TheoremBase

The Open Sets of a Metric Space Form a Set, Given by a Predicative Formula

Being an open subset of a metric space is expressed by a predicative formula, so the open subsets form a set, the metric topology, whose elements are exactly the open subsets.

Statement

In the setting of Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices, let (X,d)(X,d) be a metric space, let << and 00 be the strict order and the zero of R\mathbb{R}, and let P(X)\mathcal{P}(X) be the power set of XX.

The condition on the set variable UU, with parameters XX, dd, R\mathbb{R}, << and 00, that U⊆XU\subseteq X and UU is open in (X,d)(X,d) is expressed by a predicative formula, namely by ω(U)\omega(U):

∀y (y∈U⇒y∈X)∧∀x (x∈U⇒∃r (r∈R∧0<r∧∀y ((y∈X∧∃z (((x,y),z)∈d∧z<r))⇒y∈U))),\forall y\,(y\in U\Rightarrow y\in X)\wedge\forall x\,\Bigl(x\in U\Rightarrow\exists r\,\bigl(r\in\mathbb{R}\wedge0<r\wedge\forall y\,\bigl((y\in X\wedge\exists z\,(((x,y),z)\in d\wedge z<r))\Rightarrow y\in U\bigr)\bigr)\Bigr),

where xx, yy, zz and rr are set variables distinct from UU and from one another, 0<r0<r and z<rz<r stand for (0,r)∈<(0,r)\in{<} and (z,r)∈<(z,r)\in{<}, and for x,y∈Xx,y\in X and r∈Rr\in\mathbb{R} the part ∃z (((x,y),z)∈d∧z<r)\exists z\,(((x,y),z)\in d\wedge z<r) holds if and only if d(x,y)<rd(x,y)<r.

There is exactly one set T\mathcal{T} such that, for every set UU, U∈TU\in\mathcal{T} if and only if U⊆XU\subseteq X and UU is open in (X,d)(X,d); it is {U∈P(X):ω(U)}\{U\in\mathcal{P}(X):\omega(U)\}, and it is the set denoted Td\mathcal{T}_{d} in Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces §topology.

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