TheoremBase

Sets and Functions Defined Componentwise on Families of Vector- and Matrix-Valued Functions

For a set of maps from a set S into RqR^q, or into the real p x q matrices, the maps all of whose components, or entries, lie in a given set form exactly one set, and applying a map to every component, or entry, defines exactly one map on that set.

Statement

In the setting of Euclidean Space and Real Matrices: Standing Notation, let p,q∈Np,q\in\mathbb{N} and let SS be a set. For a function f:S→Rqf:S\to\mathbb{R}^{q} and k∈[q]k\in[q], fk:S→Rf_{k}:S\to\mathbb{R} is its kk-th component function; for a function F:S→Mp×q(R)F:S\to\mathcal{M}_{p\times q}(\mathbb{R}), i∈[p]i\in[p] and j∈[q]j\in[q], Fij:S→RF_{ij}:S\to\mathbb{R} is its entry function in row ii and column jj. For sets JJ and BB, BJB^{J} is the set of functions from JJ to BB; for a function GG on [p]×[q][p]\times[q], G(i,j)G(i,j) means G((i,j))G((i,j)).

Componentwise subset, vector case. Let DD be a set of functions from SS to Rq\mathbb{R}^{q} and let AA be a set. There is exactly one set ZZ whose elements are exactly the f∈Df\in D with

fk∈Afor every k∈[q].f_{k}\in A\qquad\text{for every }k\in[q].

Componentwise function, vector case. Let DD, AA and ZZ be as in (i), let BB be a set and let Φ:A→B\Phi:A\to B be a function. There is exactly one function Ψ:Z→B[q]\Psi:Z\to B^{[q]} with

Ψ(f)(k)=Φ(fk)(f∈Z, k∈[q]).\Psi(f)(k)=\Phi(f_{k})\qquad(f\in Z,\ k\in[q]).

Componentwise subset, matrix case. Let DD be a set of functions from SS to Mp×q(R)\mathcal{M}_{p\times q}(\mathbb{R}) and let AA be a set. There is exactly one set ZZ whose elements are exactly the F∈DF\in D with

Fij∈Afor all i∈[p] and j∈[q].F_{ij}\in A\qquad\text{for all }i\in[p]\text{ and }j\in[q].

Componentwise function, matrix case. Let DD, AA and ZZ be as in (iii), let BB be a set and let Φ:A→B\Phi:A\to B be a function. There is exactly one function Ψ:Z→B[p]×[q]\Psi:Z\to B^{[p]\times[q]} with

Ψ(F)(i,j)=Φ(Fij)(F∈Z, i∈[p], j∈[q]).\Psi(F)(i,j)=\Phi(F_{ij})\qquad(F\in Z,\ i\in[p],\ j\in[q]).

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