The Supremum Axiom

Every bounded-above set has a least upper bound sup⁡S\sup S — the axiom filling the gaps in the rationals, twin of the rational-holes scene.

sup⁡{1−1n}=1\sup\left\{1-\frac{1}{n}\right\} = 1
S≠∅ bounded above ⇒sup⁡S existsS \ne \varnothing \text{ bounded above } \Rightarrow \sup S \text{ exists}

Part of these learning paths

Examples & Exercises

Prove sup⁡S=1\sup S = 1 for S={1−1n:n≥1}S = \left\{1 - \frac{1}{n} : n \geq 1\right\}. (Note 1∉S1 \notin S.)

sup⁡{1−1n}=1∉S\sup\left\{1-\frac{1}{n}\right\} = 1 \notin S
Candidate 1.21.2 — 0.20.2 above the true supremum 11.
Show proof
  1. Since 1−1n<11 - \frac{1}{n} < 1 for every nn, 11 is an upper bound of SS.
  2. Take any real u<1u < 1. The inequality 1−1n>u1 - \frac{1}{n} > u is equivalent to 1n<1−u\frac{1}{n} < 1 - u, i.e. n>11−un > \frac{1}{1-u}. By the Archimedean property such an nn exists, so uu is not an upper bound.
  3. So there is no upper bound below 11: 11 is the least upper bound, sup⁡S=1\sup S = 1. For reference, the n=1n = 1 term is 00, so 1∉S1 \notin S — a supremum need not belong to the set.

Prove sup⁡(A+B)=sup⁡A+sup⁡B\sup(A+B) = \sup A + \sup B for nonempty bounded-above A,B⊂RA, B \subset \mathbb{R}.

Tightening the candidate 1.51.5 down to the true supremum.
Show proof
  1. (≤) For all a∈Aa \in A and b∈Bb \in B, a+b≤sup⁡A+sup⁡Ba + b \leq \sup A + \sup B, so sup⁡A+sup⁡B\sup A + \sup B is an upper bound of A+BA + B. Hence sup⁡(A+B)≤sup⁡A+sup⁡B\sup(A+B) \leq \sup A + \sup B.
  2. (≥) Let ε>0\varepsilon > 0. We can choose a∈Aa \in A with a>sup⁡A−ε2a > \sup A - \frac{\varepsilon}{2} and b∈Bb \in B with b>sup⁡B−ε2b > \sup B - \frac{\varepsilon}{2} (definition of supremum). Then a+b>sup⁡A+sup⁡B−εa + b > \sup A + \sup B - \varepsilon, so sup⁡(A+B)≥sup⁡A+sup⁡B−ε\sup(A+B) \geq \sup A + \sup B - \varepsilon.
  3. Since ε\varepsilon was arbitrary, sup⁡(A+B)≥sup⁡A+sup⁡B\sup(A+B) \geq \sup A + \sup B. Combining the two inequalities gives sup⁡(A+B)=sup⁡A+sup⁡B\sup(A+B) = \sup A + \sup B.

Accept from completeness that s=sup⁡{x∈Q:x≥0,x2<2}s = \sup\{x \in \mathbb{Q} : x \geq 0, x^2 < 2\} exists, and verify s2=2s^2 = 2. (The existence step is idea-level.)

s2<2⇒contradiction,s2>2⇒contradictions^2 < 2 \Rightarrow \text{contradiction},\quad s^2 > 2 \Rightarrow \text{contradiction}
2≈1.4142\sqrt{2} \approx 1.4142 — rational truncations approach the hole.
Show proof
  1. (Lemma — cited) The rationals Q\mathbb{Q} are dense in R\mathbb{R}: every open interval of reals contains a rational number.
  2. S={x∈Q:x≥0,x2<2}S = \{x \in \mathbb{Q} : x \geq 0, x^2 < 2\} is nonempty (1∈S1 \in S) and bounded above (e.g. 22 is an upper bound). By the completeness axiom, s=sup⁡Ss = \sup S exists.
  3. Suppose s2<2s^2 < 2. Put ε=2−s2>0\varepsilon = 2 - s^2 > 0. By the lemma there is a rational qq with s<q<s+ε2(s+1)s < q < s + \frac{\varepsilon}{2(s+1)}. Then q2−s2=(q−s)(q+s)<εq^2 - s^2 = (q-s)(q+s) < \varepsilon, so q2<s2+ε=2q^2 < s^2 + \varepsilon = 2 — i.e. q∈Sq \in S with q>sq > s, contradicting that ss is an upper bound.
  4. Suppose s2>2s^2 > 2. Put δ=s2−2>0\delta = s^2 - 2 > 0. By the lemma there is a rational rr with s−δ2s<r<ss - \frac{\delta}{2s} < r < s. Then s2−r2=(s−r)(s+r)<δs^2 - r^2 = (s-r)(s+r) < \delta, so r2>s2−δ=2r^2 > s^2 - \delta = 2 — i.e. rr is an upper bound of SS smaller than ss, contradicting that ss is the least upper bound.
  5. Hence s2=2s^2 = 2. The completeness axiom fills the hole in the rationals, producing the real number that is 2\sqrt{2}.

1. Number Systems