Lecture # 5
Countable Sets
In this lecture, we shall discuss countable and un-countable sets and their
properties.
Denumerable set
Set 𝐷 is said to be denumerable if it is equivalent to set 𝑁 of natural numbers.
Example 1: Let 𝐴 = {2, 4, 6, … }. Then 𝐴 is denumerable.
Solution: Let 𝑓: 𝑁 → 𝐴 be defined as:
𝑓(𝑛) = 2𝑛.
𝑓 is one-one
Let 𝑓(𝑛1 ) = 𝑓(𝑛2 ) ⇒ 2𝑛1 = 2𝑛2 ⇒ 𝑛1 = 𝑛2 .
𝑓 is onto.
Since 2𝑛 ∈ 𝐴 is the image of some 𝑛 ∈ 𝑁 under 𝑓, so 𝑓 is onto.
Thus 𝑓 is both one-one and onto, so 𝐴~𝑁. Therefore, 𝐴 is denumerable.
Example 2: Show that the set of integers is denumerable set.
Solution: : Let 𝑓: 𝑁 → 𝑍 be defined as:
𝑛
𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
2
𝑓(𝑛) = {−𝑛 + 1
𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑
2
𝑓 is one-one
Let
𝑓(𝑛1 ) = 𝑓(𝑛2 ).
(1)
Case1 When 𝑛1 and 𝑛2 both are even
Equation (1) implies,
𝑛1
2
=
𝑛2
2
⇒ 𝑛1 = 𝑛2 .
Case2 When 𝑛1 and 𝑛2 both are odd
Equation (1) implies,
−𝑛1 +1
2
=
−𝑛2 +1
2
⇒ 𝑛1 = 𝑛2 .
S0, 𝑓 is one-one.
𝑓 is onto.
Clearly 𝑓 is onto.
Thus 𝑓 is both one-one and onto, so 𝑍~𝑁. Therefore, 𝑍 is denumerable.
Theorem: Show that an infinite sequence of distinct elements is denumerable.
Proof. Let 𝑎1 , 𝑎2 , 𝑎3 , … be an infinite sequence of distinct elements. Let 𝐴 =
{𝑎1 , 𝑎2 , 𝑎3 , … } be the set of points of the sequence. Define 𝑓: 𝑁 → 𝐴 by
𝑓(𝑛) = 𝑎𝑛
Then clearly 𝑓 is bijective. This completes the proof.
Theorem: Every infinite set contains a subset which is denumerable.
Proof. Let 𝐴 be an infinite set and 𝑎1 ∈ 𝐴 ⇒ {𝑎1 } ⊆ 𝐴. Let 𝑎2 ∈ 𝐴\{𝑎1 } ⇒
{𝑎1 , 𝑎2 } ⊆ 𝐴. Let 𝑎3 ∈ 𝐴\{𝑎1 , 𝑎2 } ⇒ {𝑎1 , 𝑎2 , 𝑎3 } ⊆ 𝐴. Continuing in this way,
we obtain a subset 𝐵 = {𝑎1 , 𝑎2 , 𝑎3 , … } of 𝐴 which is denumerable.
Remark
If A and B are denumerable sets then A x B is denumerable.
Theorem: A subset of a denumerable set is either finite or denumerable.
Proof. Left as an exercise.
Countable set
A set is said to be countable if it is either finite or denumerable.
Remark:
 All finite and denumerable sets are countable.
 A subset of a countable set is countable.
Theorem: Let 𝐴1 , 𝐴2 , 𝐴3 , …. be a denumerable family of pairwise disjoint sets.
Then ∪ 𝐴𝑖 is denumerable.
Proof. Let 𝐴𝑖 = {𝑎𝑖1 , 𝑎𝑖2 , … . }. Then
∪ 𝐴𝑖 = {𝑎11 , 𝑎12 , ….
𝑎21 , 𝑎22 , ….
… … … … … }.
Define 𝑓:∪ 𝐴𝑖 → 𝑁 × 𝑁 by 𝑓(𝑎𝑖𝑗 ) = (𝑖, 𝑗). Clearly 𝑓 is bijective. As 𝑁 × 𝑁 is
denumerable, so ∪ 𝐴𝑖 is denumerable.
Example 2: Show that the set 𝑄 is denumerable.
𝑄+ ~𝑁 × 𝑁.
Solution: As
𝑄 = 𝑄+ ∪ 𝑄− ∪ {0}.
(5.1)
𝑝
Define 𝑓: 𝑄 + → 𝑁 × 𝑁 by 𝑓 ( ) = (𝑝, 𝑞).
𝑞
𝑓 is one-one
Let
This implies that
𝑓 is onto
𝑝
𝑝
𝑓 ( 1) = 𝑓 ( 2) ⟹ (𝑝1 , 𝑞1 ) = (𝑝2 , 𝑞2 ).
𝑞
𝑞
1
2
𝑝1
𝑞1
=
𝑝2
𝑞2
.
𝑝
Since for each (𝑝, 𝑞) ∈ 𝑁 × 𝑁 there exists ∈ 𝑁 × 𝑁 under 𝑓, therefore, 𝑓 is
𝑞
onto.
So, 𝑄+ ~𝑁 × 𝑁. As 𝑁 × 𝑁 is denumerable, so is countable and this implies that
𝑄+ is countable.
𝑝
𝑝
Now define, 𝑔: 𝑄 + → 𝑄− by 𝑔 ( ) = − . Clearly, 𝑔 is bijective, so , 𝑄− is
𝑞
𝑞
countable. Also {0} is countable, so (5.1) implies that 𝑄 is countable. As 𝑄 is
not finite so is denumerable.
Example: Show that the set [0, 1] is non-denumerable.
Solution: Left as an exercise.
Example: Show that the set [a, b] is non-denumerable.
Solution: Define 𝑓: [0,1] → [𝑎, 𝑏] by 𝑓(𝑥) = 𝑎 + (𝑏 − 𝑎)𝑥. It can be easily
seen that 𝑓 is both one-one and onto. This implies [0,1] ∼ [𝑎, 𝑏]. As [0,1] is
non-denumerable, so [𝑎, 𝑏] is non denumerable.
Example: Show that the set 𝑄′ is non-denumerable.
Solution: Left as an exercise.