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LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26
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∣d(x,A)−d(y,A)∣≤d(x,y), so the distance to a fixed nonempty set is 1-Lipschitz

Statement

Let (X,d) be a metric space (Metric space: d(x,y)=0 iff x=y, symmetry, and the triangle inequality; pseudometric and ultrametric), let A⊆X be nonempty and let x,y∈X. Then

∣d(x,A)−d(y,A)∣≤d(x,y),

with d(⋅,A) the distance to a nonempty set (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space). Thus the real-valued function u↦d(u,A) changes by at most d(u,v) between u and v: it is 1-Lipschitz.

Facts & Assumptions

Given: A metric space (X,d), a nonempty A⊆X, and points x,y∈X; write E(u):={ d(u,a):a∈A } for u∈X.

[A1]

The triangle inequality (M3) of Metric space: d(x,y)=0 iff x=y, symmetry, and the triangle inequality; pseudometric and ultrametric: d(u,a)≤d(u,v)+d(v,a) for all u,v∈X and a∈A.

[L2]

The infimum is a lower bound of its set and is the greatest such: inf⁡S≤s for every s∈S, and ℓ≤inf⁡S for every lower bound ℓ of S (Greatest lower bound (infimum)).

[L3]

Adding a constant to an inequality: if a≤b then a+c≤b+c. Order is preserved by adding a constant and by adding inequalities states the strict form only; the nonstrict form used here is that form together with the case a=b, settled by totality of the order (Ordered field, Complete ordered field (least-upper-bound property)).

[L4]

For every real u, ∣u∣ equals u or −u (Basic properties of the absolute value, Absolute value in an ordered field).

Proof

technique · direct
1.1givenL1

Both d(x,A) and d(y,A) are defined real numbers, since A is nonempty.

1.2A1

For every a∈A: d(x,a)≤d(x,y)+d(y,a).

1.3A1

For every a∈A: d(y,a)≤d(y,x)+d(x,a), and d(y,x)=d(x,y) by symmetry (M2), so d(y,a)≤d(x,y)+d(x,a).

2.1step 1.1step 1.2L2L3

For every a∈A: d(x,A)≤d(x,a), since d(x,A) is a lower bound of E(x) and d(x,a)∈E(x); combining with step 1.2 gives d(x,A)≤d(x,y)+d(y,a), hence d(x,A)−d(x,y)≤d(y,a).

2.2step 1.1step 1.3L2L3

For every a∈A: d(y,A)≤d(y,a)≤d(x,y)+d(x,a) by the same reasoning with the roles of x and y exchanged, hence d(y,A)−d(x,y)≤d(x,a).

3.1step 2.1L2L3

The real number d(x,A)−d(x,y) is therefore a lower bound of E(y), so it is at most the greatest lower bound: d(x,A)−d(x,y)≤d(y,A), that is d(x,A)−d(y,A)≤d(x,y).

3.2step 2.2L2L3

Symmetrically d(y,A)−d(x,y) is a lower bound of E(x), so d(y,A)−d(x,A)≤d(x,y).

4.1step 3.1step 3.2L4∎

By [L4] the value ∣d(x,A)−d(y,A)∣ is d(x,A)−d(y,A) or its negative d(y,A)−d(x,A), and steps 3.1 and 3.2 bound both by d(x,y); hence ∣d(x,A)−d(y,A)∣≤d(x,y).

Remarks

Depends on

Dependency tree · 2 levels

Sources