March25 ,2021 CARDINALITY:FINITESETS ConceptsinAbstractMathematics

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MAT246H1-S – LEC0201/9201

Concepts in Abstract Mathematics

CARDINALITY: FINITE SETS

March 25^th, 2021

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Mar 25, 2021 1 / 11

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Finite sets – 1

Definition: finite set

We say that a set𝐸is finite if there exists𝑛 ∈ ℕand a bijection𝑓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → 𝐸.

Then we write|𝐸| = 𝑛.

Note that{𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} = {0, 1, 2, … , 𝑛 − 1}.

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Finite sets – 2

Lemma

Let𝑛, 𝑝 ∈ ℕ. If there exists an injective function𝑓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → {𝑘 ∈ ℕ ∶ 𝑘 < 𝑝}then𝑛 ≤ 𝑝.

Proof.We prove the statement by induction on𝑛.

• Base case at𝑛 = 0:for any𝑝 ∈ ℕwe have𝑛 ≤ 𝑝.

• Induction step.Assume that the statement holds for some𝑛 ∈ ℕ.

Let𝑝 ∈ ℕ. Assume that there exists an injective function𝑓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛 + 1} → {𝑘 ∈ ℕ ∶ 𝑘 < 𝑝}. Define𝑔 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → {𝑘 ∈ ℕ ∶ 𝑘 < 𝑝 − 1}as follows:𝑔(𝑥) =

{

𝑓 (𝑥) if𝑓 (𝑥) < 𝑓 (𝑛) 𝑓 (𝑥) − 1 if𝑓 (𝑥) > 𝑓 (𝑛) Note that𝑓 (𝑥) ≠ 𝑓 (𝑛)since𝑓is injective.

⋆ Claim 1:𝑔is well-defined, i.e.∀𝑥 ∈ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛}, 𝑔(𝑥) ∈ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑝 − 1}. Let𝑥 ∈ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛}.

So either,𝑓 (𝑥) < 𝑓 (𝑛)and then𝑔(𝑥) = 𝑓 (𝑥) < 𝑓 (𝑛) < 𝑝, therefore0 ≤ 𝑔(𝑥) < 𝑝 − 1. Or,𝑓 (𝑥) > 𝑓 (𝑛)and then𝑔(𝑥) = 𝑓 (𝑥) − 1 < 𝑝 − 1, therefore0 ≤ 𝑔(𝑥) < 𝑝 − 1.

⋆ Claim 2:𝑔is injective.

Let𝑥, 𝑦 ∈ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛}be such that𝑔(𝑥) = 𝑔(𝑦). First case:𝑓 (𝑥), 𝑓 (𝑦) < 𝑓 (𝑛).

Then𝑔(𝑥) = 𝑓 (𝑥)and𝑔(𝑦) = 𝑓 (𝑦). So𝑓 (𝑥) = 𝑓 (𝑦)and thus𝑥 = 𝑦since𝑓is injective. Second case:𝑓 (𝑥), 𝑓 (𝑦) > 𝑓 (𝑛).

Then𝑔(𝑥) = 𝑓 (𝑥) − 1and𝑔(𝑦) = 𝑓 (𝑦) − 1. So𝑓 (𝑥) = 𝑓 (𝑦)and thus𝑥 = 𝑦since𝑓is injective. Third case:𝑓 (𝑥) < 𝑓 (𝑛)and𝑓 (𝑦) > 𝑓 (𝑛).

Then𝑔(𝑥) = 𝑓 (𝑥) < 𝑓 (𝑛)and𝑔(𝑦) = 𝑓 (𝑦) − 1 > 𝑓 (𝑛) − 1 ≥ 𝑓 (𝑛). Therefore, this case is impossible.

Therefore, by the induction hypothesis,𝑛 ≤ 𝑝 − 1, i.e.𝑛 + 1 ≤ 𝑝. ■

0 1 2 3 4 5

0 1 2 3 4 5 6 7 𝑓

0 1 2 3 4

0 1 2 3 4 5 𝑔 6

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Finite sets – 2

Lemma

Let𝑝 ∈ ℕ. Assume that there exists an injective function𝑓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛 + 1} → {𝑘 ∈ ℕ ∶ 𝑘 < 𝑝}.

