Lecture 5: The Archimedian Property, Density of the Rationals, and Absolute Value

Main Ideas and Concepts

• Construction/characterization of the real numbers (\mathbb{R}) (from the previous lecture):

  • There exists a unique ordered field (\mathbb{R}) containing the rationals (\mathbb{Q}) such that (\mathbb{R}) satisfies the least upper bound (supremum) property:
    • For any non-empty subset of (\mathbb{R}) that is bounded above, its supremum exists in (\mathbb{R}) (even if the supremum is not an element of the subset).
  • Example showing why (\mathbb{Q}) is insufficient:
    • The set [ {q \in \mathbb{Q} : q>0,\; q^2 < 2} ] is bounded above in (\mathbb{Q}), but has no supremum in (\mathbb{Q}), because the supremum is essentially (\sqrt{2}\notin\mathbb{Q}).

• Limits as central to analysis

  • The instructor emphasizes that limits will be the “central object of study” in analysis, working inside the real numbers.

• Density of the rationals in (\mathbb{R})

  • Key property: between any two real numbers (x<y), there exists a rational (r) such that [ x<r<y. ]

  • This will be proven using the Archimedean property.

• Archimedean property of (\mathbb{R}) (first major theorem)

  • If (X>0) and (Y\in\mathbb{R}), then there exists a natural number (N) such that [ NX>Y. ]

• Three-case strategy for density proof

  • The density proof is split into cases based on the signs of (x) and (y).
  • In the excerpt, the instructor works through case B and reduces case C to case B; case A is mentioned but not detailed.

• Supremum/infimum characterizations

  • The instructor gives an equivalent (“iff”) characterization of when a number is the supremum of a set, in terms of:
    1. Being an upper bound
    2. Being approximable from below within any (\varepsilon>0)
  • An analogous characterization is stated for infimum.

• Suprema of specific sequence sets

  • Example: for the set [ \left{1-\frac{1}{n} : n\in\mathbb{N}\right}, ]

    • the supremum is (1),
    • proved using the Archimedean property to achieve the required “within (\varepsilon)” approximation.

• Supremum rules under shifting and scaling

  • For sets (A\subseteq\mathbb{R}):
    • Shift: (X+A={X+a:a\in A})
    • Scaling: (X\cdot A={Xa:a\in A})

  • Stated and proved:

    • If (A) is bounded above, then [ \sup(X+A)=X+\sup(A). ]

    • If (X>0) and (A) is bounded above, then [ \sup(X\cdot A)=X\cdot\sup(A). ]

    • A variant for negative scaling is mentioned (inequalities flip and bounded-below conditions are needed), leading to a form involving (\inf(A)).

• Absolute value and its foundational properties

  • Definition: [ |x|=\begin{cases} x,& x\ge 0,\ -x,& x\le 0. \end{cases} ]

  • Properties proven include:

    • Nonnegativity
    • (|x|=0 \iff x=0)
    • (|-x|=|x|)
    • (|xy|=|x||y|)
    • (|x^2|=|x|^2) (presented via squares)
    • An order-related equivalence involving negatives (inequality equivalence)
    • (-|x|\le x\le |x|)

  • Triangle inequality introduced: [ |x+y|\le |x|+|y| ]

    • The proof/explanation is started but not completed in the excerpt.

---

Methods / Proof Structures

A) Proof method: Archimedean property via contradiction

Goal: If (X>0), show (\exists N\in\mathbb{N}) such that (NX>Y).

1. Assume the negation (for contradiction): For all (n\in\mathbb{N}), [ n \le \frac{Y}{X}. ]

2. Conclude boundedness: Then (\mathbb{N}\subset\mathbb{R}) is bounded above by (\frac{Y}{X}).

3. Apply least upper bound property: Let (A=\sup(\mathbb{N})), which exists in (\mathbb{R}).

4. Use the defining property of supremum: Since (A-1) is not an upper bound, there exists (M\in\mathbb{N}) such that: [ A-1<M. ] Hence: [ A<M+1. ]

5. Contradiction: But (M+1\in\mathbb{N}), so there is a natural number bigger than (A), contradicting that (A) is the least upper bound.

6. Therefore: The assumption is false, so some (N) exists with [ N>\frac{Y}{X} \quad\Longleftrightarrow\quad NX>Y. ]

---

B) Proof method: Density of rationals using Archimedean property + well-ordering (case B)

Overall goal: Given (x<y), show (\exists r\in\mathbb{Q}) such that (x<r<y).

