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real analysis 3

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REAL ANALYSIS
Part III:- by Lathika Wathsara
Overview: Completeness axioms
Definitions: 𝐿𝑒𝑑 𝐴 ⊆ ℝ
1. We say that 𝐴 is bounded above, if ∃ 𝑼 ∈ ℝ s.t,
∀π‘₯πœ–π΄ π‘₯ ≤ π‘ˆ
U is called an upper bound of A.
2. We say that 𝐴 is bounded below, if ∃ 𝑉 ∈ ℝ s.t,
∀π‘₯πœ–π΄ π‘₯ ≥ 𝑉
V is called a lower bound of A.
3. If A is both bounded above and below, we say that A is bounded.
Ex:I.
II.
III.
(-∞ , 2)
(4,89)
(4, ∞)
Definitions: a) Let 𝐴 ⊆ ℝ which is bounded above, suppose ∃ π‘ˆ ∈ ℝ with following
properties,
I. U is an upper bound of A.
II. If u is any upper bound of A , and U≤u (lowest upper bound)
Then, we say that U is the supremum of A. U=Sup(A).
b) Let 𝐴 ⊆ ℝ which is bounded below, suppose ∃ 𝐿 ∈ ℝ with following
properties,
III. L is a lower bound of A.
IV. If l is any lower bound of A, and L ≥ l, ( greatest lower bound)
Then, we say that L is the infimum of A.
L=Inf(A).
Definitions: Let 𝐴 ⊆ ℝ,
a. Max(A) → maximum of A.
Highest element in A.
b. Min(A) → minimum of A.
Lowest element in A.
max(A)∈ 𝐴.
min(A) ∈ 𝐴.
Explain: Upper bound, lower bound, maximum, minimum of
1)
2)
3)
4)
(2,5)
[2,5)
(9,12]
(4, ∞)
Completeness Property 1:• Every non-empty subset of ℝ which is bounded above has a
supremum in ℝ.
Completeness Property 2:• Every non-empty subset of ℝ which is bounded below has a infimum
in ℝ.
Theorem: I.
II.
U is an upper bound of A.
For all πœ€ ≥ 0,
∃ 𝒂 ∈ 𝐴 𝑠. 𝑑 , π‘Ž > 𝑼 − πœ€
⟺
Sup(A) = U
⟺
Inf(A) = L
Theorem: III.
IV.
L is an lower bound of A.
For all πœ€ ≥ 0,
∃ 𝒂 ∈ 𝐴 𝑠. 𝑑 , π‘Ž < 𝑳 + πœ€
Exercise:
1.
2.
3.
4.
5.
6.
Prove that sup (π‘Ž, 𝑏) = 𝑏 and inf (π‘Ž, 𝑏) = π‘Ž.
Which of the following sets have the completeness axiom property β„€, β„š, β„šπ‘
Show that β„€ is unbounded.
Show that for every π‘Ž ∈ ℝ there is 𝑛 ∈ β„€ such that 𝑛 > π‘Ž
Show that for given π‘Ž, 𝑏 ∈ ℝ with 𝑏 > π‘Ž, there exists 𝑛 ∈ β„€ such that π‘›π‘Ž > 𝑏
Show that for π‘Ž, 𝑏 ∈ β„€+ such that π‘Ž < 𝑏, there exists unique π‘₯, 𝑦 ∈ β„€+ such that 𝑏 =
π‘₯π‘Ž + 𝑦 with 0 ≤ 𝑦 < π‘Ž
7. Prove the existence of 𝑖𝑛𝑓 using the existence of 𝑠𝑒𝑝 with suitable conditions.
8. Suppose we have ∀π‘Ž ∈ 𝐴, ∀𝑏 ∈ 𝐡; π‘Ž < 𝑏. Show that sup 𝐴 ≤ sup 𝐡. (A,B are nonempty subsets of ℝ)
9. Show that ∀πœ– > 0, ∃π‘Ž ∈ 𝐴; π‘Ž + πœ– > sup 𝐴
10. Show that ∀πœ– > 0, ∃π‘Ž ∈ 𝐴; π‘Ž − πœ– < inf 𝐴
11. Show that if ∃π‘Ž, ∀πœ– > 0; 0 ≤ π‘Ž < πœ– then π‘Ž = 0
12. Define 𝐴 + 𝐡 = {π‘Ž + 𝑏|π‘Ž ∈ 𝐴, 𝑏 ∈ 𝐡}. Show that sup (𝐴 + 𝐡) = sup 𝐴 + sup 𝐡
13. Show that there is a rational number and an irrational number between any two
real numbers.
From 8th:
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