The Timpson Lyceum · Real Analysis (2020)

Lecture 3 (Real Analysis 2020)

Overview

This lecture has two halves. The first reviews several proof-by-induction exercises on the natural numbers and integers: closed-form sums ($\sum i^2$, $\sum i^3$) and the closure of $\mathbb{N}$ and $\mathbb{Z}$ under addition and subtraction. The recurring lesson is to phrase each claim as an "if–then" statement, define an inductive set $A$, and exploit the fact that $\mathbb{N}$ is a subset of every inductive set. The second half opens Section 1.2 and builds toward two headline results: every half-open unit interval $(c, c+1]$ contains exactly one integer, and the rationals (and irrationals) are dense in $\mathbb{R}$. The key tool unlocking all of this is the Archimedean property, proved from the completeness axiom.

Key Concepts

Inductive sets and the induction template. A set is inductive if it contains $1$ and is closed under $+1$ (i.e. $k \in A \Rightarrow k+1 \in A$). $\mathbb{N}$ is a subset of every inductive set. To prove a statement holds for all naturals, let $A$ be the set of naturals making it true, show $A$ is inductive, and conclude $A = \mathbb{N}$.

• Phrase every proof as "if $P$ then $Q$." Naming the antecedent and consequent tells you exactly what to assume and what to induct on. You must induct on a quantity that has a smallest value ($1$) — inducting on "$q = m+n$" fails because no sum of two naturals equals $1$.

Standard closed-form sums (by induction):

• $\displaystyle\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$
• $\displaystyle\sum_{i=1}^{n} i^3 = \left(\sum_{i=1}^{n} i\right)^2 = \frac{n^2(n+1)^2}{4}$. Reuse the known value of $\sum i = \frac{n(n+1)}{2}$ rather than reproving it.

Closure of $\mathbb{N}$ and $\mathbb{Z}$. Proved by induction, each reusing the previous one:

• Sum: if $m, n \in \mathbb{N}$ then $m + n \in \mathbb{N}$ (induct on $n$, using "$\mathbb{N}$ is inductive" for the step).
• Predecessor: if $n \in \mathbb{N}$ and $n > 1$, then $n - 1 \in \mathbb{N}$. Start the induction effectively at $2$ by putting $1$ into $A$ "for free": let $A = \{\,n : n = 1 \text{ or } n-1 \in \mathbb{N}\,\}$.
• Difference: if $m, n \in \mathbb{N}$ and $n > m$ then $n - m \in \mathbb{N}$.
• The integer versions ($\mathbb{Z}$ closed under $+$ and $-$) follow by splitting into sign cases (zero, both positive, mixed signs, both negative) and applying the natural-number results, since $\mathbb{Z} = \mathbb{N} \cup \{0\} \cup \{-\mathbb{N}\}$.

Starting induction "above 1." To prove a claim only from $n = 2$ (or $5$, etc.) onward, define $A$ to also contain the small excluded values outright. Then proving $A = \mathbb{N}$ still works, and the genuinely-true cases are exactly the ones you care about.

Completeness axiom. Every nonempty set of reals bounded above has a least upper bound (supremum). Unlike a maximum, the supremum need not lie in the set — e.g. $\sup(0,1) = 1 \notin (0,1)$, and $(0,1)$ has no maximum.

Archimedean property. For every real $c$ there exists $n \in \mathbb{N}$ with $n > c$; equivalently, $\mathbb{N}$ is not bounded above. Proved by contradiction: if some $c$ bounded $\mathbb{N}$, then $b = \sup\mathbb{N}$ exists, $b - \tfrac{1}{2}$ is not an upper bound, so some $n > b - \tfrac{1}{2}$, whence $n+1 > b + \tfrac{1}{2} > b$ — a natural exceeding the supremum.

• Corollary. For every $\varepsilon > 0$ there is $n \in \mathbb{N}$ with $\tfrac{1}{n} < \varepsilon$. (Apply the property to $\tfrac{1}{\varepsilon}$.) This is the workhorse for limit proofs.

Integers and intervals.

• No integer lies strictly between consecutive integers: $(n, n+1)$ contains no integer. (Reduce to "no integer lies in $(0,1)$," since the smallest positive integer is $1$.)
• A nonempty set of integers bounded above contains a maximum (its supremum lands inside it).
• Exactly one integer in $(c, c+1]$: for every real $c$ there is one and only one integer in $\{x : c < x \le c+1\}$.

Density. A set is dense in $\mathbb{R}$ if every open interval $(a,b)$ contains a point of it.

• $\mathbb{Q}$ is dense in $\mathbb{R}$.
• The irrationals are dense in $\mathbb{R}$, using that a (nonzero) rational times an irrational is irrational.

Worked Example

Claim. $\mathbb{Q}$ is dense in $\mathbb{R}$: every open interval $(a,b)$ contains a rational.

Idea: pick a step $\tfrac{1}{n}$ smaller than the gap $b-a$, scale up by $n$ so the window $(bn-1,\,bn]$ is guaranteed to capture an integer $m$, then scale back down so $\tfrac{m}{n}$ lands in $(a,b)$.

Thus the rational $\tfrac{m}{n} \in (a,b)$. For the irrationals, apply this to the interval $\left(\tfrac{a}{\sqrt{2}}, \tfrac{b}{\sqrt{2}}\right)$ to get a rational $q$ there; then $q\sqrt{2} \in (a,b)$ and is irrational.

Summary

• Frame inductive proofs as "if $P$ then $Q$," define $A = \{$naturals making it true$\}$, show $A$ is inductive, and conclude $A = \mathbb{N}$; induct on something with a least value.
• Key sums: $\sum_{i=1}^n i^2 = \tfrac{n(n+1)(2n+1)}{6}$ and $\sum_{i=1}^n i^3 = \big(\tfrac{n(n+1)}{2}\big)^2$; reuse earlier results instead of reproving.
• $\mathbb{N}$ and $\mathbb{Z}$ are closed under addition and (legal) subtraction, each proved by induction reusing the prior closure fact.
• The completeness axiom (every set bounded above has a supremum) gives the Archimedean property: $\mathbb{N}$ is unbounded, so for any $\varepsilon > 0$ some $\tfrac{1}{n} < \varepsilon$.
• Every half-open unit interval $(c, c+1]$ contains exactly one integer; a bounded set of integers has a maximum.
• $\mathbb{Q}$ and the irrationals are both dense in $\mathbb{R}$ — every open interval, no matter how small, contains each.