False statementConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)

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FALSE: the supremum of a set belongs to the set

Statement

False claim: if $S \subseteq R$ is nonempty and bounded above, then $\sup S \in S$ .

Equivalently, the false claim asserts that every nonempty set bounded above has a maximum (Maximum and minimum of a set). It is refuted below by the open unit interval, whose supremum exists, is unique, and lies outside the set.

Facts & Assumptions

Given: The set $S : = {x \in R : 0 < x < 1}$ inside the complete ordered field $R$ , and the abbreviation $2 : = 1 + 1$ .

[L1]

Epsilon characterisation of the supremum: for a nonempty $X \subseteq R$ bounded above and an upper bound $w$ of $X$ , one has $w = \sup X$ if and only if for every $ε > 0$ there is $x \in X$ with $w - ε < x$ (Epsilon characterisation of the supremum).

[L2]

Maximum: $m = \max X$ means $m \in X$ and $x \leq m$ for every $x \in X$ (Maximum and minimum of a set).

[L3]

Order: trichotomy holds, so exactly one of $a < b$ , $a = b$ , $b < a$ is true, the negation of $a \leq b$ is $b < a$ , and $a < a$ is impossible; the order is transitive; and adding a constant preserves it, so $a < b$ if and only if $a + c < b + c$ (Complete ordered field (least-upper-bound property), Ordered field, Order is preserved by adding a constant and by adding inequalities).

[L4]

Positivity and multiplication: $0 < 1$ (The multiplicative identity is positive); sums and products of positive elements are positive (axiom O2 of Ordered field); every nonzero element has a multiplicative inverse (Field); $0 \cdot c = 0$ for every $c$ (Multiplication by zero: $0 \cdot a = 0$ ); and for a positive multiplier $c$ one has $x < y$ if and only if $x c < y c$ (claim 4 of Sign rules for products and monotonicity of multiplication).

Refutation

technique · direct

1.1

Since $0 < 1$ , the element $2 = 1 + 1$ is positive, hence nonzero, so $2^{- 1}$ exists; from $2^{- 1} \cdot 2 = 1 > 0 = 0 \cdot 2$ we get $2^{- 1} > 0$ , and from $2^{- 1} \cdot 2 = 1 < 2 = 1 \cdot 2$ (the inequality $1 < 2$ holding because $2 - 1 = 1 > 0$ ) we get $2^{- 1} < 1$ ; therefore $2^{- 1} \in S$ and $S \neq \emptyset$ .

L3L4algebra

1.2

Every $x \in S$ satisfies $x < 1$ and hence $x \leq 1$ , so $1$ is an upper bound of $S$ and $S$ is bounded above.

1.3

Let $ε > 0$ . Then $1 + ε > 0$ is nonzero, so we may put $η : = ε (1 + ε)^{- 1}$ , which satisfies $η (1 + ε) = ε$ ; multiplying by the positive $1 + ε$ is an equivalence, so $η > 0$ follows from $η (1 + ε) = ε > 0 = 0 \cdot (1 + ε)$ , next $η < 1$ follows from $η (1 + ε) = ε < 1 + ε = 1 \cdot (1 + ε)$ , and finally $η < ε$ follows from $η (1 + ε) = ε < ε + ε^{2} = ε (1 + ε)$ , the last inequality holding because $ε^{2} > 0$ .

L3L4algebra

2.1

Put $s : = 1 - η$ . From $0 < η$ we get $s = 1 - η < 1$ , from $η < 1$ we get $0 = 1 - 1 < 1 - η = s$ , so $s \in S$ ; and from $η < ε$ we get $1 - ε < 1 - η = s$ . Since $ε > 0$ was arbitrary, for every $ε > 0$ there is an element of $S$ strictly greater than $1 - ε$ .

step 1.3L3algebra

2.2

The number $1$ is not an element of $S$ , because membership in $S$ requires $x < 1$ and $1 < 1$ is impossible by trichotomy.

step 1.1L3

2.3

The set $S$ has no maximum: if $m$ were one then $m \in S$ , so $0 < m < 1$ ; putting $m^{'} : = (m + 1) 2^{- 1}$ , so that $m^{'} \cdot 2 = m + 1$ , the inequality $m \cdot 2 = m + m < m + 1 = m^{'} \cdot 2$ gives $m < m^{'}$ , the inequality $m^{'} \cdot 2 = m + 1 < 1 + 1 = 1 \cdot 2$ gives $m^{'} < 1$ , and $m^{'} \cdot 2 = m + 1 > 0 = 0 \cdot 2$ gives $m^{'} > 0$ ; hence $m^{'} \in S$ with $m < m^{'}$ , contradicting the requirement $m^{'} \leq m$ for a maximum.

step 1.1L2L3L4algebra

3.1

The set $S$ is nonempty and bounded above with upper bound $1$ , and every $1 - ε$ with $ε > 0$ is exceeded by some element of $S$ , so the epsilon characterisation gives $\sup S = 1$ .

step 1.1step 1.2step 2.1L1

4.1

Thus $S$ is a nonempty subset of $R$ that is bounded above, its supremum exists and equals $1$ , and $1 \notin S$ ; the claim that the supremum of a set belongs to the set is therefore false, and correspondingly $S$ has no maximum, so no element of $S$ could have served as its supremum.

step 3.1step 2.2step 2.3∎

Remarks

The refutation is self-contained: the witness $S$ , the value of $\sup S$ and the failure of membership are all verified here from the complete-ordered-field axioms and the items this page has already proved.
What is true is the corrected statement The supremum is attained exactly when a maximum exists: for a set whose supremum exists, $\sup S \in S$ exactly when $S$ has a maximum, and then $\sup S = \max S$ . Being nonempty and finite is a sufficient condition for having a maximum (Every nonempty finite set of reals has a maximum and a minimum); nonemptiness cannot be dropped there, since $\emptyset$ is finite and has no maximum. Being nonempty and bounded above is not sufficient, which is exactly what the witness above shows.
The error is a common one because it is harmless on finite sets, which is where intuition is trained. The whole point of the supremum is to name a boundary that the set approaches without reaching.

Depends on

Used by

A supremum need not belong to its set: sup(0,1) = 1 ∉ (0,1) Counterexample

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Infimum and supremum (Wikipedia) (standard reference, not scraped)
Maximum and minimum (Wikipedia) (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)