TheoremBase

Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces

Open sets, closed sets, points and closed balls of a metric space are Borel; the open sets of a subspace are the traces of open sets of the space; traces of Borel sets on a subspace are Borel in it and generate its Borel sets, and for a Borel subspace these are exactly the Borel subsets of the space it contains.

Statement

In the setting of Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices, let (X,d)(X,d) be a metric space, with the topology Td\mathcal{T}_{d} of Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces (iii) and the Borel σ\sigma-algebra B(X)\mathcal{B}(X).

Every subset of XX that is open in (X,d)(X,d) belongs to B(X)\mathcal{B}(X).

Every subset of XX that is closed in (X,Td)(X,\mathcal{T}_{d}) belongs to B(X)\mathcal{B}(X).

For every x∈Xx\in X, {x}∈B(X)\{x\}\in\mathcal{B}(X).

For every x∈Xx\in X and every real r≥0r\ge0, the closed ball Bˉd(x,r)\bar{B}_{d}(x,r) belongs to B(X)\mathcal{B}(X).

Let Y⊆XY\subseteq X, let dY=d∣Y×Yd_{Y}=d|_{Y\times Y} be the restriction of dd, a metric on YY by Subsets of a Metric Space: the Restricted Metric, Sequences, Limits in Closed Subsets and Completeness of Closed Subsets (i), and let B(Y)=B(Y,dY)\mathcal{B}(Y)=\mathcal{B}(Y,d_{Y}). Let {B∩Y:B∈B(X)}\{B\cap Y:B\in\mathcal{B}(X)\} be the image of B(X)\mathcal{B}(X) under the map B↦B∩YB\mapsto B\cap Y from B(X)\mathcal{B}(X) to the power set P(Y)\mathcal{P}(Y) given by Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (ii).

A subset UU of YY is open in (Y,dY)(Y,d_{Y}) if and only if U=V∩YU=V\cap Y for some VV open in (X,d)(X,d).

B∩Y∈B(Y)B\cap Y\in\mathcal{B}(Y) for every B∈B(X)B\in\mathcal{B}(X), and B(Y)\mathcal{B}(Y) is the σ\sigma-algebra on YY generated by {B∩Y:B∈B(X)}\{B\cap Y:B\in\mathcal{B}(X)\}.

If Y∈B(X)Y\in\mathcal{B}(X), then B(Y)={B∩Y:B∈B(X)}={B∈B(X):B⊆Y}\mathcal{B}(Y)=\{B\cap Y:B\in\mathcal{B}(X)\}=\{B\in\mathcal{B}(X):B\subseteq Y\}.

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