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4.1Source-backedEstimated reading time: 4 min

4.1 Order, bounds, and completeness

Use total order and least-upper-bound language to explain precisely why R is complete and Q is not.

Note collections

MATH1090: Set theory

Rigorous course notes on logic, sets, and number construction, written in short linked sections with careful proofs and examples.

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All note collectionsSeries overview

Chapter 1

Logic

Reasoning tools for statements, connectives, and quantifiers.

1.1

1.1 Propositional logic

1.2

1.2 Truth tables and equivalence

1.3

1.3 Quantifiers and negation

Chapter 2

Sets and relations

Basic set language, functions, and relations.

2.1

2.1 Sets and set operations

2.2

2.2 Functions and relations

Chapter 3

Numbers by construction

How natural numbers, integers, and rationals are built, and where Q still falls short.

3.1

3.1 Natural numbers and Peano axioms

3.2

3.2 Induction and recursive arithmetic

3.3

3.3 Integers from equivalence classes

3.4

3.4 Rationals and well-defined operations

3.5

3.5 Gaps in Q and why sqrt(2) is not rational

Chapter 4

Order and completeness

Total order, bounds, supremum and infimum, and the completeness gap between Q and R.

4.1

4.1 Order, bounds, and completeness

Why this chapter is the turning point

Earlier chapters built number systems by construction. This section changes the question: not only what numbers are, but what order structure those numbers support. The concepts of upper bounds, lower bounds, supremum, and infimum are not vocabulary decoration. They are the technical bridge from algebraic manipulation to analysis.

Ordered sets and total order

Definition

Total order

A set SS with relation ≤\le is totally ordered if for all a,b,c∈Sa,b,c\in S:

  1. a≤aa\le a (reflexive),
  2. a≤ba\le b and b≤ab\le a imply a=ba=b (antisymmetric),
  3. a≤ba\le b and b≤cb\le c imply a≤ca\le c (transitive),
  4. either a≤ba\le b or b≤ab\le a (comparability).

Comparability is the extra ingredient that separates total order from partial order. In RR, any two real numbers can be compared.

Upper/lower bounds and bounded sets

Definition

Upper bound and lower bound

Let A⊆RA\subseteq R.

  • u is an upper bound of AA if x≤ux\le u for all x∈Ax\in A.
  • ℓ\ell is a lower bound of AA if ℓ≤x\ell\le x for all x∈Ax\in A.

If upper bounds exist, AA is bounded above. If lower bounds exist, it is bounded below.

Worked example

A set with many upper bounds

Take A=(0,1)A=(0,1). Every number u≥1u\ge 1 is an upper bound. The value 1 is special because it is the smallest upper bound.

Common mistake

Maximum is not the same as supremum

Students often think a supremum must belong to the set. Not true. For A=(0,1)A=(0,1), sup⁡A=1\sup A=1, but 1∉A1\notin A.

Supremum and infimum

Definition

Supremum and infimum

Let A⊆RA\subseteq R be nonempty.

  • s=sup⁡As=\sup A means: (i) s is an upper bound of AA; (ii) any upper bound u satisfies s≤us\le u.
  • t=inf⁡At=\inf A means: (i) t is a lower bound of AA; (ii) any lower bound ℓ\ell satisfies ℓ≤t\ell\le t.

Equivalent approximation property for supremum:

s=sup⁡As=\sup A iff s is an upper bound and for every ε>0\varepsilon>0 there exists a∈Aa\in A with s−ε<a≤ss-\varepsilon<a\le s.

This turns order language into a usable proof tool.

Proof

Why the approximation property is equivalent

Completeness axiom

Theorem

Least-upper-bound property of R

Every nonempty subset A⊆RA\subseteq R that is bounded above has a supremum in RR.

This is what “RR is complete” means in order language. It is not true in QQ.

Worked example

A rational set without rational supremum

Let

A={q∈Q:q>0, q2<2}.A=\{q\in Q : q>0,\ q^2<2\}.

Inside RR, sup⁡A=2\sup A=\sqrt2. But 2∉Q\sqrt2\notin Q, so AA has no supremum in QQ. Therefore QQ fails the least-upper-bound property.

How this controls future analysis

  • Monotone convergence arguments use bounded monotone sequences and supremum.
  • Continuity proofs use interval and order completeness repeatedly.
  • Existence theorems are often hidden completeness arguments.

So this chapter is foundational, not optional.

Quick checks

Quick check

If A has a maximum m, what is sup A?

Use the definition directly.

Solution

Answer

Quick check

Can a nonempty set be bounded above but not bounded below? Give one example.

Think of half-lines.

Solution

Answer

Exercises

Quick check

Find sup and inf of A={1-1/n : n in N}. Does A have a maximum?

Write first few terms and identify the limit behavior.

Solution

Guided solution

Quick check

Show that if B is nonempty and bounded below, then inf B = -sup(-B).

Translate lower-bound statements into upper-bound statements for −B-B.

Solution

Guided solution

Prerequisite links

Read this after 3.4 Rationals and well-defined operations and 3.5 Gaps in Q and sqrt(2).

Prerequisites

3.5

3.5 Gaps in Q and why sqrt(2) is not rational

Key terms in this unit

More notes in this series

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3.5 Gaps in Q and why sqrt(2) is not rational

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