Proof: Existence of Supremum in R

This proof establishes that $\mathbb{R}$ satisfies the completeness axiom (least upper bound property) when constructed via Dedekind cuts. It shows that every nonempty bounded-above set of real numbers has a supremum, making $\mathbb{R}$ a complete ordered field. This is the cornerstone property distinguishing $\mathbb{R}$ from $\mathbb{Q}$ .

Statement to prove

Theorem1.1Completeness of R (via Dedekind Cuts)

Let $\mathbb{R}$ be the set of Dedekind cuts (as defined in Dedekind Cuts). Then every nonempty subset $A \subseteq \mathbb{R}$ that is bounded above has a least upper bound in $\mathbb{R}$ .

Proof

Let $A \subseteq \mathbb{R}$ be nonempty and bounded above. Recall that each element of $\mathbb{R}$ is a Dedekind cut: a subset $\alpha \subseteq \mathbb{Q}$ satisfying:

$\alpha \neq \varnothing$ and $\alpha \neq \mathbb{Q}$ .
$\alpha$ is downward closed: if $p \in \alpha$ and $q < p$ , then $q \in \alpha$ .
$\alpha$ has no greatest element: if $p \in \alpha$ , there exists $r \in \alpha$ with $r > p$ .

We construct the supremum of $A$ explicitly.

Construction of the supremum

Define

$\gamma = \bigcup_{\alpha \in A} \alpha.$

In other words, $\gamma$ consists of all rationals $q \in \mathbb{Q}$ such that $q \in \alpha$ for some $\alpha \in A$ . We claim that $\gamma$ is a Dedekind cut, and $\gamma = \sup A$ .

Step 1: γ is nonempty

Since $A$ is nonempty, there exists $\alpha \in A$ . Since $\alpha$ is a Dedekind cut, $\alpha \neq \varnothing$ . Thus $\gamma = \bigcup_{\alpha \in A} \alpha \supseteq \alpha \neq \varnothing$ , so $\gamma$ is nonempty.

Step 2: γ ≠ Q

Since $A$ is bounded above, there exists a Dedekind cut $\beta \in \mathbb{R}$ such that $\alpha \leq \beta$ (i.e., $\alpha \subseteq \beta$ ) for all $\alpha \in A$ . In particular, every $\alpha \in A$ is a subset of $\beta$ .

Since $\beta$ is a cut, $\beta \neq \mathbb{Q}$ : there exists $q_0 \in \mathbb{Q}$ with $q_0 \notin \beta$ . We claim $q_0 \notin \gamma$ .

Indeed, if $q_0 \in \gamma$ , then $q_0 \in \alpha$ for some $\alpha \in A$ . But $\alpha \subseteq \beta$ , so $q_0 \in \beta$ , contradiction. Thus $q_0 \notin \gamma$ , so $\gamma \neq \mathbb{Q}$ .

Step 3: γ is downward closed

Let $q \in \gamma$ and $p < q$ . Then $q \in \alpha$ for some $\alpha \in A$ . Since $\alpha$ is downward closed and $p < q \in \alpha$ , we have $p \in \alpha$ . Thus $p \in \gamma$ .

Step 4: γ has no greatest element

Let $q \in \gamma$ . Then $q \in \alpha$ for some $\alpha \in A$ . Since $\alpha$ has no greatest element, there exists $r \in \alpha$ with $r > q$ . Thus $r \in \gamma$ and $r > q$ . So $\gamma$ has no greatest element.

Conclusion: γ is a Dedekind cut

By Steps 1–4, $\gamma$ satisfies all three conditions to be a Dedekind cut. Therefore, $\gamma \in \mathbb{R}$ .

Step 5: γ is an upper bound of A

For any $\alpha \in A$ , we have $\alpha \subseteq \gamma$ (by definition of $\gamma$ as the union). In the ordering on $\mathbb{R}$ (where $\alpha \leq \beta$ means $\alpha \subseteq \beta$ ), this means $\alpha \leq \gamma$ . Thus $\gamma$ is an upper bound of $A$ .

Step 6: γ is the least upper bound

Let $\beta$ be any upper bound of $A$ . Then $\alpha \subseteq \beta$ for all $\alpha \in A$ . By definition,

$\gamma = \bigcup_{\alpha \in A} \alpha \subseteq \beta.$

Thus $\gamma \leq \beta$ . So $\gamma$ is the least upper bound of $A$ .

Conclusion

We have shown that $\gamma = \sup A$ exists in $\mathbb{R}$ . This completes the proof that $\mathbb{R}$ is complete.

Remarks and corollaries

RemarkUniqueness of supremum

The supremum is unique: if $\gamma$ and $\gamma'$ are both least upper bounds of $A$ , then $\gamma \leq \gamma'$ and $\gamma' \leq \gamma$ , so $\gamma = \gamma'$ .