Define𝑔 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → {𝑘 ∈ ℕ ∶ 𝑘 < 𝑝 − 1}as follows:𝑔(𝑥) = {

⋆ Claim 1:𝑔is well-defined, i.e.∀𝑥 ∈ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛}, 𝑔(𝑥) ∈ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑝 − 1}. Let𝑥 ∈ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛}.

So either,𝑓 (𝑥) < 𝑓 (𝑛)and then𝑔(𝑥) = 𝑓 (𝑥) < 𝑓 (𝑛) < 𝑝, therefore0 ≤ 𝑔(𝑥) < 𝑝 − 1. Or,𝑓 (𝑥) > 𝑓 (𝑛)and then𝑔(𝑥) = 𝑓 (𝑥) − 1 < 𝑝 − 1, therefore0 ≤ 𝑔(𝑥) < 𝑝 − 1.

Let𝑥, 𝑦 ∈ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛}be such that𝑔(𝑥) = 𝑔(𝑦). First case:𝑓 (𝑥), 𝑓 (𝑦) < 𝑓 (𝑛).

Then𝑔(𝑥) = 𝑓 (𝑥)and𝑔(𝑦) = 𝑓 (𝑦). So𝑓 (𝑥) = 𝑓 (𝑦)and thus𝑥 = 𝑦since𝑓is injective. Second case:𝑓 (𝑥), 𝑓 (𝑦) > 𝑓 (𝑛).

Then𝑔(𝑥) = 𝑓 (𝑥) − 1and𝑔(𝑦) = 𝑓 (𝑦) − 1. So𝑓 (𝑥) = 𝑓 (𝑦)and thus𝑥 = 𝑦since𝑓is injective. Third case:𝑓 (𝑥) < 𝑓 (𝑛)and𝑓 (𝑦) > 𝑓 (𝑛).

0 1 2 3 4 5

0 1 2 3 4 5 6 7 𝑓

0 1 2 3 4

0 1 2 3 4 5 𝑔 6

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Finite sets – 2

Lemma

Let𝑝 ∈ ℕ. Assume that there exists an injective function𝑓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛 + 1} → {𝑘 ∈ ℕ ∶ 𝑘 < 𝑝}.

Define𝑔 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → {𝑘 ∈ ℕ ∶ 𝑘 < 𝑝 − 1}as follows:𝑔(𝑥) = {

⋆ Claim 1:𝑔is well-defined, i.e.∀𝑥 ∈ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛}, 𝑔(𝑥) ∈ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑝 − 1}.

Let𝑥 ∈ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛}.

So either,𝑓 (𝑥) < 𝑓 (𝑛)and then𝑔(𝑥) = 𝑓 (𝑥) < 𝑓 (𝑛) < 𝑝, therefore0 ≤ 𝑔(𝑥) < 𝑝 − 1.

Or,𝑓 (𝑥) > 𝑓 (𝑛)and then𝑔(𝑥) = 𝑓 (𝑥) − 1 < 𝑝 − 1, therefore0 ≤ 𝑔(𝑥) < 𝑝 − 1.

Let𝑥, 𝑦 ∈ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛}be such that𝑔(𝑥) = 𝑔(𝑦).

First case:𝑓 (𝑥), 𝑓 (𝑦) < 𝑓 (𝑛).

Then𝑔(𝑥) = 𝑓 (𝑥)and𝑔(𝑦) = 𝑓 (𝑦). So𝑓 (𝑥) = 𝑓 (𝑦)and thus𝑥 = 𝑦since𝑓is injective.

Second case:𝑓 (𝑥), 𝑓 (𝑦) > 𝑓 (𝑛).

Then𝑔(𝑥) = 𝑓 (𝑥) − 1and𝑔(𝑦) = 𝑓 (𝑦) − 1. So𝑓 (𝑥) = 𝑓 (𝑦)and thus𝑥 = 𝑦since𝑓is injective.

