Foundations: The Real Number System & Completeness
Foundations: The Real Number System & Completeness
Real analysis is the rigorous study of the real numbers, sequences, and functions on $\mathbb{R}$. The central property distinguishing $\mathbb{R}$ from $\mathbb{Q}$ is completeness: every Cauchy sequence converges, and every non-empty bounded set has a supremum. This axiom is the bedrock of all convergence theorems, the existence of integrals, and the intermediate value theorem.
Supremum & Completeness Axiom
Let $S\subseteq\mathbb{R}$ be non-empty and bounded above. An upper bound $M$ is the least upper bound (supremum) $\sup S$ if $M\leq M'$ for every upper bound $M'$. The completeness axiom (Dedekind): every non-empty bounded-above subset of $\mathbb{R}$ has a supremum in $\mathbb{R}$. Equivalently: every Cauchy sequence $(a_n)$ (where $|a_n-a_m|\to 0$) converges to some $L\in\mathbb{R}$. Rationals lack completeness: $\{x\in\mathbb{Q}:x^2<2\}$ has no rational supremum.
Archimedean Property & Density
(1) For any $x\in\mathbb{R}$, $\exists n\in\mathbb{N}$ with $n>x$. Proof: if not, $\mathbb{N}$ is bounded above, so $\sup\mathbb{N}$ exists; then $\sup\mathbb{N}-1$ is not an upper bound, contradiction. (2) $\mathbb{Q}$ is dense in $\mathbb{R}$: for any $a
Example 1
Prove $\sup\{1-1/n:n\in\mathbb{N}\}=1$.
Solution: Claim 1 is an upper bound: $1-1/n<1$ for all $n\geq 1$. Claim it's least: for any $\epsilon>0$, by Archimedean property $\exists N$ with $1/N<\epsilon$, so $1-1/N>1-\epsilon$ is in the set. Hence $1-\epsilon$ is not an upper bound for any $\epsilon>0$. So $\sup=1$.
Example 2
Show $\sqrt{2}\in\mathbb{R}$ using the completeness axiom. Define $S=\{x>0:x^2<2\}$.
Solution: $S$ is non-empty ($1\in S$) and bounded above ($2\in\mathbb{R}$, and if $x>2$ then $x^2>4>2$, so 2 is an upper bound). Let $s=\sup S$. Show $s^2=2$: if $s^2<2$, pick $\delta$ small so $(s+\delta)^2<2$, then $s+\delta\in S$ contradicting $s=\sup S$; if $s^2>2$, pick $\delta$ small so $(s-\delta)^2>2$, then $s-\delta$ is an upper bound contradicting minimality. Hence $s^2=2$ and $\sqrt{2}=s\in\mathbb{R}$.
Practice
- Prove: if $a=\sup A$ and $b=\sup B$, then $a+b=\sup(A+B)$ where $A+B=\{x+y:x\in A,y\in B\}$.
- Show $\inf S=-\sup(-S)$ for any bounded-below set $S$.
- Prove every bounded sequence has a convergent subsequence (Bolzano-Weierstrass theorem).
- Show $\mathbb{R}\setminus\mathbb{Q}$ (the irrationals) is dense in $\mathbb{R}$.
Show Answer Key
1. Let $s\in A+B$, so $s=a+b$ with $a\le\sup A$, $b\le\sup B$, hence $s\le\sup A+\sup B$. Conversely, for any $\epsilon>0$, $\exists a\in A, b\in B$ with $a>\sup A-\epsilon/2$ and $b>\sup B-\epsilon/2$, so $a+b>\sup A+\sup B-\epsilon$. Thus $\sup(A+B)=\sup A+\sup B$.
2. $-S=\{-s:s\in S\}$. If $m=\inf S$, then $m\le s$ for all $s\in S$, so $-m\ge -s$ for all $s$, meaning $-m$ is an upper bound of $-S$. For any upper bound $u$ of $-S$: $u\ge -s$ for all $s$, so $-u\le s$ for all $s$, giving $-u\le m$, hence $u\ge -m$. So $\sup(-S)=-m=-\inf S$.
3. Proof sketch: Bisect the interval containing the sequence into two halves; at least one contains infinitely many terms. Repeat to get nested intervals whose lengths $\to0$. By completeness, they shrink to a single point $L$, and a subsequence converging to $L$ can be extracted.
4. For any $x\in\mathbb{R}$ and $\epsilon>0$, by the Archimedean property choose $n$ with $1/n<\epsilon$. Among the $n$ intervals $(k/n,(k+1)/n)$, at least one lies in $(x-\epsilon,x+\epsilon)$; each such interval contains an irrational (e.g., $k/n+\sqrt{2}/(2n)$). So irrationals are dense.