April6 ,2021 COUNTABLESETS&CANTOR’SDIAGONALARGUMENT ConceptsinAbstractMathematics

COUNTABLE SETS & CANTOR’S DIAGONAL ARGUMENT

April 6^th, 2021

Notation

In what follows, we setℵ₀≔ |ℕ|(pronouncedaleph nought).

Definition

A set𝐸is countable if either𝐸is finite or|𝐸| = ℵ₀.

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Apr 6, 2021 2 / 13

Proposition

1 |ℕ ⧵ {0}| = ℵ₀

2 |{𝑛 ∈ ℕ ∶ 𝑛 ≡ 0 mod2}| = ℵ₀

3 |ℕ × ℕ| = ℵ₀

Proof.

1 The function𝑓 ∶ ℕ → ℕ ⧵ {0}defined by𝑓 (𝑛) = 𝑛 + 1is bijective with inverse 𝑓⁻¹∶ ℕ ⧵ {0} → ℕdefined by𝑓⁻¹(𝑛) = 𝑛 − 1.

2 The function𝑓 ∶ ℕ → {𝑛 ∈ ℕ ∶ 𝑛 ≡ 0 mod2}defined by𝑓 (𝑛) = 2𝑛is bijective.

3 Define𝑓 ∶ ℕ × ℕ → ℕby𝑓 (𝑎, 𝑏) = 2^𝑎3^𝑏.

Then𝑓 is injective by uniqueness of the prime decomposition. Thus|ℕ × ℕ| ≤ ℵ₀. Besides{0} × ℕ ⊂ ℕ × ℕ, thusℵ₀= |{0} × ℕ| ≤ |ℕ × ℕ|.

Hence|ℕ × ℕ| = ℵ₀by Cantor–Schröder–Bernstein theorem. ■

Proposition

If𝑆 ⊂ ℕis infinite then|𝑆| = ℵ₀.

Proof.Let’s define the function𝑓 ∶ ℕ → 𝑆by induction as follows.

Set𝑓 (0) =min𝑆(which is well-defined by the well-ordering principle since𝑆 ≠ ∅as it is infinite).

And then, assuming that𝑓 (𝑛)is already defined, we set𝑓 (𝑛 + 1) =min{𝑘 ∈ 𝑆 ∶ 𝑘 > 𝑓 (𝑛)}

(which is well-defined by the well-ordering principle: the involved set is non-empty since otherwise𝑆would be finite).

It is easy to check that𝑓 is injective (note that∀𝑛 ∈ ℕ, 𝑓 (𝑛 + 1) > 𝑓 (𝑛)), thereforeℵ₀≤ |𝑆|.

But since𝑆 ⊂ ℕ, we also have|𝑆| ≤ ℵ₀.

Thus, by Cantor–Schröder–Bernstein theorem,|𝑆| = ℵ₀. ■

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Apr 6, 2021 4 / 13

Proposition

A set𝐸is countable if and only if|𝐸| ≤ ℵ₀(i.e. there exists an injection𝑓 ∶ 𝐸 → ℕ),

otherwise stated𝐸is countable if and only if there exists a bijection between𝐸and a subset ofℕ.

Proof.

⇒Assume that𝐸is countable.

• Either𝐸is finite and then there exists𝑛 ∈ ℕand a bijection𝑔 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → 𝐸.

We define𝑓 ∶ 𝐸 → ℕby𝑓 (𝑥) = 𝑔⁻¹(𝑥)(which is well-defined since{𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} ⊂ ℕ).

And𝑓 is an injection since𝑔⁻¹is.

• Or|𝐸| = ℵ₀, i.e. there exists a bijection𝑓 ∶ 𝐸 → ℕ.

⇐Assume there exists an injection𝑓 ∶ 𝐸 → ℕ.

Assume that𝐸is infinite. Then|𝐸| = |𝑓 (𝐸)| = ℵ₀.

Thus either𝐸is finite or|𝐸| = ℵ₀. In both cases𝐸is countable. ■

Theorem

A countable union of countable sets is countable,

i.e. if𝐼 is countable and if for every𝑖 ∈ 𝐼,𝐸_𝑖 is countable then⋃_𝑖∈𝐼𝐸_𝑖is countable.

Proof.

WLOG we may now assume that𝐼 ⊂ ℕ.

Let𝑖 ∈ 𝐼. Since𝐸_𝑖is countable, there exists an injection𝑓_𝑖∶ 𝐸_𝑖→ ℕ¹.

We define𝜑 ∶ ⋃_𝑖∈𝐼𝐸_𝑖 → ℕ × ℕby𝜑(𝑥) = (𝑛, 𝑓_𝑛(𝑥))where𝑛 =min{𝑖 ∈ 𝐼 ∶ 𝑥 ∈ 𝐸_𝑖}(which exists by the well-ordering principle).