Third case:𝑓 (𝑥) < 𝑓 (𝑛)and𝑓 (𝑦) > 𝑓 (𝑛).

0 1 2 3 4 5

0 1 2 3 4 5 6 7 𝑓

0 1 2 3 4

0 1 2 3 4 5 𝑔 6

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Finite sets – 3

Definition: finite set

Corollary

Let𝐸be a finite set. If|𝐸| = 𝑛and|𝐸| = 𝑚, then𝑚 = 𝑛.

Then we say that|𝐸|is thecardinalof𝐸, which is uniquely defined.

Proof.Assume there exists a bijection𝑓₁∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → 𝐸and a bijection𝑓₂∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑚} → 𝐸. Then𝑓₂⁻¹∘ 𝑓₁∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → {𝑘 ∈ ℕ ∶ 𝑘 < 𝑚}is a bijection, so by the above lemma,𝑛 ≤ 𝑚.

Similarly,𝑓₁⁻¹∘ 𝑓₂∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑚} → {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛}is a bijection and thus𝑚 ≤ 𝑛.

Therefore𝑛 = 𝑚. ■

Remark: the empty set

|𝐸| = 0 ⇔ 𝐸 = ∅

Indeed, if𝐸 = ∅then𝑓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 0} → 𝐸is always bijective: injectiveness and surjectiveness are vacuously true. So|𝐸| = 0.

Otherwise, if𝐸 ≠ ∅then𝑓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 0} → 𝐸is never surjective (thus never bijective), so|𝐸| ≠ 0.

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Finite sets – 3

Corollary

Proof.Assume there exists a bijection𝑓₁∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → 𝐸and a bijection𝑓₂∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑚} → 𝐸.

Then𝑓₂⁻¹∘ 𝑓₁∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → {𝑘 ∈ ℕ ∶ 𝑘 < 𝑚}is a bijection, so by the above lemma,𝑛 ≤ 𝑚.

|𝐸| = 0 ⇔ 𝐸 = ∅

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Finite sets – 3

Corollary

Proof.Assume there exists a bijection𝑓₁∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → 𝐸and a bijection𝑓₂∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑚} → 𝐸.

Then𝑓₂⁻¹∘ 𝑓₁∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → {𝑘 ∈ ℕ ∶ 𝑘 < 𝑚}is a bijection, so by the above lemma,𝑛 ≤ 𝑚.

|𝐸| = 0 ⇔ 𝐸 = ∅

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Finite sets – 4

Proposition

If𝐸 ⊂ 𝐹 and𝐹 is finite then𝐸is finite too, besides,|𝐸| ≤ |𝐹 |.

Proof.Let’s prove by induction on𝑛 = |𝐹 |that if𝐸 ⊂ 𝐹 then𝐸is finite and|𝐸| ≤ 𝑛.

• Base case at𝑛 = 0:then𝐹 = ∅, so the only possible subset is𝐸 = ∅and then|𝐸| = 0.

• Induction step.Assume that the statement holds for some𝑛 ∈ ℕ. Let𝐹 be a set such that|𝐹 | = 𝑛 + 1.

• First case:𝐸 = 𝐹.Then the statement is obvious.

• Second case: 𝐸 ≠ 𝐹.Then there exists𝑥 ∈ 𝐹 ⧵ 𝐸. There exists a bijection𝑓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛 + 1} → 𝐹.

Since𝑓 is bijective, there exists a unique𝑚 < 𝑛 + 1such that𝑓 (𝑚) = 𝑥. Define𝑔 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → 𝐹 ⧵ {𝑥}by𝑔(𝑘) = 𝑓 (𝑘)for𝑘 ≠ 𝑚, and, if𝑚 ≠ 𝑛,𝑔(𝑚) = 𝑓 (𝑛).

Then𝑔is a bijection, so𝐹 ⧵ {𝑥}is finite and|𝐹 ⧵ {𝑥}| = 𝑛.

Since𝐸 ⊂ 𝐹 ⧵ {𝑥}, by the induction hypothesis,𝐸is finite and|𝐸| ≤ 𝑛 < 𝑛 + 1.

