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The function equal to q at a rational p/q in lowest terms and to 0 at every irrational is finite at every point and unbounded on every nondegenerate interval

Example

Throughout, N⊆Z⊆Q⊆R denotes the canonical copies. Natural numbers, integers and rationals are written directly when used in real arithmetic.

Let q(x) be the least denominator of a rational x (The Dirichlet function 1Q, and Thomae's function t with t(x)=1/q at a rational x=p/q in lowest terms with q≥1 and t(x)=0 at every irrational x) and define h:R→R by

h(x):=q(x)  for x∈Q,h(x):=0  for x∉Q,

where a natural q is identified with its canonical real value (The canonical natural ι(n)=n⋅1F of a field) and Q is the canonical copy of the rationals inside R (The rationals embed densely in the reals). Equivalently h(x)=1/t(x) at a rational x, where t is Thomae's function. Then:

  1. h(x) is a real number for every real x: h is finite at every point;
  2. h is unbounded on every nondegenerate interval (Lower bound, bounded below, bounded set, Intervals of R: the nine order-convex forms, nondegeneracy, and length): for all reals a<b and every real M there is x∈(a,b) with h(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 h is bounded on no neighbourhood of any point.

Facts & Assumptions

Given: The function h above, with q(x)=min⁡{ q∈N:q≥1 and qx∈Z } for x∈Q.

[A1]

q(x)≥1 is a natural with q(x) x∈Z, and q(x)≤q for every natural q≥1 with qx∈Z (The Dirichlet function 1Q, and Thomae's function t with t(x)=1/q at a rational x=p/q in lowest terms with q≥1 and t(x)=0 at every irrational x).

[L2]

A nonzero integer has absolute value at least 1, since no integer lies strictly between 0 and 1 (Integer part: for every real x there is exactly one integer m with m≤x<m+1, Basic properties of the absolute value).

[L3]

For every real η>0 there is a natural n≥1 with 1/n<η, and for every real x a natural n≥1 with x<n; the natural numbers ≥1 are positive in R and their order is preserved there (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε, Every complete ordered field is Archimedean, Canonical naturals are positive and strictly increasing, The canonical natural ι(n)=n⋅1F of a field).

Verification

technique · direct
1.1A1

Claim 1: for a rational x the value h(x)=q(x) is a canonical natural, hence a real number, and for an irrational x the value is 0. Every real falls under exactly one clause, so h is a function R→R.

1.2A1L2L4

Separation of rationals by their denominators. Let x≠y be rationals. Then ∣x−y∣≥1/(q(x) q(y)). Indeed, put q1:=q(x), q2:=q(y), p1:=q1x and p2:=q2y, all integers; then x−y=(p1q2−p2q1)/(q1q2), the numerator is an integer, and it is nonzero because x≠y; so its absolute value is at least 1.

2.1step 1.2L1L3L4

Claim 2: let a<b be reals and let M be real. Take a rational x1 with a<x1<b and put q1:=q(x1). Take a natural N≥1 with M<N, and put η:=min⁡{ 1q1 N, b−x1 }>0.

3.1step 2.1L1

With x1, q1, N and η as in step 2.1, take a rational y with x1<y<x1+η. Then a<x1<y<b, so y∈(a,b); and y≠x1 with ∣y−x1∣<η≤1/(q1N).

4.1step 1.2step 3.1L3

Hence q(y)>N. If instead q(y)≤N then q(y)≤N, and step 1.2 would give ∣y−x1∣≥1/(q1q(y))≥1/(q1N), contradicting step 3.1.

5.1step 2.1step 3.1step 4.1L3∎

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

Remarks

Depends on

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Sources