactly one question: "does this object belong to the set, or not?" That yes/no relationship between an object and a set is called membership. If an object is in the set, it is a member (or element) of the set; if it isn't, it simply is not.

A set is only "well-defined" if this question always has a clear answer for every possible object — that's what makes membership testable.

The notation

For a set $A$ and an object $x$ :

$x \in A$ reads " $x$ belongs to $A$ " or " $x$ is an element of $A$ " — TRUE membership.
$x \notin A$ reads " $x$ does not belong to $A$ " — FALSE membership.

Example. Let $A = \{2, 4, 6, 8\}$ .

4 \in A \qquad \text{(4 is listed inside A)}

5 \notin A \qquad \text{(5 is not listed inside A)}

Testing membership: roster form vs. set-builder form

Roster form — just look for the object in the list.

B = \{1, 3, 5, 7\}, \quad 3 \in B, \quad 4 \notin B

Set-builder form — plug the candidate into the defining rule and check if it's satisfied.

C = \{x \mid x \text{ is a prime number less than } 10\}

Is $7 \in C$ ? Check: is 7 prime and less than 10? Yes → $7 \in C$ .

Is $9 \in C$ ? Check: 9 is less than 10 but not prime ( $9 = 3 \times 3$ ) → $9 \notin C$ .

Properties every student must know

Each element is either in or out — never "partly in." Membership is binary, not a matter of degree.
Repetition doesn't affect membership. $\{1, 1, 2\} = \{1, 2\}$ — asking "is 1 a member?" gives the same YES either way.
Order never affects membership. $2 \in \{1, 2, 3\}$ is exactly the same fact as $2 \in \{3, 2, 1\}$ .
Nothing belongs to the empty set. For any $x$ , $x \notin \emptyset$ — there is nothing inside to belong to.
A set can be an element of another set. If $D = \{1, \{2, 3\}\}$ , then $1 \in D$ is true, and $\{2,3\} \in D$ is true, but $2 \in D$ is false — 2 is not directly listed in $D$ ; it is only inside the set that is listed.

A common slip

Students often confuse $x \in A$ (membership: is the object present?) with $B \subseteq A$ (subset: is every element of $B$ also in $A$ ?). The two symbols compare different kinds of things:

$\in$ / $\notin$ compares an element to a set.
$\subseteq$ / $\not\subseteq$ compares a set to a set. …

Why this formula?

Let's break down the definition of a set — not as a formula to memorise, but as a fundamental idea that underpins all of mathematics.

1. What is a Set? (The Core Idea)

A set is a well-defined collection of distinct objects.

The "why" here is about clarity and precision — we need to know exactly what belongs and what does not.

Well-defined: For any object, we can say yes or no — no ambiguity.
Distinct: No duplicates — each object appears only once.

Why? Because if we couldn't decide membership, we couldn't do any logical operations. Sets are the building blocks of all mathematical structures.

2. The Key "Formula": Set-Builder Notation

The most common way to define a set is:

S = \{ x \mid P(x) \}

This reads: "S is the set of all objects $x$ such that property $P(x)$ is true."

Why does this work?

$x$ is a placeholder for any object.
$P(x)$ is a logical condition (a predicate) that is either true or false for each $x$ .
The vertical bar $\mid$ means "such that".

Example:

A = \{ n \mid n \in \mathbb{N}, n \text{ is even} \}

Here, $P(n)$ is " $n$ is a natural number and $n$ is even".

Only those $n$ that satisfy both conditions are included.

Why this form? It avoids listing infinitely many elements. It gives a rule — a decision procedure — for membership.

3. The Two Fundamental Properties (Axioms)

Every set definition relies on two intuitive truths:

(a) Extensionality — Two sets are equal if they have the same elements.

A = B \iff (\forall x)(x \in A \iff x \in B)

Why? A set is completely determined by its members. There is no other hidden property.

If you know what's inside, you know the set.

(b) Membership — The only relation is $\in$ (belongs to).

x \in S \quad \text{or} \quad x \notin S

Why? Because a set is just a container. The only question we can ask is: "Is this object inside?"

4. Why Can't We Just List Everything?

For small sets, listing works:

\{1, 2, 3\}

But for infinite sets (like all natural numbers), listing is impossible.

Set-builder notation solves this by giving a rule instead of a list.

Example:

\mathbb{N} = \{ n \mid n \text{ is a positive integer} \}

This is not a formula to memorise — it's a definition by property.

5. The "Empty Set" — Why It Exists

The empty set $\emptyset$ (or $\{\}$ ) is the set with no elements. …

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The set-builder form describes a set by a rule its elements obey instead of listing them. Every element here is a fraction whose numerator is one less than its denominator, i.e. of the form $\dfrac{n}{n+1}$ , with $n$ ranging over the natural numbers $1$ through $6$ .

Concept

The set-builder form of a set writes it as $\{x : x \text{ satisfies some property } P\}$ — we replace an explicit list (roster form) by a defining rule. The task is to spot the pattern common to all six elements and encode it as a formula together with the range of the index.

Spotting the pattern

Write the elements in order and label the position $n$ :

\underbrace{\tfrac{1}{2}}_{n=1},\ \underbrace{\tfrac{2}{3}}_{n=2},\ \underbrace{\tfrac{3}{4}}_{n=3},\ \underbrace{\tfrac{4}{5}}_{n=4},\ \underbrace{\tfrac{5}{6}}_{n=5},\ \underbrace{\tfrac{6}{7}}_{n=6}

In each fraction:

the numerator equals the position $n$ ;
the denominator is one more than the numerator, i.e. $n+1$ .

So the general element is

x = \frac{n}{n+1}.

Fixing the range of $n$ …

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←Previous question · 1.2Example 3 — Write the set A = 1, 4, 9, 16, 25, . . . in set-builder form →Next question · 1.2Example 5 — Match each of the set on the left described in the roster form with th…

Q.Write the set {12,23,34,45,56,67}\left\{\dfrac{1}{2}, \dfrac{2}{3}, \dfrac{3}{4}, \dfrac{4}{5}, \dfrac{5}{6}, \dfrac{6}{7}\right\}{21​,32​,43​,54​,65​,76​} in the set-builder form.

Concept

Spotting the pattern

Fixing the range of nnn …

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Q.Write the set $\left\{\dfrac{1}{2}, \dfrac{2}{3}, \dfrac{3}{4}, \dfrac{4}{5}, \dfrac{5}{6}, \dfrac{6}{7}\right\}$ in the set-builder form.

Fixing the range of $n$ …

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