0 1 2 3 4 5

𝑥

𝑓

0 1 2 3 4

𝑔

■

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Finite sets – 4

Proposition

If𝐸 ⊂ 𝐹 and𝐹 is finite then𝐸is finite too, besides,|𝐸| ≤ |𝐹 |.

Proof.Let’s prove by induction on𝑛 = |𝐹 |that if𝐸 ⊂ 𝐹 then𝐸is finite and|𝐸| ≤ 𝑛.

• Base case at𝑛 = 0:then𝐹 = ∅, so the only possible subset is𝐸 = ∅and then|𝐸| = 0.

Let𝐹 be a set such that|𝐹 | = 𝑛 + 1.

• First case:𝐸 = 𝐹.Then the statement is obvious.

• Second case: 𝐸 ≠ 𝐹.Then there exists𝑥 ∈ 𝐹 ⧵ 𝐸.

There exists a bijection𝑓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛 + 1} → 𝐹.

Since𝑓 is bijective, there exists a unique𝑚 < 𝑛 + 1such that𝑓 (𝑚) = 𝑥.

Define𝑔 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → 𝐹 ⧵ {𝑥}by𝑔(𝑘) = 𝑓 (𝑘)for𝑘 ≠ 𝑚, and, if𝑚 ≠ 𝑛,𝑔(𝑚) = 𝑓 (𝑛).

Then𝑔is a bijection, so𝐹 ⧵ {𝑥}is finite and|𝐹 ⧵ {𝑥}| = 𝑛.

Since𝐸 ⊂ 𝐹 ⧵ {𝑥}, by the induction hypothesis,𝐸is finite and|𝐸| ≤ 𝑛 < 𝑛 + 1.

0 1 2 3 4 5

𝑥

𝑓

0 1 2 3 4

𝑔

■

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Finite sets – 5

Proposition

Let𝐸 ⊂ 𝐹 with𝐹 finite. Then|𝐹 | = |𝐸| + |𝐹 ⧵ 𝐸|.

Proof.Since𝐹 ⧵ 𝐸 ⊂ 𝐹 and𝐸 ⊂ 𝐹, we know that𝐸and𝐹 ⧵ 𝐸are finite. Denote𝑟 = |𝐸|and𝑠 = |𝐹 ⧵ 𝐸|.

There exist bijections𝑓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑟} → 𝐸and𝑔 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑠} → 𝐹 ⧵ 𝐸. Defineℎ ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑟 + 𝑠} → 𝐹 byℎ(𝑘) =

{

𝑓 (𝑘) if𝑘 < 𝑟 𝑔(𝑘 − 𝑟) if𝑘 ≥ 𝑟 .

• ℎis well-defined:

Indeed, if0 ≤ 𝑘 < 𝑟then𝑓 (𝑘)is well-defined and𝑓 (𝑘) ∈ 𝐸 ⊂ 𝐹.

If𝑟 ≤ 𝑘 < 𝑟 + 𝑠then0 ≤ 𝑘 − 𝑟 < 𝑠so that𝑔(𝑘 − 𝑟)is well-defined and𝑔(𝑘 − 𝑟) ∈ 𝐹 ⧵ 𝐸 ⊂ 𝐹.

• ℎis a bijection:

• ℎis injective: let𝑥, 𝑦 ∈ {0, 1, … , 𝑟 + 𝑠 − 1}be such thatℎ(𝑥) = ℎ(𝑦).

Eitherℎ(𝑥) = ℎ(𝑦) ∈ 𝐸and then𝑓 (𝑥) = ℎ(𝑥) = ℎ(𝑦) = 𝑓 (𝑦)thus𝑥 = 𝑦since𝑓is injective.

Orℎ(𝑥) = ℎ(𝑦) ∈ 𝐹 ⧵ 𝐸and then𝑔(𝑥 − 𝑟) = ℎ(𝑥) = ℎ(𝑦) = 𝑔(𝑦 − 𝑟)thus𝑥 − 𝑟 = 𝑦 − 𝑟since𝑔is injective, hence𝑥 = 𝑦.

• ℎis surjective: let𝑦 ∈ 𝐹.

Either𝑦 ∈ 𝐸, and then there exists𝑥 ∈ {0, 1, … , 𝑟 − 1}such that𝑓 (𝑥) = 𝑦, since𝑓is surjective. Thenℎ(𝑥) = 𝑓 (𝑥) = 𝑦. Or𝑦 ∈ 𝐹 ⧵ 𝐸, and then there exists𝑥 ∈ {0, 1, … , 𝑠 − 1}such that𝑔(𝑥) = 𝑦since𝑔is surjective. Thenℎ(𝑥 + 𝑟) = 𝑔(𝑥) = 𝑦.

Therefore|𝐹 | = 𝑟 + 𝑠 = |𝐸| + |𝐹 ⧵ 𝐸|. ■

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Finite sets – 5

Proposition

Let𝐸 ⊂ 𝐹 with𝐹 finite. Then|𝐹 | = |𝐸| + |𝐹 ⧵ 𝐸|.

Proof.Since𝐹 ⧵ 𝐸 ⊂ 𝐹 and𝐸 ⊂ 𝐹, we know that𝐸and𝐹 ⧵ 𝐸are finite.

Denote𝑟 = |𝐸|and𝑠 = |𝐹 ⧵ 𝐸|.

There exist bijections𝑓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑟} → 𝐸and𝑔 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑠} → 𝐹 ⧵ 𝐸.

Defineℎ ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑟 + 𝑠} → 𝐹 byℎ(𝑘) = {

𝑓 (𝑘) if𝑘 < 𝑟 𝑔(𝑘 − 𝑟) if𝑘 ≥ 𝑟 .

• ℎis well-defined:

Indeed, if0 ≤ 𝑘 < 𝑟then𝑓 (𝑘)is well-defined and𝑓 (𝑘) ∈ 𝐸 ⊂ 𝐹.

If𝑟 ≤ 𝑘 < 𝑟 + 𝑠then0 ≤ 𝑘 − 𝑟 < 𝑠so that𝑔(𝑘 − 𝑟)is well-defined and𝑔(𝑘 − 𝑟) ∈ 𝐹 ⧵ 𝐸 ⊂ 𝐹.

• ℎis a bijection:

• ℎis injective: let𝑥, 𝑦 ∈ {0, 1, … , 𝑟 + 𝑠 − 1}be such thatℎ(𝑥) = ℎ(𝑦).

Eitherℎ(𝑥) = ℎ(𝑦) ∈ 𝐸and then𝑓 (𝑥) = ℎ(𝑥) = ℎ(𝑦) = 𝑓 (𝑦)thus𝑥 = 𝑦since𝑓is injective.

Orℎ(𝑥) = ℎ(𝑦) ∈ 𝐹 ⧵ 𝐸and then𝑔(𝑥 − 𝑟) = ℎ(𝑥) = ℎ(𝑦) = 𝑔(𝑦 − 𝑟)thus𝑥 − 𝑟 = 𝑦 − 𝑟since𝑔is injective, hence𝑥 = 𝑦.

• ℎis surjective: let𝑦 ∈ 𝐹.

Either𝑦 ∈ 𝐸, and then there exists𝑥 ∈ {0, 1, … , 𝑟 − 1}such that𝑓 (𝑥) = 𝑦, since𝑓is surjective. Thenℎ(𝑥) = 𝑓 (𝑥) = 𝑦.

Or𝑦 ∈ 𝐹 ⧵ 𝐸, and then there exists𝑥 ∈ {0, 1, … , 𝑠 − 1}such that𝑔(𝑥) = 𝑦since𝑔is surjective. Thenℎ(𝑥 + 𝑟) = 𝑔(𝑥) = 𝑦.

Therefore|𝐹 | = 𝑟 + 𝑠 = |𝐸| + |𝐹 ⧵ 𝐸|. ■

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Finite sets – 6

Proposition

Let𝐸and𝐹 be two finite sets. Then

1 |𝐸 ∪ 𝐹 | = |𝐸| + |𝐹 | − |𝐸 ∩ 𝐹 |

2 |𝐸 × 𝐹 | = |𝐸| × |𝐹 |

Proof.

1 Using the previous proposition twice, we get

|𝐸 ∪ 𝐹 | = |𝐸 ⊔ (𝐹 ⧵ (𝐸 ∩ 𝐹 ))| = |𝐸| + |𝐹 ⧵ (𝐸 ∩ 𝐹 )| = |𝐸| + |𝐹 | − |𝐸 ∩ 𝐹 |

2 We prove this proposition by induction on𝑛 = |𝐹 | ∈ ℕ.

• Base case at𝑛 = 0:then𝐹 = ∅so𝐸 × 𝐹 = ∅too and|𝐸 × 𝐹 | = 0 = |𝐸| × 0 = |𝐸| × |𝐹 |.

• Case𝑛 = 1:we will use this special case later in the proof.

Assume that𝐹 = {∗}and that|𝐸| = 𝑛. Then there exists a bijection𝑓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → 𝐸. Note that𝑔 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → 𝐸 × 𝐹defined by𝑔(𝑘) = (𝑓 (𝑘), ∗)is a bijection.

Therefore|𝐸 × 𝐹 | = 𝑛 = 𝑛 × 1 = |𝐸| × |𝐹 |.

• Induction step.Assume that the statement holds for some𝑛 ∈ ℕ. Let𝐹be a set such that|𝐹 | = 𝑛 + 1.

Since|𝐹 | > 0, there exists𝑥 ∈ 𝐹and|𝐹 ⧵ {𝑥}| = |𝐹 | − |{𝑥}| = 𝑛 + 1 − 1 = 𝑛. Then

|𝐸 × 𝐹 | = |(𝐸 × (𝐹 ⧵ {𝑥})) ⊔ (𝐸 × {𝑥})| = |𝐸 × (𝐹 ⧵ {𝑥})| + |𝐸 × {𝑥}|

= |𝐸| × |𝐹 ⧵ {𝑥}| + |𝐸|using the induction hypothesis and the case𝑛 = 1

= |𝐸| × (|𝐹 | − 1) + |𝐸| = |𝐸| × |𝐹 | ■

(14)

Finite sets – 6

Proposition

Let𝐸and𝐹 be two finite sets. Then

1 |𝐸 ∪ 𝐹 | = |𝐸| + |𝐹 | − |𝐸 ∩ 𝐹 |

2 |𝐸 × 𝐹 | = |𝐸| × |𝐹 |

Proof.

1 Using the previous proposition twice, we get

|𝐸 ∪ 𝐹 | = |𝐸 ⊔ (𝐹 ⧵ (𝐸 ∩ 𝐹 ))| = |𝐸| + |𝐹 ⧵ (𝐸 ∩ 𝐹 )| = |𝐸| + |𝐹 | − |𝐸 ∩ 𝐹 |

2 We prove this proposition by induction on𝑛 = |𝐹 | ∈ ℕ.

• Base case at𝑛 = 0:then𝐹 = ∅so𝐸 × 𝐹 = ∅too and|𝐸 × 𝐹 | = 0 = |𝐸| × 0 = |𝐸| × |𝐹 |.

• Case𝑛 = 1:we will use this special case later in the proof.

Assume that𝐹 = {∗}and that|𝐸| = 𝑛. Then there exists a bijection𝑓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → 𝐸.

Note that𝑔 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → 𝐸 × 𝐹defined by𝑔(𝑘) = (𝑓 (𝑘), ∗)is a bijection.

Therefore|𝐸 × 𝐹 | = 𝑛 = 𝑛 × 1 = |𝐸| × |𝐹 |.

Let𝐹be a set such that|𝐹 | = 𝑛 + 1.

