ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)

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$(0,1) + (2,3) = (2,4)$ , with supremum $4 = \sup(0,1) + \sup(2,3)$

Example

Take $S = (0,1)$ and $T = (2,3)$ and form the sumset $S + T = \{\, s + t : s \in S,\ t \in T \,\}$ . Then $S + T = (2,4), \qquad \sup(S+T) = 4 = 1 + 3 = \sup S + \sup T,$ which is the identity of Supremum of a sumset: $\sup(S + T) = \sup S + \sup T$ on a concrete pair of intervals.

Two things are actually computed here, and it is worth keeping them apart. The set identity $S + T = (2,4)$ is not a supremum statement at all: the inclusion $\subseteq$ comes from adding inequalities, and the inclusion $\supseteq$ needs a construction, namely a decomposition $x = s_x + t_x$ of each $x \in (2,4)$ with $s_x \in S$ and $t_x \in T$ . The supremum statement is then supplied by Supremum of a sumset: $\sup(S + T) = \sup S + \sup T$ together with $\sup S = 1$ ( $\sup(0,1) = 1$ and $\inf(0,1) = 0$ , with neither attained) and $\sup T = 3$ , the latter obtained from $\sup S$ by translation (Supremum of a translate: $\sup(a + S) = a + \sup S$ ) rather than by a second epsilon argument.

The value $4$ is not attained: it fails the strict inequality $x < 4$ that defines $(2,4)$ , so $S + T$ has no maximum (The supremum is attained exactly when a maximum exists, Maximum and minimum of a set). The identity $\sup(S+T) = \sup S + \sup T$ therefore holds with the supremum on the left unattained, just as $\sup S = 1$ is unattained in $\sup(0,1) = 1$ and $\inf(0,1) = 0$ , with neither attained.

Facts & Assumptions

Given: The complete ordered field $\mathbb{R}$ ; the abbreviations $2 := 1 + 1$ , $3 := 2 + 1$ , $4 := 3 + 1$ ; the sets $S := \{x \in \mathbb{R} : 0 < x < 1\}$ , $T := \{x \in \mathbb{R} : 2 < x < 3\}$ , $U := \{x \in \mathbb{R} : 2 < x < 4\}$ ; the translate $2 + S := \{2 + s : s \in S\}$ and the sumset $S + T := \{s + t : s \in S,\ t \in T\}$ .

[L1]

The open unit interval: $S$ is nonempty and bounded above, and $\sup S = 1$ ( $\sup(0,1) = 1$ and $\inf(0,1) = 0$ , with neither attained).

[L2]

Translation: for nonempty $X \subseteq \mathbb{R}$ bounded above and $a \in \mathbb{R}$ , the translate $a + X$ is nonempty and bounded above and $\sup(a + X) = a + \sup X$ (Supremum of a translate: $\sup(a + S) = a + \sup S$ ).

[L3]

Sumset: for nonempty $X, Y \subseteq \mathbb{R}$ both bounded above, the sumset $X + Y$ is nonempty and bounded above and $\sup(X + Y) = \sup X + \sup Y$ (Supremum of a sumset: $\sup(S + T) = \sup S + \sup T$ ).

[L4]

Order and addition: $a < b$ implies $a + c < b + c$ , and strict inequalities add, so $a < b$ together with $c < d$ gives $a + c < b + d$ (claims 1 and 2 of Order is preserved by adding a constant and by adding inequalities). Applying the first claim with the constant $c$ and then with the constant $-c$ turns it into the equivalence $a < b$ if and only if $a + c < b + c$ , which is the form used below whenever a constant is subtracted from each part of a chain of inequalities (Ordered field).

[L5]

Halving: $0 < 1$ , hence $2 = 1 + 1 > 0$ , so $2 \ne 0$ and $2^{-1}$ exists with $c = (c \cdot 2^{-1}) \cdot 2$ for every $c$ ; $0 \cdot 2 = 0$ and $1 \cdot 2 = 2$ (Multiplication by zero: $0 \cdot a = 0$ , Field); and multiplying by the positive constant $2$ is an order equivalence, so $x < y$ if and only if $x \cdot 2 < y \cdot 2$ (claim 4 of Sign rules for products and monotonicity of multiplication) (The multiplicative identity is positive, Field, Ordered field).

[L6]

Arithmetic of the named constants, from the field axioms: $2 + 1 = 3$ , $3 + 1 = 4$ , $2 + 2 = 4$ , $1 + 3 = 4$ , $x + x = x \cdot 2$ for every $x$ , $2 \cdot 2 = 4$ and $3 \cdot 2 = 4 + 2$ (Field, Complete ordered field (least-upper-bound property)).

[L7]

Order: trichotomy holds in an ordered field, so $a < a$ is impossible for every $a$ (Ordered field, Complete ordered field (least-upper-bound property)).

[L8]

Maximum and attainment: $m = \max X$ means $m \in X$ and $x \le m$ for every $x \in X$ (Maximum and minimum of a set); and if a nonempty $X$ has a maximum then $\sup X$ exists and equals $\max X \in X$ , so a set whose supremum exists and does not belong to it has no maximum (The supremum is attained exactly when a maximum exists).

