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ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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The distance ψ(x)=d(x,Z)\psi(x) = d(x, \mathbb{Z}) from a real number to the integers is 11-Lipschitz, hence uniformly continuous, takes values in [0,1/2][0,1/2], and vanishes exactly on Z\mathbb{Z}

Example

Take R\mathbb{R} with its usual metric d(x,y)=xyd(x,y) = |x-y| (The absolute value makes R\mathbb{R} a metric space: d(x,y)=xyd(x,y) = |x-y| is a metric, its open balls are the intervals (xr,x+r)(x-r, x+r), and it is unbounded), identify Z\mathbb{Z} with its canonical copy inside R\mathbb{R} (The integers as equivalence classes of pairs of naturals, The naturals embed in the integers, The integers embed in the rationals, The rationals embed densely in the reals), and let

ψ(x)  :=  d(x,Z)  =  inf{xn : nZ}\psi(x) \;:=\; d(x, \mathbb{Z}) \;=\; \inf\{\, |x - n| \ : \ n \in \mathbb{Z} \,\}

be the distance from xx to the nonempty set Z\mathbb{Z} (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space). Then:

  1. ψ\psi is 11-Lipschitz on R\mathbb{R}: ψ(x)ψ(y)xy|\psi(x) - \psi(y)| \le |x-y| for all real x,yx, y. Consequently ψ\psi is uniformly continuous on R\mathbb{R} (Uniform continuity of f:ARf : A \to \mathbb{R}: one δ\delta serving every pair of points of AA) and continuous on R\mathbb{R} (Continuity of f:ARf : A \to \mathbb{R} at a point of AA and on AA: the ε\varepsilon-δ\delta condition, its agreement with limxcf(x)=f(c)\lim_{x \to c} f(x) = f(c) at a limit point, and continuity at an isolated point).
  2. The infimum is attained, and computed. Writing m:=xm := \lfloor x \rfloor for the integer part of xx (Integer part: for every real xx there is exactly one integer mm with mx<m+1m \le x < m + 1) and t:=xmt := x - m, so that 0t<10 \le t < 1, ψ(x)  =  min{t, 1t}  =  min{xm, x(m+1)},\psi(x) \;=\; \min\{\, t,\ 1-t \,\} \;=\; \min\bigl\{\, |x-m|,\ |x-(m+1)| \,\bigr\} , so ψ(x)=xn\psi(x) = |x - n| for n=mn = m or n=m+1n = m+1.
  3. Range. 0ψ(x)1/20 \le \psi(x) \le 1/2 for every real xx (Intervals of R\mathbb{R}: the nine order-convex forms, nondegeneracy, and length).
  4. Zero set. ψ(x)=0\psi(x) = 0 if and only if xZx \in \mathbb{Z}.

Why this example is here. It is the standard uniformly continuous function of this track that is not defined by a formula in the field operations, and it is obtained from the metric machinery rather than rebuilt: claim 1 is d(x,A)d(y,A)d(x,y)|d(x,A) - d(y,A)| \le d(x,y), so the distance to a fixed nonempty set is 11-Lipschitz applied to A:=ZA := \mathbb{Z} in the metric space R\mathbb{R}, transported to a statement about a real function by Dictionary: for ARA \subseteq \mathbb{R} with the metric d(x,y)=xyd(x,y) = |x-y|, continuity and uniform continuity of f:ARf : A \to \mathbb{R} agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of R\mathbb{R} is compact in the open-cover sense of R\mathbb{R} exactly when it is a compact metric subspace, clause 3. Only claims 2 to 4, which compute the value, need an argument of their own.

The same function is computed elsewhere, and nothing here depends on that. The trigonometry-free oscillator ψ(x)=infnZxn\psi(x) = \inf_{n \in \mathbb{Z}} |x - n| is well defined and attained at a nearest integer, takes values in [0,1/2][0, 1/2], vanishes exactly on Z\mathbb{Z}, equals 1/21/2 at half-integers, and is 11-periodic introduces ψ\psi on the companion page of The ε\varepsilon-δ\delta limit limxcf(x)=L\lim_{x \to c} f(x) = L of f:ARf : A \to \mathbb{R} at a limit point cc of AA and proves the same computation together with 11-periodicity and the value at half-integers. That item lives on an examples page, which is a leaf of the dependency graph, so no item may rest on it; the verification below is therefore self-contained, and the duplication is deliberate rather than an oversight.

Facts & Assumptions

Given: R\mathbb{R} with the metric d(x,y)=xyd(x,y) = |x-y|; the canonical copy of Z\mathbb{Z} inside R\mathbb{R}; a real xx, the integer m:=xm := \lfloor x \rfloor and the real t:=xmt := x - m; and ψ(x):=d(x,Z)\psi(x) := d(x,\mathbb{Z}).

[L1]

Distance to a nonempty set: for nonempty AXA \subseteq X in a metric space, d(x,A):=inf{d(x,a):aA}d(x,A) := \inf\{\, d(x,a) : a \in A \,\}, the infimum existing because the set of distances is nonempty and bounded below by 00 (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Greatest lower bound (infimum), Metric space: d(x,y)=0d(x,y) = 0 iff x=yx = y, symmetry, and the triangle inequality; pseudometric and ultrametric).

[L2]

d(x,A)d(y,A)d(x,y)|d(x,A) - d(y,A)| \le d(x,y): the distance to a fixed nonempty set is 11-Lipschitz as a map of metric spaces (d(x,A)d(y,A)d(x,y)|d(x,A) - d(y,A)| \le d(x,y), so the distance to a fixed nonempty set is 11-Lipschitz).

