Characterization of the supremum

(1)

MAT246H1-S – LEC0201/9201

Concepts in Abstract Mathematics

REAL NUMBERS – 2

March 16^th, 2021

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Mar 16, 2021 1 / 5

(2)

Characterization of the supremum

Proposition

Let𝐴 ⊂ ℝand𝑀 ∈ ℝ. Then𝑀 =sup(𝐴) ⇔ {

∀𝑥 ∈ 𝐴, 𝑥 ≤ 𝑀

∀𝜀 > 0, ∃𝑥 ∈ 𝐴, 𝑀 − 𝜀 < 𝑥

𝐴 ℝ

𝑀 𝑀 − 𝜀

𝜀

𝑥

Proof.

⇒Assume that𝑀 =sup(𝐴). Then𝑀is an upper bound of𝐴so∀𝑥 ∈ 𝐴, 𝑥 ≤ 𝑆.

We know that if𝑇is an other upper bound of𝐴then𝑀 ≤ 𝑇(since𝑀is the least upper bound).

So, by taking the contrapositive, if𝑇 < 𝑀then𝑇isn’t an upper bound of𝐴.

Let𝜀 > 0. Since𝑀 − 𝜀 < 𝑀, we know that𝑀 − 𝜀is not an upper bound of𝐴, meaning that there exists𝑥 ∈ 𝐴such that𝑀 − 𝜀 < 𝑥.

⇐Assume that {

∀𝑥 ∈ 𝐴, 𝑥 ≤ 𝑀

∀𝜀 > 0, ∃𝑥 ∈ 𝐴, 𝑀 − 𝜀 < 𝑥

Then, by the first condition,𝑀is an upper bound of𝐴. Let’s prove it is the least one.

We will show the contrapositive: if𝑇 < 𝑀then𝑇isn’t an upper bound of𝐴.

Let𝑇 ∈ ℝ. Assume that𝑇 < 𝑀. Set𝜀 = 𝑀 − 𝑇 > 0. Then there exists𝑥 ∈ 𝐴such that𝑀 − 𝜀 < 𝑥, i.e.𝑇 < 𝑥.

Hence𝑇isn’t an upper bound of𝐴. ■

(3)

Characterization of the infimum

Proposition

Let𝐴 ⊂ ℝand𝑚 ∈ ℝ. Then

𝑚 =inf(𝐴) ⇔ {

∀𝑥 ∈ 𝐴, 𝑚 ≤ 𝑥

∀𝜀 > 0, ∃𝑥 ∈ 𝐴, 𝑥 < 𝑚 + 𝜀

𝐴 ℝ

𝑚 𝑚 + 𝜀

𝜀

𝑥

(4)

ℝ is archimedean

Theorem: ℝ is archimedean

∀𝜀 > 0, ∀𝐴 > 0, ∃𝑛 ∈ ℕ, 𝑛𝜀 > 𝐴 Proof.

Let𝜀 > 0and𝐴 > 0.

Assume by contradiction that∀𝑛 ∈ ℕ, 𝑛𝜀 ≤ 𝐴.

Then𝐸 = {𝑛𝜀 ∶ 𝑛 ∈ ℕ}is non-empty and bounded from above so it admits a supremum 𝑀 =sup𝐸by the least upper bound principle.

Since𝑀 − 𝜀 < 𝑀,𝑀 − 𝜀is not an upper bound of𝐸, so there exists𝑛 ∈ ℕsuch that𝑛𝜀 > 𝑀 − 𝜀.

Therefore(𝑛 + 1)𝜀 > 𝑀, hence a contradiction. ■

The above theorem means that lim

𝑛→+∞

1

𝑛 = 0, or equivalently thatℝdoesn’t contain infinitesimal elements (i.e. there is not infinitely large or infinitely small elements).

(5)

Floor function

Proposition: floor function

For every𝑥 ∈ ℝ, there exists a unique𝑛 ∈ ℤsuch that𝑛 ≤ 𝑥 < 𝑛 + 1.

We say that𝑛is the integer part (or the floor function value) of𝑥and we denote it by⌊𝑥⌋.

Proof.Let𝑥 ∈ ℝ.

Existence.

• First case: if𝑥 ≥ 0.

We set𝐸 = {𝑛 ∈ ℕ ∶ 𝑥 < 𝑛}.

By the archimedean property (with𝜀 = 1), there exists𝑚 ∈ ℕsuch that𝑚 > 𝑥. Hence𝐸 ≠ ∅.

By the well-ordering principle,𝐸admits a least element𝑝.

We have that𝑥 < 𝑝since𝑝 ∈ 𝐸and that𝑝 − 1 ≤ 𝑥since𝑝 − 1 ∉ 𝐸.

Therefore𝑛 = 𝑝 − 1satisfies𝑛 ≤ 𝑥 < 𝑛 + 1.

• Second case: if𝑥 < 0.Then we apply the first case to−𝑥.

Uniqueness.Assume that𝑛, 𝑛^′∈ ℤare two suitable integers. Then(1) 𝑛 ≤ 𝑥 < 𝑛 + 1and𝑛^′≤ 𝑥 < 𝑛^′+ 1.

We deduce from the last inequality that(2) − 𝑛^′− 1 < −𝑥 ≤ −𝑛^′. Summing(1)and(2), we get that𝑛 − 𝑛^′− 1 < 0 < 𝑛 − 𝑛^′+ 1.

Hence𝑛 − 𝑛^′< 1, i.e.𝑛 − 𝑛^′≤ 0, and−1 < 𝑛 − 𝑛^′i.e.0 ≤ 𝑛 − 𝑛^′.

Therefore𝑛 = 𝑛^′. ■