Verification

technique · direct

1.1

$T = 2 + S$ : for $s \in \mathbb{R}$ , adding $2$ to each part of $0 < s < 1$ gives $2 < 2 + s < 2 + 1 = 3$ , and conversely each $y$ with $2 < y < 3$ equals $2 + (y - 2)$ where subtracting $2$ from each part gives $0 < y - 2 < 1$ ; the two directions are equivalences because adding a constant is one.

L4L6

1.2

$S + T \subseteq U$ : for $s \in S$ and $t \in T$ , adding $0 < s$ to $2 < t$ gives $2 = 0 + 2 < s + t$ , and adding $s < 1$ to $t < 3$ gives $s + t < 1 + 3 = 4$ .

L4L6

1.3

Let $x \in U$ be arbitrary, so $2 < x < 4$ , and put $s_x := (x - 2) \cdot 2^{-1}$ and $t_x := (x + 2) \cdot 2^{-1}$ .

assume-hypL5choose

2.1

$s_x \in S$ and $s_x + t_x = x$ : subtracting $2$ from each part of $2 < x < 4$ gives $0 < x - 2 < 2$ , that is $0 \cdot 2 < s_x \cdot 2 < 1 \cdot 2$ , so $0 < s_x < 1$ by [L5]; and $s_x + t_x = \big((x - 2) + (x + 2)\big) \cdot 2^{-1} = (x + x) \cdot 2^{-1} = (x \cdot 2) \cdot 2^{-1} = x$ .

step 1.3L4L5L6

2.2

$t_x \in T$ : adding $2$ to each part of $2 < x < 4$ gives $2 \cdot 2 = 4 < x + 2 < 4 + 2 = 3 \cdot 2$ , and $t_x \cdot 2 = x + 2$ , so $2 < t_x < 3$ by [L5].

step 1.3L4L5L6

2.3

$\sup T = 3$ , and $T$ is nonempty and bounded above: applying the translation identity to the nonempty bounded-above set $S$ with $a = 2$ gives that $2 + S$ is nonempty and bounded above with $\sup(2 + S) = 2 + \sup S = 2 + 1 = 3$ , and $2 + S = T$ by 1.1.

step 1.1L1L2L6

3.1

$U \subseteq S + T$ , hence $S + T = U$ : for arbitrary $x \in U$ the elements $s_x \in S$ and $t_x \in T$ satisfy $s_x + t_x = x$ , so $x \in S + T$ ; combined with the inclusion of 1.2 this gives $S + T = U$ , that is $(0,1) + (2,3) = (2,4)$ .

step 2.1step 2.2step 1.2step 1.3

3.2

$\sup(S + T) = 4$ : both $S$ and $T$ are nonempty and bounded above, so the sumset identity applies and gives $\sup(S + T) = \sup S + \sup T = 1 + 3 = 4$ .

step 2.3L1L3L6

4.1

Therefore $(0,1) + (2,3)$ is exactly the interval $(2,4)$ , and its supremum is $4$ , which is the sum $\sup(0,1) + \sup(2,3)$ of the two suprema; in particular $\sup U = 4$ .

step 3.1step 3.2L3

4.2

The value $4$ does not belong to $S + T$ : by 3.1 one has $S + T = U$ , membership in $U$ requires $x < 4$ , and $4 < 4$ is impossible.

step 3.1L7

5.1

Hence $S + T$ is nonempty with $\sup(S + T) = 4 \notin S + T$ , so by the attainment criterion $S + T$ has no maximum: the identity $\sup(S+T) = \sup S + \sup T$ holds with the left-hand supremum unattained.

step 4.1step 4.2L3L8∎

Remarks

The inclusion $S + T \subseteq U$ alone already gives $\sup(S+T) \le 4$ , which is the easy half of Supremum of a sumset: $\sup(S + T) = \sup S + \sup T$ . What the explicit decomposition $x = s_x + t_x$ buys is the reverse inclusion, and with it the fact that the sumset is the whole interval rather than a proper subset of it. The lemma proves the reverse inequality without any such decomposition, by combining two epsilon approximations at $\varepsilon/2$ each.
$\sup T$ was obtained by translating $S$ rather than by repeating the epsilon computation. This is the normal division of labour: the general lemmas (Supremum of a translate: $\sup(a + S) = a + \sup S$ , Supremum of a scalar multiple, Supremum of a sumset: $\sup(S + T) = \sup S + \sup T$ ) are proved once, and concrete suprema are then transported rather than recomputed.
No analogue holds for products of sets. The sumset identity depends on the order being translation invariant, which multiplication is not: scaling by a negative number exchanges suprema and infima (Supremum of a scalar multiple).

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Sources

Infimum and supremum (Wikipedia) (standard reference, not scraped)
Minkowski addition (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (standard reference, not scraped)
Peter J. Olver, Continuous Calculus (standard reference, not scraped)

(0,1)+(2,3)=(2,4)(0,1) + (2,3) = (2,4)(0,1)+(2,3)=(2,4), with supremum 4=sup⁡(0,1)+sup⁡(2,3)4 = \sup(0,1) + \sup(2,3)4=sup(0,1)+sup(2,3)