MAT158H1/Unit 02

Unit 02 · Analysis 1A

The Real Numbers & Least Upper Bounds

What separates $\mathbb{R}$ from $\mathbb{Q}$ is completeness — no gaps. The least upper bound axiom is the single foundation every deep theorem of analysis stands on. This unit is where the course stops taking the real numbers for granted.

By the end you can

• State the ordered-field axioms of $\mathbb{R}$ and what they cover.
• Define upper bound, supremum, and infimum, and compute them.
• State the Least Upper Bound axiom and use the ε-characterization of sup.
• Prove the Archimedean property and the density of $\mathbb{Q}$.

Why it matters & where it comes from

Why it exists

$\mathbb{Q}$ has holes — there is no rational whose square is $2$. Analysis needs a number line with no gaps so that limits actually land somewhere. Completeness is exactly that guarantee.

In the real world

Floating-point arithmetic and numerical analysis constantly wrestle with the gap between true real numbers and what a computer can store. Understanding completeness clarifies why rounding errors and convergence behave as they do.

Story & open questions

Dedekind (with "cuts") and Cantor (with Cauchy sequences) independently constructed $\mathbb{R}$ in the 1870s — finally giving a precise definition to the objects calculus had used for two hundred years.

What the axioms say

$\mathbb{R}$ is an ordered field: the field axioms govern $+$ and $\cdot$ (associativity, commutativity, identities $0,1$, inverses, distributivity), and the order axioms make $<$ compatible with them ($a<b\Rightarrow a+c<b+c$; and $a<b$ with $c>0$ gives $ac<bc$). Everything from high-school algebra follows from these. But $\mathbb{Q}$ satisfies all of them too — so one more axiom is needed to tell the two apart.

Bounds and the supremum

Definition Supremum & infimum

$M$ is an upper bound of $S$ if $x\le M$ for all $x\in S$. The supremum $\sup S$ is the least upper bound: an upper bound that is $\le$ every other upper bound. The infimum $\inf S$ is the greatest lower bound.

Example

$S=\{1-\tfrac1n:n\in\mathbb{N}\}$ has $\sup S=1$, even though $1\notin S$. The sup need not be attained; when it is, it is called the maximum.

Theorem ε-characterization

Let $s$ be an upper bound of $S$. Then $s=\sup S$ if and only if for every $\varepsilon>0$ there is some $x\in S$ with $x>s-\varepsilon$.

Proof

If $s=\sup S$ and some $\varepsilon>0$ had no such $x$, then $s-\varepsilon$ would be an upper bound smaller than the least one — impossible.

Conversely, if every $s-\varepsilon$ fails to be an upper bound, no number below $s$ is an upper bound, so $s$ is the least one. ∎

The completeness axiom

Axiom Least Upper Bound

Every nonempty subset of $\mathbb{R}$ that is bounded above has a supremum in $\mathbb{R}$.

This is exactly what $\mathbb{Q}$ lacks: $\{x\in\mathbb{Q}:x^2<2\}$ is nonempty and bounded above, but has no least upper bound inside $\mathbb{Q}$ — its upper bounds pile up against the hole where $\sqrt2$ should be. The axiom is the engine behind the Intermediate Value Theorem, the Extreme Value Theorem, and the convergence of monotone sequences.

Example $\sqrt2$ exists in $\mathbb{R}$

Let $s=\sup\{x>0:x^2<2\}$ — the set contains $1$ and is bounded above by $2$, so $s$ exists by completeness.

If $s^2<2$: for small $h\in(0,1)$, $(s+h)^2=s^2+2sh+h^2<s^2+(2s+1)h$, which is $<2$ once $h<\tfrac{2-s^2}{2s+1}$ — so $s+h$ is in the set, contradicting "upper bound."

If $s^2>2$: similarly $(s-h)^2>2$ for small $h>0$, making $s-h$ a smaller upper bound, contradicting "least." So $s^2=2$. ∎

Sup versus max A maximum is a supremum that belongs to the set: $\sup(0,1)=1$ with no maximum, while $\sup[0,1]=1=\max$. To prove $s=\sup S$, always prove two things — $s$ is an upper bound, and nothing smaller is.

Two consequences you will use constantly

Theorem Archimedean property

For any real $x$ there is $n\in\mathbb{N}$ with $n>x$. Equivalently: for any $\varepsilon>0$ there is $n$ with $\tfrac1n<\varepsilon$.

Proof

If not, $\mathbb{N}$ is bounded above, so $s=\sup\mathbb{N}$ exists. Then $s-1$ is not an upper bound, so some $n\in\mathbb{N}$ has $n>s-1$.

But then $n+1>s$ with $n+1\in\mathbb{N}$ — contradicting that $s$ is an upper bound of $\mathbb{N}$. ∎

Theorem Density of $\mathbb{Q}$

Between any two reals $a<b$ there is a rational number.

Proof idea

Pick $q\in\mathbb{N}$ with $\tfrac1q<b-a$ (Archimedean property), and let $p$ be the smallest integer with $p>qa$.

Then $qa<p\le qa+1$, so $a<\tfrac pq\le a+\tfrac1q<b$. ∎ (The irrationals are dense too — shift this argument by $\sqrt2$.)

Common pitfalls

• The supremum may not belong to the set; "max" requires it to be attained.
• Completeness says the sup exists as a real number — existence is the whole content of the axiom.
• A set with no upper bound has no supremum; writing $\sup S=\infty$ is shorthand, not a real number.
• To show $s=\sup S$, both halves are needed: upper bound, and least such.