It is not difficult to check that𝜑is injective.

Therefore⋃_𝑖∈𝐼𝐸_𝑖is countable. ■

1We use the axiom of countable choice here.

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Apr 6, 2021 6 / 13

Theorem

If𝐸is an infinite set then there exists𝑇 ⊂ 𝐸such that|𝑇 | = ℵ₀, i.e. ℵ₀is the least infinite cardinal.

Proof.

For𝑛 ∈ ℕ, set𝐸_𝑛= {𝑆 ∈ 𝒫(𝐸) ∶ |𝑆| = 𝑛}.

Since𝐸is infinite, it contains a subset of cardinal𝑛, therefore𝐸_𝑛≠ ∅.

So for every𝑛 ∈ ℕ, we can pick² 𝑆_𝑛∈ 𝐸_𝑛.

Then𝑇 ≔ ⋃_𝑛∈ℕ𝑆_𝑛is countable as a countable union of countable sets.

Besides,∀𝑛 ∈ ℕ, 𝑆_𝑛⊂ 𝑇 and|𝑆_𝑛| = 𝑛.

Therefore𝑇 is infinite since for every𝑛 ∈ ℕit contains a subset of cardinal𝑛.

Thus|𝑇 | = ℵ₀as an infinite countable set. ■

2We use the axiom of countable choice here.

Theorem

|ℤ| = ℵ₀

Proof 1. Sinceℕ ⊂ ℤ, we have|ℕ| ≤ |ℤ|.

Define𝑓 ∶ ℤ → ℕby𝑓 (𝑛) = {

2^𝑛 if𝑛 ≥ 0 3^−𝑛 if𝑛 < 0 .

Then𝑓 is injective by uniqueness of the prime factorization. Therefore|ℤ| ≤ |ℕ|.

Hence|ℤ| = |ℕ|by Cantor–Schröder–Bernstein theorem. ■

Proof 2.

Define𝑓 ∶ ℤ → ℕby𝑓 (𝑛) = {

2𝑛 if𝑛 ≥ 0

−(2𝑛 + 1) if𝑛 < 0 . Then𝑓 is bijective with inverse𝑓⁻¹(𝑚) =

{

𝑘 if∃𝑘 ∈ ℕ, 𝑚 = 2𝑘

−𝑘 − 1 if∃𝑘 ∈ ℕ, 𝑚 = 2𝑘 + 1 .

Therefore|ℤ| = |ℕ|. ■

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Apr 6, 2021 8 / 13

Theorem

|ℚ| = ℵ₀

Proof 1.Note thatℕ ⊂ ℚ, thereforeℵ₀≤ |ℚ|.

Define𝑓 ∶ ℚ → ℤ × ℤby𝑓 (^𝑎𝑏) = (𝑎, 𝑏)where ^𝑎

𝑏 is in lowest form.

Then𝑓is injective and thus|ℚ| ≤ |ℤ × ℤ|. Since|ℤ| = |ℕ|, we get|ℤ × ℤ| = |ℕ × ℕ| = ℵ₀.

We conclude using Cantor–Schröder–Bernstein theorem. ■

Proof 2.Note thatℕ ⊂ ℚ, thereforeℵ₀≤ |ℚ|.

Moreover𝑓 ∶ ℤ × ℕ ⧵ {0} → ℚdefined by𝑓 (𝑎, 𝑏) = ^𝑎_𝑏 is surjective. Thus|ℚ| ≤ |ℤ × ℕ ⧵ {0}|.

Since|ℤ| = |ℕ|and|ℕ ⧵ {0}| = |ℕ|, we get|ℤ × ℕ ⧵ {0}| = |ℕ × ℕ| = ℵ₀.

We conclude using Cantor–Schröder–Bernstein theorem. ■

Proof 3.Note thatℕ ⊂ ℚ, thereforeℵ₀≤ |ℚ|.

Sinceℚ = ⋃

(𝑎,𝑏)∈ℤ×ℕ⧵{0}{𝑎

𝑏 },ℚis countable as a countable union of countable sets. So|ℚ| ≤ ℵ₀.

We conclude using Cantor–Schröder–Bernstein theorem. ■

Theorem: ℝ is not countable

Proof.We are going to prove that there is no surjectionℕ → ℝ(and hence no such bijection).

Let𝑓 ∶ ℕ → ℝbe a function. Given𝑛 ∈ ℕ, we know that𝑓 (𝑛)has a unique proper decimal expansion𝑓 (𝑛) =

+∞

∑𝑘=0

𝑎_𝑛𝑘10^−𝑘 where𝑎_𝑛0∈ ℤand𝑎_𝑛𝑘∈ {0, 1, … , 9}for𝑘 ≥ 1, i.e.

