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ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The distance from a point to a nonempty compact set is attained at a point of that set, and two disjoint compact sets are at positive distance

Example

Let (X,d)(X,d) be a metric space (Metric space: d(x,y)=0d(x,y) = 0 iff x=yx = y, symmetry, and the triangle inequality; pseudometric and ultrametric), with d(x,A)d(x,A) and d(A,B)d(A,B) the distances from a point to a nonempty set and between two nonempty sets (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space). Then:

  1. If KXK \subseteq X is nonempty and compact (Open cover, subcover, compact metric space, and compact subset of a metric space) and xXx \in X, there is aKa \in K with d(x,a)=d(x,K)d(x,a) = d(x,K): the infimum defining the distance is attained.
  2. If K,LXK, L \subseteq X are nonempty, compact and disjoint, then d(K,L)>0d(K,L) > 0, and again the value is attained at a point of KK.

Neither statement holds for arbitrary closed sets, and neither uses a choice principle.

Facts & Assumptions

Given: A metric space (X,d)(X,d), nonempty compact subsets KK and LL of XX, and a point xXx \in X.

[L1]

For nonempty AXA \subseteq X, d(x,A)=inf{d(x,y):yA}d(x,A) = \inf\{d(x,y) : y \in A\} and d(A,B)=inf{d(u,v):uA, vB}d(A,B) = \inf\{d(u,v) : u \in A,\ v \in B\}; an infimum is a lower bound of its set and is at least every lower bound (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Greatest lower bound (infimum), Epsilon characterisation of the infimum).

[L5]

A minimum of a set of reals is a member of it and bounds it below (Maximum and minimum of a set).

Verification

technique · direct
1.1

The map f:KRf : K \to \mathbb{R}, f(u):=d(x,{u})=d(x,u)f(u) := d(x,\{u\}) = d(x,u), is the restriction to KK of ud(u,{x})u \mapsto d(u,\{x\}), which is Lipschitz with constant 11 and hence continuous; KK with the restricted metric is a nonempty compact metric space.

L2L3
2.1

So ff attains a least value at some aKa \in K: d(x,a)d(x,u)d(x,a) \le d(x,u) for every uKu \in K.

L3L5step 1.1
3.1

Hence d(x,a)d(x,a) is a lower bound of {d(x,u):uK}\{d(x,u) : u \in K\} that belongs to the set, so it is the infimum: d(x,a)=d(x,K)d(x,a) = d(x,K), which is claim 1.

L1L5step 2.1
4.1

For claim 2, the map g:KRg : K \to \mathbb{R}, g(u):=d(u,L)g(u) := d(u,L), is continuous by the same argument, so it attains a least value at some bKb \in K.

L2L3step 3.1
5.1

g(b)>0g(b) > 0: otherwise d(b,L)=0d(b,L) = 0 would put bb in the closure of LL, which equals LL because LL is compact and hence closed, contradicting KL=K \cap L = \emptyset.

L4step 4.1
6.1

d(K,L)=g(b)d(K,L) = g(b): every uKu \in K and vLv \in L satisfy d(u,v)d(u,L)g(b)d(u,v) \ge d(u,L) \ge g(b), so g(b)g(b) is a lower bound of {d(u,v):uK,vL}\{d(u,v) : u \in K, v \in L\}; and d(K,L)d(b,v)d(K,L) \le d(b,v) for every vLv \in L, so d(K,L)d(K,L) is a lower bound of {d(b,v):vL}\{d(b,v) : v \in L\} and therefore d(K,L)d(b,L)=g(b)d(K,L) \le d(b,L) = g(b).

L1L5step 4.1step 5.1
7.1

Combining, d(K,L)=g(b)>0d(K,L) = g(b) > 0 and the value is attained at bKb \in K: claim 2.

step 5.1step 6.1

Remarks

Compactness is what makes the infimum a minimum. For a merely closed set the infimum need not be attained and disjoint closed sets can be at distance zero; what claim 1 uses is the extreme value theorem, and claim 2 additionally uses that a compact set is closed (A compact subset of a metric space is closed and bounded).

Only one of the two sets has to be compact for the attainment in claim 1, the point xx playing the role of a one-point compact set. In claim 2 compactness of KK gives the attainment and closedness of LL gives the positivity, which is why the proof calls on the two properties in different places.

Depends on

Used by

Nothing in the library uses this result yet.

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Sources