Practice

Original problems written for this guide in the style of Spivak's exercises. Write a full solution before opening the answer.

Practice 1 · Compute

Find $\sup$ and $\inf$ of $S=\{\tfrac1n:n\in\mathbb{N}\}$ and say which are attained.

Answer

$\sup S=1$, attained at $n=1$; $\inf S=0$, not attained.

Practice 2 · ε-characterization

Prove: if $s=\sup S$ then for every $\varepsilon>0$ there is $x\in S$ with $x>s-\varepsilon$.

Solution

Otherwise $s-\varepsilon$ would be an upper bound of $S$ smaller than $s$, contradicting that $s$ is the least upper bound.

Practice 3 · Archimedes at work

Use the Archimedean property to prove $\inf\{\tfrac1n:n\in\mathbb{N}\}=0$.

Solution

$0$ is a lower bound. For any $\varepsilon>0$, some $\tfrac1n<\varepsilon$, so no number above $0$ is a lower bound. Hence $0$ is the greatest lower bound.

Practice 4 · Compute

Find $\sup$ and $\inf$ of $S=\{\tfrac{n}{n+1}:n\in\mathbb{N}\}$.

Answer

The terms $\tfrac12,\tfrac23,\tfrac34,\dots$ increase toward $1$: $\inf S=\tfrac12$ (attained), $\sup S=1$ (not attained — $\tfrac{n}{n+1}<1$ always, yet $\tfrac{n}{n+1}>1-\varepsilon$ once $n>\tfrac1\varepsilon$).

Practice 5 · Sup of a sum

For nonempty bounded sets, define $A+B=\{a+b:a\in A,\ b\in B\}$. Prove $\sup(A+B)=\sup A+\sup B$.

Solution

Every $a+b\le\sup A+\sup B$, so it is an upper bound. Given $\varepsilon>0$, pick $a>\sup A-\tfrac\varepsilon2$ and $b>\sup B-\tfrac\varepsilon2$; then $a+b>\sup A+\sup B-\varepsilon$. The ε-characterization finishes it.

Practice 6 · Duality

For nonempty $S$ bounded below, let $-S=\{-x:x\in S\}$. Prove $\inf S=-\sup(-S)$, so completeness gives infima for free.

Solution

$m$ is a lower bound of $S$ iff $-m$ is an upper bound of $-S$. The greatest lower bound therefore matches the least upper bound with a sign flip.

Practice 7 · A sup proof

Let $S=\{x\in\mathbb{R}:x^2<3\}$. Prove $\sup S=\sqrt3$.

Solution

If $x\in S$ then $x\le|x|<\sqrt3$ (since $x^2<3$), so $\sqrt3$ is an upper bound. For any $\varepsilon\in(0,\sqrt3)$, $(\sqrt3-\varepsilon)^2<3$, so $\sqrt3-\varepsilon\in S$ — nothing below $\sqrt3$ is an upper bound.

Practice 8 · An ε-trick to memorize

Prove: if $a\le b+\varepsilon$ for every $\varepsilon>0$, then $a\le b$.

Solution

If instead $a>b$, take $\varepsilon=\tfrac{a-b}2>0$. The hypothesis gives $a\le b+\tfrac{a-b}2=\tfrac{a+b}2<a$ — a contradiction. This one-line lemma gets used all over analysis.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT158H1 topic map. It should cover the following items before you move on:

• Ordered field. Use field laws, order laws, absolute value, and inequalities inside proofs.
• Completeness. Apply the least-upper-bound property as the key axiom distinguishing $\mathbb{R}$ from $\mathbb{Q}$.
• Supremum and infimum. Prove a number is a supremum using upper-bound and approximation clauses.
• Archimedean tools. Use Archimedean property and density arguments when bounds need integers or rationals.

Textbook/source alignment

Use the current UofT MAT158H1 calendar topics as the coverage spine, and pair lecture notes with a Spivak-style proof-calculus text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

The spine is: ordered-field algebra gives inequalities, completeness gives suprema, and suprema prove the major existence theorems of calculus.

Study routine. When proving $s=\sup A$, always prove both: every $a\in A$ satisfies $a\le s$, and for every $\epsilon>0$ some $a\in A$ satisfies $s-\epsilon<a$.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Supremum

Find and prove the supremum of $A=\{1-1/n:n\in\mathbb{N}\}$.

Guide

The supremum is $1$. Every term is less than $1$, and for any $\epsilon>0$, choose $n>1/\epsilon$ so $1-1/n>1-\epsilon$.

Extension 2 · No supremum in Q

Explain why $\{q\in\mathbb{Q}:q^2<2\}$ has no supremum in $\mathbb{Q}$.

Guide

Its real supremum is $\sqrt2$, not rational. Any rational upper bound above it can be lowered, and any rational below it is not an upper bound.

Extension 3 · Density

Use the Archimedean property to show there is an integer $n$ with $n>x$.

Guide

Apply Archimedean property directly, or use contradiction: if all integers were $\le x$, then the natural numbers would be bounded, contradicting completeness consequences.

Extension 4 · Absolute value

Prove $|x+y|\le |x|+|y|$.

Guide

Square both sides or use $-|x|\le x\le |x|$ and $-|y|\le y\le |y|$, then add inequalities.

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