MAT246H1-S – LEC0201/9201
Concepts in Abstract Mathematics
REAL NUMBERS – 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. ■
Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Mar 16, 2021 2 / 5
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.
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
Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Mar 16, 2021 3 / 5
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).
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).
Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Mar 16, 2021 4 / 5
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
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′. ■
Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Mar 16, 2021 5 / 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′.
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