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ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28
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The function equal to qq at a rational p/qp/q in lowest terms and to 00 at every irrational is finite at every point and unbounded on every nondegenerate interval

Example

Let q(x)q(x) be the least denominator of a rational xx (The Dirichlet function 1Q1_{\mathbb{Q}}, and Thomae's function tt with t(x)=1/qt(x) = 1/q at a rational x=p/qx = p/q in lowest terms with q1q \ge 1 and t(x)=0t(x) = 0 at every irrational xx) and define h:RRh : \mathbb{R} \to \mathbb{R} by

h(x):=ι(q(x))  for xQ,h(x):=0  for xQ,h(x) := \iota(q(x)) \ \text{ for } x \in \mathbb{Q}, \qquad h(x) := 0 \ \text{ for } x \notin \mathbb{Q},

where ι(q)\iota(q) is the canonical natural (The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field) and Q\mathbb{Q} is the canonical copy of the rationals inside R\mathbb{R} (The rationals embed densely in the reals). Equivalently h(x)=1/t(x)h(x) = 1/t(x) at a rational xx, where tt is Thomae's function. Then:

  1. h(x)h(x) is a real number for every real xx: hh is finite at every point;
  2. hh is unbounded on every nondegenerate interval (Lower bound, bounded below, bounded set, Intervals of R\mathbb{R}: the nine order-convex forms, nondegeneracy, and length): for all reals a<ba < b and every real MM there is x(a,b)x \in (a,b) with h(x)>Mh(x) > M.

So a function may be finite at every single point and yet fail to be bounded on every interval, however short. In particular hh is bounded on no neighbourhood of any point.

Facts & Assumptions

Given: The function hh above, with q(x)=min{qN:q1 and ι(q)xZ}q(x) = \min\{\, q \in \mathbb{N} : q \ge 1 \text{ and } \iota(q)x \in \mathbb{Z} \,\} for xQx \in \mathbb{Q}.

[A1]

q(x)1q(x) \ge 1 is a natural with ι(q(x))xZ\iota(q(x))\,x \in \mathbb{Z}, and q(x)qq(x) \le q for every natural q1q \ge 1 with ι(q)xZ\iota(q)x \in \mathbb{Z} (The Dirichlet function 1Q1_{\mathbb{Q}}, and Thomae's function tt with t(x)=1/qt(x) = 1/q at a rational x=p/qx = p/q in lowest terms with q1q \ge 1 and t(x)=0t(x) = 0 at every irrational xx).

[L2]

A nonzero integer has absolute value at least 11, since no integer lies strictly between 00 and 11 (Integer part: for every real xx there is exactly one integer mm with mx<m+1m \le x < m + 1, Basic properties of the absolute value).

[L3]

For every real η>0\eta > 0 there is a natural n1n \ge 1 with 1/ι(n)<η1/\iota(n) < \eta, and for every real xx a natural n1n \ge 1 with x<ι(n)x < \iota(n); ι\iota is positive and strictly increasing on the naturals 1\ge 1 (For every ε>0\varepsilon > 0 in a complete ordered field there is a natural n1n \ge 1 with 1/n<ε1/n < \varepsilon, Every complete ordered field is Archimedean, Canonical naturals are positive and strictly increasing, The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field).

[L4]

R\mathbb{R} is an ordered field (Complete ordered field (least-upper-bound property)).

Verification

technique · direct
1.1

Claim 1: for a rational xx the value h(x)=ι(q(x))h(x) = \iota(q(x)) is a canonical natural, hence a real number, and for an irrational xx the value is 00. Every real falls under exactly one clause, so hh is a function RR\mathbb{R} \to \mathbb{R}.

A1
1.2

Separation of rationals by their denominators. Let xyx \ne y be rationals. Then xy1/(ι(q(x))ι(q(y)))|x - y| \ge 1/(\iota(q(x))\,\iota(q(y))). Indeed, put q1:=q(x)q_{1} := q(x), q2:=q(y)q_{2} := q(y), p1:=ι(q1)xp_{1} := \iota(q_{1})x and p2:=ι(q2)yp_{2} := \iota(q_{2})y, all integers; then xy=(p1ι(q2)p2ι(q1))/(ι(q1)ι(q2))x - y = (p_{1}\iota(q_{2}) - p_{2}\iota(q_{1}))/(\iota(q_{1})\iota(q_{2})), the numerator is an integer, and it is nonzero because xyx \ne y; so its absolute value is at least 11.

A1L2L4
2.1

Claim 2: let a<ba < b be reals and let MM be real. Take a rational x1x_{1} with a<x1<ba < x_{1} < b and put q1:=q(x1)q_{1} := q(x_{1}). Take a natural N1N \ge 1 with M<ι(N)M < \iota(N), and put η:=min{1ι(q1)ι(N), bx1}>0.\eta := \min\Bigl\{\, \frac{1}{\iota(q_{1})\,\iota(N)},\ b - x_{1} \,\Bigr\} > 0 .

step 1.2L1L3L4
3.1

With x1x_{1}, q1q_{1}, NN and η\eta as in step 2.1, take a rational yy with x1<y<x1+ηx_{1} < y < x_{1} + \eta. Then a<x1<y<ba < x_{1} < y < b, so y(a,b)y \in (a,b); and yx1y \ne x_{1} with yx1<η1/(ι(q1)ι(N))|y - x_{1}| < \eta \le 1/(\iota(q_{1})\iota(N)).

step 2.1L1
4.1

Hence q(y)>Nq(y) > N. If instead q(y)Nq(y) \le N then ι(q(y))ι(N)\iota(q(y)) \le \iota(N), and step 1.2 would give yx11/(ι(q1)ι(q(y)))1/(ι(q1)ι(N))|y - x_{1}| \ge 1/(\iota(q_{1})\iota(q(y))) \ge 1/(\iota(q_{1})\iota(N)), contradicting step 3.1.

step 1.2step 3.1L3
5.1

Therefore h(y)=ι(q(y))>ι(N)>Mh(y) = \iota(q(y)) > \iota(N) > M, and y(a,b)y \in (a,b): the values of hh on (a,b)(a,b) exceed every real, so hh is unbounded on (a,b)(a,b), and hence on every set containing it.

step 2.1step 3.1step 4.1L3

Remarks

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