Since|𝐹 | > 0, there exists𝑥 ∈ 𝐹and|𝐹 ⧵ {𝑥}| = |𝐹 | − |{𝑥}| = 𝑛 + 1 − 1 = 𝑛. Then

|𝐸 × 𝐹 | = |(𝐸 × (𝐹 ⧵ {𝑥})) ⊔ (𝐸 × {𝑥})| = |𝐸 × (𝐹 ⧵ {𝑥})| + |𝐸 × {𝑥}|

= |𝐸| × |𝐹 ⧵ {𝑥}| + |𝐸|using the induction hypothesis and the case𝑛 = 1

= |𝐸| × (|𝐹 | − 1) + |𝐸| = |𝐸| × |𝐹 | ■

(15)

Finite sets – 7

Proposition

Assume that𝐸 ⊂ 𝐹 with𝐹 finite. Then𝐸 = 𝐹 ⇔ |𝐸| = |𝐹 |.

Proof.

⇒It is obvious.

⇐Assume that|𝐸| = |𝐹 |. Then|𝐹 ⧵ 𝐸| = |𝐹 | − |𝐸| = 0. Thus𝐹 ⧵ 𝐸 = ∅, i.e. 𝐸 = 𝐹. ■

Proposition

Let𝐸a finite set. Then𝐹 is finite and|𝐸| = |𝐹 |if and only if there exists a bijection𝑓 ∶ 𝐸 → 𝐹. Proof.

⇒Assume that𝐹 is finite and that|𝐸| = |𝐹 | = 𝑛.

Then there exist bijections𝜑 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → 𝐸 and𝜓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → 𝐹. Therefore𝑓 = 𝜓 ∘ 𝜑⁻¹ ∶ 𝐸 → 𝐹 is a bijection.

⇐Assume that there exists a bijection𝑓 ∶ 𝐸 → 𝐹.

Since𝐸is finite there exists a bijection𝜑 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < |𝐸|} → 𝐸.

Thus𝑓 ∘ 𝜑 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < |𝐸|} → 𝐹 is a bijection. Therefore𝐹 is finite and|𝐹 | = |𝐸|. ■

(16)

Finite sets – 7

Proposition

Proof.

⇒It is obvious.

Proposition

(17)

Finite sets – 7

Proposition

Proof.

⇒It is obvious.

Proposition

Let𝐸a finite set. Then𝐹 is finite and|𝐸| = |𝐹 |if and only if there exists a bijection𝑓 ∶ 𝐸 → 𝐹.

Proof.

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Finite sets – 7

Proposition

Proof.

⇒It is obvious.

Proposition

Then there exist bijections𝜑 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → 𝐸and𝜓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < 𝑛} → 𝐹. Therefore𝑓 = 𝜓 ∘ 𝜑⁻¹ ∶ 𝐸 → 𝐹 is a bijection.

(19)

Finite sets – 8

Proposition

Let𝐸, 𝐹 be two finite sets such that|𝐸| = |𝐹 |. Let𝑓 ∶ 𝐸 → 𝐹. Then TFAE:

1 𝑓 is injective,

2 𝑓 is surjective,

3 𝑓 is bijective.

Proof.

Assume that𝑓is injective.

There exists a bijection𝜑 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < |𝐸|} → 𝐸.

Then𝑓 ∘ 𝜑 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < |𝐸|} → 𝑓 (𝐸)is a bijection. Thus|𝑓 (𝐸)| = |𝐸| = |𝐹 |. Since𝑓 (𝐸) ⊂ 𝐹 and|𝑓 (𝐸)| = |𝐹 |, we get𝑓 (𝐸) = 𝐹, i.e.𝑓 is surjective. Assume that𝑓is surjective.

Then for every𝑦 ∈ 𝐹,𝑓⁻¹(𝑦) ⊂ 𝐸is finite and non-empty, i.e.|𝑓⁻¹(𝑦)| ≥ 1. Assume by contradiction that there exists𝑦 ∈ 𝐹 such that|𝑓⁻¹(𝑦)| > 1. Thus|𝐸| =

|⨆_𝑦∈𝐹𝑓⁻¹(𝑦)

|=∑

𝑦∈𝐹

|𝑓⁻¹(𝑦)| > |𝐹 | = |𝐸|. Hence a contradiction.

