LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)

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Every nonempty finite set of reals has a maximum and a minimum

Statement

For every $n \in N$ and all $a_{0}, a_{1}, \dots, a_{n} \in R$ , the set ${a_{0}, a_{1}, \dots, a_{n}}$ has a maximum and a minimum (Maximum and minimum of a set).

What is proved below is exactly the displayed statement, by induction on $n$ .

The usual reading, that every nonempty finite subset of $R$ has a maximum and a minimum, follows once one identifies the nonempty finite subsets of $R$ with the sets listable as ${a_{0}, \dots, a_{n}}$ . That identification is recorded as a stipulation in the Given below, because this page has no definition of finiteness to prove it against. It is discharged, not merely assumed: The nonempty finite subsets of $R$ are exactly the listable ones proves that the two descriptions of a nonempty finite subset of $R$ agree. That lemma is recorded in justified_by rather than in deps, since it is about the sets this lemma quantifies over and therefore depends on this one. This is what licenses the notation $\max {a_{1}, \dots, a_{n}}$ and $\min {a_{1}, \dots, a_{n}}$ for finite sets of real numbers from this page onwards.

Statement references

Maximum and minimum of a set

The nonempty finite subsets of $R$ are exactly the listable ones ↗

Facts & Assumptions

Given: Real numbers $a_{0}, a_{1}, a_{2}, \dots$ ; for $n \in N$ write $F_{n} : = {a_{0}, \dots, a_{n}}$ , so that $F_{n + 1} = F_{n} \cup {a_{n + 1}}$ . A subset of $R$ is nonempty and finite exactly when it equals $F_{n}$ for some $n \in N$ and some choice of $a_{0}, \dots, a_{n}$ .

[A1]

$P (n)$ denotes the statement: for all $a_{0}, \dots, a_{n} \in R$ , the set $F_{n}$ has a maximum and a minimum.

[L1]

Maximum and minimum: $m = \max X$ means $m \in X$ and $x \leq m$ for all $x \in X$ ; $m = \min X$ means $m \in X$ and $m \leq x$ for all $x \in X$ ; each is unique when it exists (Maximum and minimum of a set).

[L2]

Induction principle: if $P (0)$ holds and $P (n)$ implies $P (n + 1)$ for every $n \in N$ , then $P (n)$ holds for every $n \in N$ , where $n + 1$ denotes the successor $σ (n)$ (The principle of mathematical induction, Addition of natural numbers).

[L3]

The order on $R$ is reflexive, total and transitive: $a \leq a$ ; for all $a, b$ exactly one of $a < b$ , $a = b$ , $b < a$ holds, so at least one of $a \leq b$ and $b \leq a$ holds; and $a \leq b$ with $b \leq c$ gives $a \leq c$ (Complete ordered field (least-upper-bound property), Ordered field).

Proof

technique · induction

1.1baseA1L1L3

Base case: $F_{0} = {a_{0}}$ , and $a_{0} \in F_{0}$ with $a_{0} \leq a_{0}$ by reflexivity, so $a_{0}$ is both a maximum and a minimum of $F_{0}$ ; hence $P (0)$ holds.

1.2ihA1

Inductive hypothesis: fix $n \in N$ and assume $P (n)$ , that is, for all reals $a_{0}, \dots, a_{n}$ the set $F_{n}$ has a maximum and a minimum.

2.1step 1.2L1

Let $a_{0}, \dots, a_{n + 1} \in R$ be arbitrary; by the inductive hypothesis the set $F_{n}$ has a maximum $M$ and a minimum $m$ , and $F_{n + 1} = F_{n} \cup {a_{n + 1}}$ .

3.1step 2.1L1L3

By totality at least one of $a_{n + 1} \leq M$ and $M \leq a_{n + 1}$ holds. If $a_{n + 1} \leq M$ , then $M \in F_{n} \subseteq F_{n + 1}$ , every element of $F_{n}$ is $\leq M$ because $M = \max F_{n}$ , and $a_{n + 1} \leq M$ as well, so $M$ is a maximum of $F_{n + 1}$ . If $M \leq a_{n + 1}$ , then $a_{n + 1} \in F_{n + 1}$ , every $x \in F_{n}$ satisfies $x \leq M \leq a_{n + 1}$ hence $x \leq a_{n + 1}$ by transitivity, and $a_{n + 1} \leq a_{n + 1}$ , so $a_{n + 1}$ is a maximum of $F_{n + 1}$ . Either way $F_{n + 1}$ has a maximum.

3.2step 2.1L1L3

Dually, at least one of $m \leq a_{n + 1}$ and $a_{n + 1} \leq m$ holds. If $m \leq a_{n + 1}$ , then $m \in F_{n + 1}$ and every element of $F_{n + 1}$ is $\geq m$ , so $m$ is a minimum of $F_{n + 1}$ . If $a_{n + 1} \leq m$ , then $a_{n + 1} \in F_{n + 1}$ and every $x \in F_{n}$ satisfies $a_{n + 1} \leq m \leq x$ hence $a_{n + 1} \leq x$ by transitivity, so $a_{n + 1}$ is a minimum of $F_{n + 1}$ . Either way $F_{n + 1}$ has a minimum.

4.1step 3.1step 3.2A1

Since $a_{0}, \dots, a_{n + 1}$ were arbitrary, $F_{n + 1}$ has a maximum and a minimum for every such list, that is, $P (n)$ implies $P (n + 1)$ .

5.1step 1.1step 4.1givenL2discharge-induction∎

The base case and the inductive step give $P (n)$ for every $n \in N$ by the induction principle; since a nonempty finite subset of $R$ is exactly a set of the form $F_{n}$ , every nonempty finite subset of $R$ has both a maximum and a minimum.

Remarks

Where the stipulation is discharged. Finiteness itself is defined later, in Finite, countably infinite, countable, uncountable ↗, as equinumerosity with a von Neumann natural; with that definition in hand The nonempty finite subsets of $R$ are exactly the listable ones ↗ proves that a subset of $R$ is nonempty and finite exactly when it is listable as ${a_{0}, \dots, a_{n}}$ , which is the Given below. So nothing on this page rests on an assumption that is never paid for; it is paid for later, and the payment is recorded in justified_by.
Only the total order is used, never completeness. The base case needs reflexivity, the inductive step needs totality and transitivity, and the induction itself runs over $N$ . The same induction works in any totally ordered field; what is recorded here is its specialisation to $R$ .
Nonemptiness is essential: $\emptyset$ is finite and has no maximum (Maximum and minimum of a set). Finiteness is essential too: ${x \in R : 0 < x < 1}$ is bounded and has no maximum (FALSE: the supremum of a set belongs to the set).
Combined with claim 1 of The supremum is attained exactly when a maximum exists, this says every nonempty finite subset of $R$ has a supremum, and that the supremum is attained, because it equals the maximum. The infimum half is not part of The supremum is attained exactly when a maximum exists, which speaks only of maxima and suprema; it follows from the minimum proved here together with the reflection identity $\inf X = - \sup (- X)$ (Reflection through zero exchanges upper and lower bounds, Every nonempty set bounded below has an infimum).

Every nonempty finite set of reals has a maximum and a minimum

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources