Real Number (By Rational Number Closure)

Every sequence of rational approximations to $\sqrt{2}$ gets arbitrarily close to the “right answer” but never lands on one. The rational numbers $\mathbb{Q}$ have a hole exactly there — and holes everywhere else an irrational should be. The real numbers $\mathbb{R}$ are what you get by sealing every hole: forming the closure of $\mathbb{Q}$.

The gap problem in $\mathbb{Q}$

Consider the sequence

$$1,\; 1.4,\; 1.41,\; 1.414,\; 1.4142,\;\ldots$$

whose $n$-th term is the decimal expansion of $\sqrt{2}$ truncated after $n$ digits. Every term is rational, and the terms cluster tighter and tighter — yet no rational number is their limit. It can be shown that no fraction $p/q$ satisfies $(p/q)^2 = 2$, so $\sqrt{2} \notin \mathbb{Q}$ and the sequence has no home in $\mathbb{Q}$.

The same phenomenon occurs for $\sqrt{3}$, $\pi$, $e$, and countless other targets: you can zero in on them with rationals but never arrive. To build a number system where every such sequence has a home, you need a precise way to say “converging to something” before you know what that something is.

Cauchy sequences: converging without a named destination

The limit definition asks you to name the limit $L$ upfront. When $L$ is the missing number you are trying to construct, that is not possible. Instead, use a condition on the terms alone.

A sequence $(q_n)$ of rationals is a Cauchy sequence if for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that

$$m,\; n \geq N \implies |q_m - q_n| < \varepsilon. \tag{1}$$

Where convergence asks “do the terms eventually stay close to a fixed point $L$?”, Cauchiness asks only “do the terms eventually stay close to each other?” — no external target required.

Fact. Every convergent sequence is Cauchy. Proof sketch: if $q_n \to L$, then for large $m$ and $n$ both $|q_m - L|$ and $|q_n - L|$ are small; the triangle inequality gives $|q_m - q_n| \leq |q_m - L| + |L - q_n|$, which is also small. $\square$

The converse fails in $\mathbb{Q}$: the sequence $1, 1.4, 1.41, \ldots$ is Cauchy yet converges to nothing in $\mathbb{Q}$. The rationals are not complete.

The guiding idea: $\mathbb{R} = \overline{\mathbb{Q}}$

From Closure you know that the closure $\overline{A}$ of a set $A$ equals $A$ together with all its accumulation points. In a metric space, $x$ is an accumulation point of $A$ when some sequence in $A$ converges to $x$.

Closing $\mathbb{Q}$ means adjoining every point that some sequence of rationals converges to. The aspiration is

$$\mathbb{R} \;\coloneqq\; \overline{\mathbb{Q}}. \tag{2}$$

There is a subtlety: to take a closure you need an ambient space to take it in, but that ambient space is what you are trying to build. The way out is to construct $\mathbb{R}$ concretely from Cauchy sequences and then verify that $(2)$ holds afterwards.

Constructing $\mathbb{R}$

Equivalence classes of Cauchy sequences

Let $\mathcal{C}$ be the set of all Cauchy sequences in $\mathbb{Q}$. Declare two sequences $(p_n), (q_n) \in \mathcal{C}$ equivalent, written $(p_n) \sim (q_n)$, when

$$|p_n - q_n| \to 0 \quad \text{as } n \to \infty. \tag{3}$$

Intuitively: they are aiming at the same target. Condition $(3)$ is an equivalence relation (it is reflexive, symmetric, and transitive — check each), so it partitions $\mathcal{C}$ into disjoint equivalence classes. Define

$$\mathbb{R} \;\coloneqq\; \mathcal{C}/{\sim}.$$

Each class $[(q_n)]$ represents the unique “intended limit” that all sequences in it share. The class of $1, 1.4, 1.41, \ldots$ is what we will call $\sqrt{2}$. The class of the constant sequence $(3, 3, 3, \ldots)$ is the rational $3$.

Embedding $\mathbb{Q}$ into $\mathbb{R}$

Send each rational $q$ to the class of the constant sequence:

$$\iota \colon \mathbb{Q} \to \mathbb{R}, \quad q \;\mapsto\; [(q,\, q,\, q,\, \ldots)].$$

Two distinct rationals give inequivalent constant sequences, so $\iota$ is injective: $\mathbb{Q}$ sits inside $\mathbb{R}$ without collision. From now on, identify each $q \in \mathbb{Q}$ with $\iota(q) \in \mathbb{R}$, writing $\mathbb{Q} \subset \mathbb{R}$.

Arithmetic and order

Operations are defined term-wise on representatives:

$$[(p_n)] + [(q_n)] \;\coloneqq\; [(p_n + q_n)], \qquad [(p_n)] \cdot [(q_n)] \;\coloneqq\; [(p_n \cdot q_n)]. \tag{4}$$

Both are well-defined: term-wise sums and products of Cauchy sequences are Cauchy, and swapping a representative for an equivalent one does not change the resulting class. Division is defined similarly, excluding the class of sequences whose terms tend to $0$. With these operations, $\mathbb{R}$ is a field extending $\mathbb{Q}$.

Declare $[(p_n)] < [(q_n)]$ when $q_n - p_n \geq \delta$ for some fixed $\delta > 0$ and all sufficiently large $n$. This makes $\mathbb{R}$ an ordered field, with the order extending that of $\mathbb{Q}$.

The metric on $\mathbb{R}$

Set $|[(q_n)]| \coloneqq [(|q_n|)]$ and $d(x, y) \coloneqq |x - y|$. This gives $\mathbb{R}$ a metric space structure that extends the one on $\mathbb{Q}$.

Key properties of $\mathbb{R}$

$\mathbb{Q}$ is dense: $\overline{\mathbb{Q}} = \mathbb{R}$

Theorem. Every real number is the limit of a sequence of rationals.

Proof. Let $x = [(q_n)] \in \mathbb{R}$. Embed each rational $q_n$ into $\mathbb{R}$ via $\iota$ and regard it as an element of $\mathbb{R}$. Fix $\varepsilon > 0$. The Cauchy condition $(1)$ gives $N \in \mathbb{N}$ such that $|q_m - q_n| < \varepsilon$ for all $m, n \geq N$. For any fixed $n \geq N$, the real number $|x - q_n| = [(|q_m - q_n|)_m]$ satisfies $|q_m - q_n| < \varepsilon$ for every $m \geq N$, so — by the order on $\mathbb{R}$ — $|x - q_n| \leq \varepsilon$. Since $\varepsilon$ was arbitrary, $q_n \to x$. Since $x$ was arbitrary, every real is a limit of rationals, and $\overline{\mathbb{Q}} = \mathbb{R}$. $\square$

This confirms $(2)$: the construction delivers exactly the closure of $\mathbb{Q}$ we aimed for.

Completeness

Theorem. Every Cauchy sequence in $\mathbb{R}$ converges in $\mathbb{R}$.

Proof sketch. Let $(x_k)$ be a Cauchy sequence in $\mathbb{R}$. By density, choose a rational $r_k$ with $|r_k - x_k| < 1/k$ for each $k$. The sequence $(r_k)$ is Cauchy in $\mathbb{Q}$:

$$|r_j - r_k| \;\leq\; |r_j - x_j| + |x_j - x_k| + |x_k - r_k| \;<\; \frac{1}{j} + |x_j - x_k| + \frac{1}{k},$$

which is small for large $j$ and $k$ since $(x_k)$ is Cauchy. Define $x \coloneqq [(r_k)] \in \mathbb{R}$. Then $|x_k - x| \leq |x_k - r_k| + |r_k - x| < 1/k + |r_k - x|$, and $r_k \to x$ by density applied to the sequence $(r_k)$, so $x_k \to x$. $\square$

Completeness is the payoff of the whole construction: every Cauchy sequence in $\mathbb{R}$ converges in $\mathbb{R}$, with no holes remaining.

Uniqueness

Up to isomorphism of ordered fields, $\mathbb{R}$ is the unique complete ordered field. Any other construction — Dedekind cuts, for instance — produces the same mathematical object, related to this one by an order-preserving field isomorphism. This is why different approaches to defining the reals are interchangeable in practice.

Summary

• $\mathbb{Q}$ has holes: Cauchy sequences of rationals need not converge within $\mathbb{Q}$.
• A Cauchy sequence $(q_n)$ satisfies $|q_m - q_n| \to 0$ as $m, n \to \infty$ — internal clustering with no external target.
• The real numbers $\mathbb{R} = \mathcal{C}/{\sim}$ are equivalence classes of Cauchy sequences in $\mathbb{Q}$, where two sequences are equivalent when their term-wise difference tends to $0$.
• $\mathbb{Q}$ is dense in $\mathbb{R}$: $\overline{\mathbb{Q}} = \mathbb{R}$, so every real number is the limit of a sequence of rationals — exactly the closure $(2)$ we set out to achieve.
• $\mathbb{R}$ is complete: every Cauchy sequence of reals converges in $\mathbb{R}$.
• Together, these properties make $\mathbb{R}$ the unique complete ordered field up to isomorphism.