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The compact binding target exists and has a unique lexicographic tie-break

Statement

Let Θ⊆Rp be the nonempty compact parameter set of Binding function and indirect-inference target, and fix an admissible s for which its objective qs:Θ→R is continuous. Then Ms:=arg min⁡θ∈Θqs(θ) is a nonempty compact set and has a unique lexicographically least element. Thus the target selector specified in Binding function and indirect-inference target exists.

Facts & Assumptions

Given: A nonempty compact metric space Θ with its Euclidean metric, and a continuous real-valued objective qs.

[F1]

The binding-target definition specifies a nonempty compact Θ⊆Rp with its Euclidean subspace metric, continuity of qs, and lexicographic coordinate tie-breaking. Binding function and indirect-inference target

[F2]

A continuous real-valued function on a nonempty compact metric space attains its minimum. A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value

[F3]

A closed subset of a compact metric space is compact. A closed subset of a compact metric space is compact

[F4]

Continuity at θ means that for every ε>0 there is δ>0 such that d(η,θ)<δ implies ∣qs(η)−qs(θ)∣<ε. Continuity of a map between metric spaces, at a point and globally, in the ε-δ form

Proof

technique · direct
1.1F1F2given

By [F2], qs attains a minimum c on Θ. Hence Ms={θ∈Θ:qs(θ)=c} is nonempty.

2.1F1F4step 1.1algebra

The set Ms is closed in Θ. Indeed, if θ∉Ms, then qs(θ)>c; take ε=(qs(θ)−c)/2>0. By [F4], some δ>0 makes d(η,θ)<δ imply ∣qs(η)−qs(θ)∣<ε, and therefore qs(η)>c. This ball misses Ms, so every point of its complement has a neighborhood in that complement.

3.1F1F3step 2.1

By [F3], Ms is compact.

4.1F2F3F5step 3.1givenconstructinduction

Set M0:=Ms. For k=1,…,p, suppose Mk−1 is nonempty compact. The coordinate map πk(x)=xk is continuous because ∣πk(x)−πk(y)∣≤∥x−y∥2. By [F2], it attains a minimum ak on Mk−1. Define Mk:={x∈Mk−1:xk=ak}; this set is nonempty. It is closed in Mk−1: for xk≠ak, a ball of radius ∣xk−ak∣/2 around x misses Mk by the same coordinate inequality. Hence [F3] makes Mk compact. Finite induction constructs nonempty compact M1,…,Mp.

5.1F1step 4.1algebra∎

The nonempty set Mp contains a vector whose kth coordinate is ak for each k, so it contains at most one vector and is a singleton. For any x∈Ms, take the first coordinate k at which xk≠ak, if one exists. Then x∈Mk−1 and the definition of ak gives xk>ak. Thus the unique element of Mp is lexicographically no larger than every member of Ms, and is the unique lexicographically least minimizer.

Depends on

Used by

Cited to discharge well-definedness by Binding function and indirect-inference target.

Dependency tree · two levels

38 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources