Construction of the Real Numbers

The hole in the rationals gets filled and a real number is actually built. One real number is defined as a bundle of rational sequences.

Lengthen the decimal expansion and the rational terms march into the hole. Sequences whose difference vanishes count as the same number. That bundle itself is one real number. Built this way, the number line ends up with no holes. Completeness is exactly what this construction buys.

For example, climb a 🪜 staircase one step at a time toward the top. The numbers 1, 1.4, 1.41, 1.414 are footprints heading upward. The whole parade of footprints is the destination itself: the real number.

(xn)∼(yn)  ⟺  xn−yn→0(x_n) \sim (y_n) \iff x_n - y_n \to 0
2=[(1,1.4,1.41,1.414,… )]\sqrt{2} = [(1, 1.4, 1.41, 1.414, \dots)]

Part of these learning paths

Examples & Exercises

Let xn=⌊2⋅10n⌋10nx_n = \frac{\lfloor \sqrt{2}\cdot 10^n \rfloor}{10^n} be 2\sqrt{2} cut off after nn decimal places.

Prove that (xn)(x_n) is a Cauchy sequence of rationals.

∣xm−xn∣<10−N(m,n≥N)|x_m - x_n| < 10^{-N} \quad (m, n \geq N)
Decimal truncations, n=4n = 4 — more digits pull the terms into the hole.
Show proof

  1. Each xnx_n is a fraction with denominator 10n10^n, hence rational. With tn=⌊2⋅10n⌋t_n = \lfloor \sqrt{2}\cdot 10^n \rfloor, for m≥nm \geq n we have tm≥10m−ntnt_m \geq 10^{m-n} t_n, so xm≥xnx_m \geq x_n — adding digits never decreases the value.

  2. Putting y=2⋅10ny = \sqrt{2}\cdot 10^n into ⌊y⌋≤y<⌊y⌋+1\lfloor y \rfloor \leq y < \lfloor y \rfloor + 1 gives xn≤2<xn+10−nx_n \leq \sqrt{2} < x_n + 10^{-n}. So the error satisfies ∣xn−2∣<10−n\lvert x_n - \sqrt{2}\rvert < 10^{-n}.

  3. For m≥nm \geq n we get 0≤xm−xn<10−n0 \leq x_m - x_n < 10^{-n} (combining xm≤2x_m \leq \sqrt{2} with the bound above). Hence ∣xm−xn∣<10−N→0\lvert x_m - x_n\rvert < 10^{-N} \to 0 whenever m,n≥Nm, n \geq N — (xn)(x_n) is a Cauchy sequence of rationals.

Prove that an=1+1na_n = 1 + \frac1n and bn=1b_n = 1 are different rational Cauchy sequences defining the same real number.

an−bn=1n→0⟹[(an)]=[(bn)]=1a_n - b_n = \frac1n \to 0 \quad\Longrightarrow\quad [(a_n)] = [(b_n)] = 1
The Babylonian sequence — a different sequence marching into the same hole 2\sqrt{2}.
Show proof

  1. First verify both are rational Cauchy sequences (the notation [(an)]=[(bn)][(a_n)] = [(b_n)] presupposes this).

    The constant sequence (bn)(b_n) is trivially Cauchy: ∣bm−bn∣=0|b_m - b_n| = 0.

    For (an)(a_n), ∣am−an∣=∣1/m−1/n∣≤1/m+1/n|a_m - a_n| = |1/m - 1/n| \le 1/m + 1/n, so given ε>0\varepsilon > 0, taking N>2/εN > 2/\varepsilon gives ∣am−an∣<ε|a_m - a_n| < \varepsilon for m,n≥Nm, n \ge N — (an)(a_n) is Cauchy too.

  2. The two sequences are different: an−bn=1n≠0a_n - b_n = \frac1n \ne 0 for every nn, i.e. an≠bna_n \ne b_n (they differ term by term as sequences). But their difference 1n→0\frac1n \to 0, so (an)∼(bn)(a_n) \sim (b_n) by the definition of the equivalence relation.

  3. Equivalent Cauchy sequences define the same class, so [(an)]=[(bn)][(a_n)] = [(b_n)] — different on the surface, one real number underneath.

  4. The class of the constant sequence (1,1,1,… )(1, 1, 1, \dots) is the rational 11 sitting inside the reals. Hence [(an)]=1[(a_n)] = 1 — the thing (1+1n)(1+\frac1n) "converges to" is the real number 11.

Prove that the termwise sum (xn+yn)(x_n + y_n) of two rational Cauchy sequences is again Cauchy.

∣(xm+ym)−(xn+yn)∣≤∣xm−xn∣+∣ym−yn∣|(x_m+y_m)-(x_n+y_n)| \leq |x_m-x_n| + |y_m-y_n|
Decimal truncations, n=5n = 5 — sums of Cauchy sequences stay Cauchy, so addition is well-defined.
Show proof

  1. Let ε>0\varepsilon > 0. Since (xn)(x_n) is Cauchy, there is N1N_1 with ∣xm−xn∣<ε2\lvert x_m - x_n\rvert < \frac{\varepsilon}{2} for m,n≥N1m, n \geq N_1, and likewise N2N_2 for (yn)(y_n).

  2. By the triangle inequality, ∣(xm+ym)−(xn+yn)∣≤∣xm−xn∣+∣ym−yn∣\lvert (x_m+y_m) - (x_n+y_n)\rvert \leq \lvert x_m - x_n\rvert + \lvert y_m - y_n\rvert.

  3. With N=max⁡(N1,N2)N = \max(N_1, N_2), the right-hand side is below ε2+ε2=ε\frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon for m,n≥Nm, n \geq N.

    So (xn+yn)(x_n + y_n) is Cauchy.

    Changing representatives to (xn′)∼(xn)(x'_n) \sim (x_n) still gives (xn+yn)−(xn′+yn)=(xn−xn′)→0(x_n+y_n) - (x'_n+y_n) = (x_n - x'_n) \to 0, hence the same class — addition is well-defined, independent of the chosen representatives.

3. Sequences