[L4]

R\mathbb{R} with d(x,y)=xyd(x,y) = |x-y| is a metric space (The absolute value makes R\mathbb{R} a metric space: d(x,y)=xyd(x,y) = |x-y| is a metric, its open balls are the intervals (xr,x+r)(x-r, x+r), and it is unbounded), and Z\mathbb{Z} sits inside R\mathbb{R} as a totally ordered subring containing 00 and 11 and closed under nn±1n \mapsto n \pm 1, with no integer strictly between nn and n+1n+1 (The integers as equivalence classes of pairs of naturals, The naturals embed in the integers, The integers embed in the rationals, The rationals embed densely in the reals).

[L5]

Integer part: for every real xx there is exactly one integer mm with mx<m+1m \le x < m+1 (Integer part: for every real xx there is exactly one integer mm with mx<m+1m \le x < m + 1).

[L6]

Infimum and minimum: a lower bound of a set that belongs to the set is its infimum and its minimum (Greatest lower bound (infimum), Maximum and minimum of a set); and the minimum of a two-element set of reals exists and is one of the two (Every nonempty finite set of reals has a maximum and a minimum).

[L7]

Absolute value and order: u0|u| \ge 0; u=0|u| = 0 exactly when u=0u = 0; u=u|u| = u for u0u \ge 0 and u=u|u| = -u for u0u \le 0; the order is total; 1/2>01/2 > 0 and 11/2=1/21 - 1/2 = 1/2 (Basic properties of the absolute value, Ordered field, Intervals of R\mathbb{R}: the nine order-convex forms, nondegeneracy, and length).

Verification

technique · direct
1.1

Z\mathbb{Z} is a nonempty subset of the metric space R\mathbb{R}, since 0Z0 \in \mathbb{Z}, so ψ(x)=d(x,Z)\psi(x) = d(x,\mathbb{Z}) is defined for every real xx by [L1], and ψ(x)0\psi(x) \ge 0.

L1L4L7
1.2

By [L5] the integer m=xm = \lfloor x \rfloor satisfies mx<m+1m \le x < m+1, so t=xmt = x - m satisfies 0t<10 \le t < 1, and (m+1)x=1t(m+1) - x = 1 - t satisfies 0<1t10 < 1 - t \le 1.

L5L7
2.1

Claim 1. By [L2], ψ(x)ψ(y)=d(x,Z)d(y,Z)d(x,y)=xy|\psi(x) - \psi(y)| = |d(x,\mathbb{Z}) - d(y,\mathbb{Z})| \le d(x,y) = |x-y| for all real x,yx,y; by [L3] this says exactly that ψ\psi is Lipschitz with constant 11 as a real function on R\mathbb{R}, hence uniformly continuous on R\mathbb{R} and continuous on R\mathbb{R}.

step 1.1L2L3
2.2

Every distance from xx to an integer is at least min{t,1t}\min\{t, 1-t\}. Let nZn \in \mathbb{Z}. By [L4] and totality either nmn \le m or m<nm < n, and in the second case m+1nm+1 \le n. If nmn \le m then xnxm=t0x - n \ge x - m = t \ge 0, so xn=xnt|x-n| = x-n \ge t. If m+1nm+1 \le n then nx(m+1)x=1t>0n - x \ge (m+1) - x = 1-t > 0, so xn=nx1t|x-n| = n-x \ge 1-t. Either way xnmin{t,1t}|x-n| \ge \min\{t, 1-t\}.

step 1.2L4L6L7
2.3

Both candidate values occur. Since t0t \ge 0 we have t=xmt = |x-m|, and since 1t>01-t > 0 we have 1t=x(m+1)1-t = |x-(m+1)|, with mm and m+1m+1 in Z\mathbb{Z}.

step 1.2L4L7
3.1

Claim 2. By steps 2.2 and 2.3 the real min{t,1t}\min\{t,1-t\} is a lower bound of {xn:nZ}\{\, |x-n| : n \in \mathbb{Z} \,\} belonging to that set, so by [L6] it is the infimum and the minimum: ψ(x)=min{t,1t}=min{xm,x(m+1)}\psi(x) = \min\{t,1-t\} = \min\{|x-m|, |x-(m+1)|\}, attained at n=mn = m or n=m+1n = m+1.

step 2.2step 2.3L1L6
4.1

Claim 3. ψ(x)0\psi(x) \ge 0 since t0t \ge 0 and 1t>01-t > 0. And ψ(x)1/2\psi(x) \le 1/2: if t1/2t \le 1/2 then ψ(x)t1/2\psi(x) \le t \le 1/2, while if t>1/2t > 1/2 then 1t<11/2=1/21-t < 1-1/2 = 1/2 and ψ(x)1t<1/2\psi(x) \le 1-t < 1/2. So 0ψ(x)1/20 \le \psi(x) \le 1/2.

step 1.2step 3.1L7
4.2

Claim 4. If ψ(x)=0\psi(x) = 0 then min{t,1t}=0\min\{t,1-t\} = 0 by step 3.1; since 1t>01-t > 0 this forces t=0t = 0, that is x=mZx = m \in \mathbb{Z}. Conversely if xZx \in \mathbb{Z} then xx=0|x-x| = 0 is a member of the set of distances and 00 is a lower bound of it by step 1.1, so ψ(x)=0\psi(x) = 0 by [L6].

step 1.1step 1.2step 3.1L6L7
5.1

Claims 1 to 4 are verified: ψ\psi is 11-Lipschitz and therefore uniformly continuous and continuous on R\mathbb{R}, its value at xx is min{t,1t}\min\{t, 1-t\} and is attained at a nearest integer, its values lie in [0,1/2][0,1/2], and it vanishes exactly on Z\mathbb{Z}.

step 2.1step 3.1step 4.1step 4.2

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