Case B handled: (0\le x<y)

1. Use Archimedean property once (choose a “scale”): Choose (n\in\mathbb{N}) such that [ n(y-x)>1 ] equivalently, [ ny>nx+1. ]

2. Use Archimedean property again (choose an integer window): Choose (L\in\mathbb{N}) such that [ L>nx. ] This ensures the set [ S={K\in\mathbb{N}:K>nx} ] is non-empty.

3. Use well-ordering: Let (m) be the least element of (S).

4. Convert minimality into inequalities:

   • Since (m) is least with (m>nx), then (nx<m).
   • Also, (m-1\notin S), so (m-1\le nx).
   • Therefore, [ m\le nx+1. ]

5. Combine inequalities: From (nx<m\le nx+1<ny), conclude [ x<\frac{m}{n}<y. ]

6. Final choice: Set (r=\frac{m}{n}\in\mathbb{Q}). Then (x<r<y).

---

C) Reduction method: case C to case B (sign flip trick)

Case C: (x<y\le 0)

• Note that [ -y \ge -x \ge 0. ]

• Apply case B to (-y) and (-x) to obtain some (\tilde r\in\mathbb{Q}) such that: [ -y<\tilde r<-x. ]

• Multiply by (-1) to reverse inequalities: [ x<-\tilde r<y. ]

• Set (r=-\tilde r).

---

D) Supremum characterization theorem (explicit “iff” structure)

For (S\subseteq\mathbb{R}), non-empty and bounded above, the instructor states:

• [ x=\sup S ] iff both:

  1. Upper bound: (x) is an upper bound of (S) (i.e. (s\le x) for all (s\in S)).
  2. Approximation from below: For every (\varepsilon>0), there exists (y\in S) such that [ x-\varepsilon<y\le x. ]

• Analogous statement for infimum (stated, not fully proved in the excerpt): (x=\inf S) iff:

  1. (x) is a lower bound of (S)
  2. For every (\varepsilon>0), there exists (y\in S) such that [ x\le y < x+\varepsilon. ]

---

E) Method: proving a supremum of a sequence set using (\varepsilon) + Archimedean property

To show [ \sup\left{1-\frac{1}{n}\right}=1, ] use the supremum characterization:

1. Let (\varepsilon>0).

2. By Archimedean property, choose (N\in\mathbb{N}) such that [ \frac{1}{N}<\varepsilon. ]

3. Then [ -\varepsilon<-\frac{1}{N}. ]

4. Add (1) to get: [ 1-\varepsilon<1-\frac{1}{N}. ]

5. Also (1-\frac{1}{N}\le 1).

6. This verifies the required “within (\varepsilon)” condition.

---

F) Supremum shift/scaling proofs (proof template)

Template: (\sup(X+A)=X+\sup(A))

Given (A) bounded above:

1. Prove (X+\sup(A)) is an upper bound of (X+A).

2. Verify the (\varepsilon)-approximation from below:

   • For any (\varepsilon>0), since (\sup(A)) is the supremum, pick (y\in A) such that [ \sup(A)-\varepsilon<y\le \sup(A). ]

   • Add (X) to each part: [ X+\sup(A)-\varepsilon < X+y \le X+\sup(A). ]

3. Conclude via the supremum characterization.

Template: (\sup(X\cdot A)=X\sup(A)) for (X>0)

1. Show (X\sup(A)) is an upper bound of (X\cdot A).

2. For (\varepsilon>0), pick (y\in A) corresponding to (\varepsilon/X) (using (X>0)): [ \sup(A)-\frac{\varepsilon}{X}<y\le \sup(A). ]

3. Multiply through by (X): [ X\sup(A)-\varepsilon< Xy \le X\sup(A). ]

---

G) Absolute value: proof-by-case method for basic properties

Many properties are proven by splitting into cases, such as:

• (x\ge 0) vs (x\le 0)
• sign combinations for (x) and (y) when proving product-related results

Order-related equivalence (one of the properties) is handled by proving both directions using cases (x\ge 0) and (x\le 0).

The triangle inequality is introduced but not fully proved in the excerpt.

---

Speakers / Sources Featured

• Speaker: Unnamed instructor/lecturer (video narrator; no personal name given in the subtitles).