LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)

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The supremum is attained exactly when a maximum exists

Statement

Let $S \subseteq R$ be nonempty.

If $S$ has a maximum (Maximum and minimum of a set), then $\sup S$ exists and $\sup S = \max S$ .
If $\sup S$ exists and $\sup S \in S$ , then $S$ has a maximum and $\max S = \sup S$ .

Hence, for a set whose supremum exists, the supremum is attained (belongs to the set) precisely when the set has a maximum, and then the two agree.

Facts & Assumptions

Given: A nonempty $S \subseteq R$ .

[L1]

Maximum: $m = \max S$ means $m \in S$ and $s \leq m$ for every $s \in S$ ; a maximum is unique (Maximum and minimum of a set).

[L2]

Supremum: $u = \sup S$ means $u$ is an upper bound of $S$ , that is $s \leq u$ for every $s \in S$ , and $u \leq u^{'}$ for every upper bound $u^{'}$ of $S$ ; it is unique when it exists (Suprema and infima are unique, Complete ordered field (least-upper-bound property)).

Proof

technique · direct

1.1assume-hypL1L2

For claim 1 assume $S$ has a maximum $m$ : then $m \in S$ and $s \leq m$ for every $s \in S$ , so $m$ is in particular an upper bound of $S$ .

1.2assume-hypL2

For claim 2 assume $\sup S$ exists and lies in $S$ , and write $u : = \sup S \in S$ .

2.1step 1.1L2

Let $u^{'}$ be an arbitrary upper bound of $S$ ; since $m \in S$ , the defining property of an upper bound applied to the element $m$ gives $m \leq u^{'}$ .

2.2step 1.2L1L2

Since $u = \sup S$ is an upper bound of $S$ we have $s \leq u$ for every $s \in S$ , and by assumption $u \in S$ ; these are exactly the two requirements for $u$ to be a maximum of $S$ , so $\max S$ exists and equals $u = \sup S$ by uniqueness of the maximum, proving claim 2.

3.1step 1.1step 2.1L1L2

Thus $m$ is an upper bound of $S$ with $m \leq u^{'}$ for every upper bound $u^{'}$ of $S$ , which is exactly the definition of a least upper bound; hence $\sup S$ exists and, by uniqueness of the least upper bound, $\sup S = m = \max S$ , proving claim 1.

4.1step 2.2step 3.1∎

Combining the two claims: when $\sup S$ exists, $\sup S \in S$ holds if and only if $S$ has a maximum, and in that case $\sup S = \max S$ .

Remarks

The dual statement, that $\inf S$ is attained exactly when $S$ has a minimum and then $\inf S = \min S$ , is not proved above. It follows by reflection: $m = \max S$ if and only if $- m = \min (- S)$ , and $\inf X = - \sup (- X)$ (Reflection through zero exchanges upper and lower bounds, Every nonempty set bounded below has an infimum), so applying the two claims proved here to $- S$ and negating gives the minimum form.
Claim 1 needs no completeness assumption: a set with a maximum has a supremum for free, since the maximum is already the least upper bound. Only claim 2 presupposes that $\sup S$ exists, which for a nonempty set bounded above is guaranteed by the least-upper-bound property (Complete ordered field (least-upper-bound property)).
The converse of "the supremum exists" is not "the maximum exists": the set ${x \in R : 0 < x < 1}$ has supremum $1$ and no maximum (FALSE: the supremum of a set belongs to the set). What forces attainment is being nonempty and finite (Every nonempty finite set of reals has a maximum and a minimum); finiteness alone does not, since $\emptyset$ is finite and has no maximum (Maximum and minimum of a set).

The supremum is attained exactly when a maximum exists

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources