LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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$|d(x,A) - d(y,A)| \le d(x,y)$ , 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 $A \subseteq X$ be nonempty and let $x, y \in X$ . Then

$|d(x,A) - d(y,A)| \le d(x,y),$

with $d(\cdot,A)$ 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 $u \mapsto d(u,A)$ 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 $A \subseteq X$ , and points $x, y \in X$ ; write $E(u) := \{\, d(u,a) : a \in A \,\}$ for $u \in X$ .

[A1]

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

[L1]

For nonempty $A$ the real number $d(u,A) = \inf E(u)$ exists, because $E(u)$ is nonempty and bounded below by $0$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Nonnegativity of a metric is a consequence of the other axioms, not an axiom, Every nonempty set bounded below has an infimum).

[L2]

The infimum is a lower bound of its set and is the greatest such: $\inf S \le s$ for every $s \in S$ , and $\ell \le \inf S$ for every lower bound $\ell$ of $S$ (Greatest lower bound (infimum)).

[L3]

Adding a constant to an inequality: if $a \le b$ then $a + c \le b + c$ . 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

1.1

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

givenL1

1.2

For every $a \in A$ : $d(x,a) \le d(x,y) + d(y,a)$ .

1.3

For every $a \in A$ : $d(y,a) \le d(y,x) + d(x,a)$ , and $d(y,x) = d(x,y)$ by symmetry (M2), so $d(y,a) \le d(x,y) + d(x,a)$ .

2.1

For every $a \in A$ : $d(x,A) \le d(x,a)$ , since $d(x,A)$ is a lower bound of $E(x)$ and $d(x,a) \in E(x)$ ; combining with step 1.2 gives $d(x,A) \le d(x,y) + d(y,a)$ , hence $d(x,A) - d(x,y) \le d(y,a)$ .

step 1.1step 1.2L2L3

2.2

For every $a \in A$ : $d(y,A) \le d(y,a) \le d(x,y) + d(x,a)$ by the same reasoning with the roles of $x$ and $y$ exchanged, hence $d(y,A) - d(x,y) \le d(x,a)$ .

step 1.1step 1.3L2L3

3.1

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: $d(x,A) - d(x,y) \le d(y,A)$ , that is $d(x,A) - d(y,A) \le d(x,y)$ .

step 2.1L2L3

3.2

Symmetrically $d(y,A) - d(x,y)$ is a lower bound of $E(x)$ , so $d(y,A) - d(x,A) \le d(x,y)$ .

step 2.2L2L3

4.1

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 $|d(x,A) - d(y,A)| \le d(x,y)$ .

step 3.1step 3.2L4∎

Remarks

Where the nonemptiness is used. Twice, and both times essentially: it is what makes $d(x,A)$ and $d(y,A)$ exist at all (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), and it is what makes $E(x)$ and $E(y)$ nonempty so that "greatest lower bound" has content. For $A = \emptyset$ the statement has no meaning in this library, since $\inf \emptyset$ is undefined.
The point case is The reverse triangle inequality $|d(x,z) - d(y,z)| \le d(x,y)$ in any metric space: taking $A = \{z\}$ gives $E(u) = \{d(u,z)\}$ , whose infimum is $d(u,z)$ , and the conclusion becomes $|d(x,z) - d(y,z)| \le d(x,y)$ .
The constant $1$ is best possible in general: on $\mathbb{R}$ with $A = \{0\}$ the function is $u \mapsto |u|$ , and $\big||x| - |y|\big| = |x-y|$ whenever $x$ and $y$ have the same sign.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 38 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Hausdorff distance (Wikipedia) (standard reference, not scraped)
Lipschitz continuity (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (standard reference, not scraped)

∣d(x,A)−d(y,A)∣≤d(x,y)|d(x,A) - d(y,A)| \le d(x,y)∣d(x,A)−d(y,A)∣≤d(x,y), so the distance to a fixed nonempty set is 111-Lipschitz