RemarkInfimum exists

By symmetry, every nonempty subset of $\mathbb{R}$ that is bounded below has a greatest lower bound (infimum). Indeed, $\inf A = -\sup(-A)$ , where $-A = \{-x \mid x \in A\}$ .

ExampleSupremum of {1 - 1/n}

Let $A = \{1 - 1/n \mid n \in \mathbb{N}\} = \{0, 1/2, 2/3, 3/4, \ldots\}$ . As a subset of $\mathbb{R}$ (thinking of each element as a Dedekind cut), the supremum is

$\sup A = 1^* = \{q \in \mathbb{Q} \mid q < 1\}.$

The union $\gamma = \bigcup_{\alpha \in A} \alpha$ consists of all rationals $q$ such that $q < 1 - 1/n$ for some $n$ . For any $q < 1$ , choosing $n$ large enough gives $1 - 1/n > q$ , so $q \in \gamma$ . Thus $\gamma = 1^*$ .

ExampleSupremum of {q ∈ Q : q² < 2}

Let $S = \{q \in \mathbb{Q} \mid q^2 < 2\}$ . As a subset of $\mathbb{Q}$ , $S$ has no supremum in $\mathbb{Q}$ (since $\sqrt{2} \notin \mathbb{Q}$ ). But in $\mathbb{R}$ , $\sup S = \sqrt{2}^*$ , the Dedekind cut corresponding to $\sqrt{2}$ .

The proof above constructs $\sqrt{2}^*$ as the union of all rationals $q$ with $q^2 < 2$ or $q \leq 0$ . This union is precisely the Dedekind cut for $\sqrt{2}$ .

ExampleUnbounded sets have no supremum

Let $A = \mathbb{N} = \{1, 2, 3, \ldots\}$ (viewed as cuts $1^*, 2^*, 3^*, \ldots$ in $\mathbb{R}$ ). Then $A$ is not bounded above in $\mathbb{R}$ , so the completeness axiom does not apply. By convention, we write $\sup A = +\infty$ , but this is not a real number.

Alternative proof via Cauchy sequences

An alternative construction of $\mathbb{R}$ uses equivalence classes of Cauchy sequences of rationals. In that framework, completeness is proved differently.

ProofSketch via Cauchy sequences

Let $A \subseteq \mathbb{R}$ be nonempty and bounded above. For each $n \in \mathbb{N}$ , let $a_n \in A$ be chosen so that

$a_n > \sup A - \frac{1}{n}$

(such $a_n$ exists by the definition of supremum). The sequence $(a_n)$ is Cauchy and converges to $\sup A$ . One verifies that the limit is indeed the least upper bound.

This approach relies on the Cauchy completeness of $\mathbb{R}$ (every Cauchy sequence converges), which is equivalent to the least upper bound property.

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RemarkEquivalence of constructions

Both Dedekind cuts and Cauchy sequences yield the same complete ordered field $\mathbb{R}$ , up to isomorphism. The choice of construction is a matter of convenience and taste. For some purposes (e.g., constructive mathematics), Cauchy sequences are preferred; for others (e.g., set-theoretic foundations), Dedekind cuts are cleaner.

Applications of completeness

The existence of suprema is used repeatedly in real analysis:

ExampleIntermediate Value Theorem

If $f : [a, b] \to \mathbb{R}$ is continuous with $f(a) < 0 < f(b)$ , define $S = \{x \in [a, b] \mid f(x) < 0\}$ . By completeness, $c = \sup S$ exists. Using continuity, one shows $f(c) = 0$ . Without completeness (e.g., over $\mathbb{Q}$ ), the IVT fails: $f(x) = x^2 - 2$ has no rational root.

ExampleExtreme Value Theorem

If $f : [a, b] \to \mathbb{R}$ is continuous, let $M = \sup\{f(x) \mid x \in [a, b]\}$ (exists by completeness). By sequential compactness, there exists $c \in [a, b]$ with $f(c) = M$ . Thus $f$ attains its maximum.

ExampleMonotone Convergence Theorem

Every bounded increasing sequence $(a_n)$ converges to $\sup\{a_n \mid n \in \mathbb{N}\}$ . This is a direct consequence of completeness and is fundamental in measure theory and integration.

Summary

This proof establishes completeness of $\mathbb{R}$ via Dedekind cuts:

The supremum of a set $A$ is constructed as $\gamma = \bigcup_{\alpha \in A} \alpha$ .
$\gamma$ is verified to be a Dedekind cut and the least upper bound of $A$ .
Completeness is built into the Dedekind construction — the union of cuts is a cut.

This makes $\mathbb{R}$ fundamentally different from $\mathbb{Q}$ and is the reason real analysis "works." See Completeness Axiom for further discussion and Monotone Convergence Theorem for an immediate application.