■

(20)

Finite sets – 8

Proposition

Let𝐸, 𝐹 be two finite sets such that|𝐸| = |𝐹 |. Let𝑓 ∶ 𝐸 → 𝐹. Then TFAE:

1 𝑓 is injective,

2 𝑓 is surjective,

3 𝑓 is bijective.

Proof.

Assume that𝑓is injective.

There exists a bijection𝜑 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < |𝐸|} → 𝐸.

Then𝑓 ∘ 𝜑 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < |𝐸|} → 𝑓 (𝐸)is a bijection. Thus|𝑓 (𝐸)| = |𝐸| = |𝐹 |.

Since𝑓 (𝐸) ⊂ 𝐹 and|𝑓 (𝐸)| = |𝐹 |, we get𝑓 (𝐸) = 𝐹, i.e.𝑓 is surjective.

Assume that𝑓is surjective.

Then for every𝑦 ∈ 𝐹,𝑓⁻¹(𝑦) ⊂ 𝐸is finite and non-empty, i.e.|𝑓⁻¹(𝑦)| ≥ 1. Assume by contradiction that there exists𝑦 ∈ 𝐹 such that|𝑓⁻¹(𝑦)| > 1.

Thus|𝐸| =

|⨆_𝑦∈𝐹𝑓⁻¹(𝑦)

|=∑

𝑦∈𝐹

|𝑓⁻¹(𝑦)| > |𝐹 | = |𝐸|. Hence a contradiction.

■

(21)

Finite sets – 9

Proposition

Let𝐸and𝐹 be two finite sets. Then|𝐸| ≤ |𝐹 |if and only if there exists an injection𝑓 ∶ 𝐸 → 𝐹.

Proof.

⇒Assume that|𝐸| ≤ |𝐹 |.

There exist bijections𝜑 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < |𝐸|} → 𝐸and𝜓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < |𝐹 |} → 𝐹. Since|𝐸| ≤ |𝐹 |,𝑓 = 𝜓 ∘ 𝜑⁻¹∶ 𝐸 → 𝐹 is well-defined and injective.

⇒Assume that there exists an injection𝑓 ∶ 𝐸 → 𝐹.

Then𝑓 induces a bijection𝑓 ∶ 𝐸 → 𝑓 (𝐸), so that|𝐸| = |𝑓 (𝐸)|.

And since𝑓 (𝐸) ⊂ 𝐹, we have|𝑓 (𝐸)| ≤ |𝐹 |. ■

Corollary: the pigeonhole principle or Dirichlet’s drawer principle

Let𝐸and𝐹 be two finite sets. If|𝐸| > |𝐹 |then there is no injective function𝐸 → 𝐹.

Examples

• There are two non-bald people in Toronto with the exact same number of hairs on their heads.

• During a post-covid party with𝑛 > 1participants, we may always find two people who shook hands to the same number of people.

(22)

Finite sets – 9

Proposition

Let𝐸and𝐹 be two finite sets. Then|𝐸| ≤ |𝐹 |if and only if there exists an injection𝑓 ∶ 𝐸 → 𝐹. Proof.

Then𝑓induces a bijection𝑓 ∶ 𝐸 → 𝑓 (𝐸), so that|𝐸| = |𝑓 (𝐸)|.

Examples

(23)

Finite sets – 9

Proposition

Examples

(24)

Finite sets – 9

Proposition

Examples

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Finite sets – 10

Remark: trichotomy principle for finite sets

Since the cardinal of a finite set is a natural number, we deduce from the fact thatℕis totally ordered, that given two finite sets𝐸and𝐹, exactly one of the followings occurs:

• either|𝐸| < |𝐹 |

i.e. there is an injection𝐸 → 𝐹 but no bijection𝐸 → 𝐹,

• or|𝐸| = |𝐹 |

i.e. there is a bijection𝐸 → 𝐹,

• or|𝐸| > |𝐹 |

i.e. there is an injection𝐹 → 𝐸but no bijection𝐸 → 𝐹.