𝑓 (0) =𝑎₀₀. 𝑎₀₁𝑎₀₂𝑎₀₃𝑎₀₄𝑎₀₅… 𝑓 (1) = 𝑎₁₀.𝑎₁₁𝑎₁₂𝑎₁₃𝑎₁₄𝑎₁₅… 𝑓 (2) = 𝑎₂₀. 𝑎₂₁𝑎₂₂𝑎₂₃𝑎₂₄𝑎₂₅… 𝑓 (3) = 𝑎₃₀. 𝑎₃₁𝑎₃₂𝑎₃₃𝑎₃₄𝑎₃₅… 𝑓 (4) = 𝑎₄₀. 𝑎₄₁𝑎₄₂𝑎₄₃𝑎₄₄𝑎₄₅…

⋮ ⋮

Given𝑘 ∈ ℕ, we set𝑏_𝑘= {

1 if𝑎_𝑘𝑘= 0 0 otherwise . Then𝑏 =

+∞

∑𝑘=0

𝑏_𝑘10^−𝑘is a real number written with its unique proper decimal expansion.

Note that for every𝑛 ∈ ℕ,𝑏 ≠ 𝑓 (𝑛)since𝑏_𝑛≠ 𝑎_𝑛𝑛(we use the uniqueness of the proper decimal expansion).

Therefore𝑏 ∉Im(𝑓 )and𝑓is not surjective. ■

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Apr 6, 2021 10 / 13

Cantor’s theorem

Given a set𝐸,|𝐸| < |𝒫(𝐸)|.

Proof.We are going to use Cantor’s diagonal argument again.

First, note that𝑔 ∶ 𝐸 → 𝒫(𝐸)defined by𝑔(𝑥) = {𝑥}is injective, therefore|𝐸| ≤ |𝒫(𝐸)|.

We are going to prove that there is no surjection𝐸 → 𝒫(𝐸)(and hence no such bijection).

Let𝑓 ∶ 𝐸 → 𝒫(𝐸)be a function. Define𝑆 = {𝑥 ∈ 𝐸 ∶ 𝑥 ∉ 𝑓 (𝑥)}.

Let𝑥 ∈ 𝐸.

• If𝑥 ∈ 𝑓 (𝑥)then𝑥 ∉ 𝑆.

• Otherwise, if𝑥 ∉ 𝑓 (𝑥)then𝑥 ∈ 𝑆.

Therefore𝑓 (𝑥) ≠ 𝑆since one contains𝑥but not the other one.

Thus𝑆 ∉Im(𝑓 )and𝑓 is not surjective. ■

Remark

There is no greatest cardinal.

We already know that|ℕ| < |ℝ|and that|ℕ| < |𝒫(ℕ)|. Actually|ℝ| = |𝒫(ℕ)|.

Theorem

|ℝ| = |𝒫(ℕ)|

Proof.

Define𝑓 ∶ 𝒫(ℕ) → ℝby𝑓 (𝑆) = ∑

𝑛∈𝑆

10^−𝑛.

Then𝑓 is injective by uniqueness of the proper decimal expansion. Thus|𝒫(ℕ)| ≤ |ℝ|.

Define𝑔 ∶ ℝ → 𝒫(ℚ)by𝑔(𝑥) = {𝑞 ∈ ℚ ∶ 𝑞 < 𝑥}.

Then𝑔 is injective. Indeed, let𝑥, 𝑦 ∈ ℝbe such that𝑥 < 𝑦. Sinceℚis dense inℝ, there exists 𝑞 ∈ ℚsuch that𝑥 < 𝑞 < 𝑦. So𝑞 ∉ 𝑔(𝑥)but𝑞 ∈ 𝑔(𝑦). Therefore𝑔(𝑥) ≠ 𝑔(𝑦).

Hence|ℝ| ≤ |𝒫(ℚ)| = |𝒫(ℕ)|(prove the last equality using that|ℚ| = |ℕ|).

We conclude thanks to Cantor–Schröder–Bernstein theorem. ■

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Apr 6, 2021 12 / 13

Theorem

There is no set containing all sets.

Proof.Assume that such a set𝑉 exists.

Then the powerset𝒫(𝑉 )exists too and𝒫(𝑉 ) ⊂ 𝑉 by definition of𝑉.

Therefore|𝒫(𝑉 )| ≤ |𝑉 |, but|𝑉 | < |𝒫(𝑉 )|by Cantor’s theorem. Hence a contradiction. ■ We may similarly prove that there is no set containing all finite sets, or even all singletons.

Theorem

There is no set containing all singletons.

Proof.Assume that the set𝑆of all singletons exists.

Define𝑓 ∶ 𝒫(𝑆) → 𝑆by𝑓 (𝑥) = {𝑥}(which is well-defined).

Since𝑓 is one-to-one, we get that|𝒫(𝑆)| ≤ |𝑆|.

Which contradicts|𝑆| < |𝒫(𝑆)|(Cantor’s theorem). ■