4.4.1. The Stone Theorem. Every open cover of a metrizable space has an open refinement which is both locally finite and σ-discrete.

Proof. Let {U_{s}}_{s∈S} be an open cover of a metrizable space X; take metric ρ on the space X and a well-ordering relation < on the set S. Define inductive families V_{i} = {V_{s,i}}_{s∈S} of subsets of X by letting

V_{s,i} = ⋃ B(c, 1/2^{i}),

where the union is taken over all points c ∈ X satisfying the following conditions:

(1)  s is the smallest element of S such that c ∈ U_{s}.

(2)  c ∉ V_{t,j} for j < i and t ∈ S.

(3)  B(c, 3/2^{i}) ⊂ U_{s}.

It follows from the definition that the sets V_{s,i} are open, and (3) implies that V_{s,i} ⊂ U_{s}. Let x be a point of X; take the smallest s ∈ S such that x ∈ U_{s} and a natural number i such that B(x, 3/2^{i}) ⊂ U_{s}. Clearly, we either have x ∈ V_{t,j} for a j < i and a t ∈ S or x ∈ V_{s,i}. Hence, the union V = ⋃_{i=1}^{∞} V_{i} is an open refinement of the cover {U_{s}}_{s∈S}.

We shall prove that for every i

(4)  if x_{1} ∈ V_{s_{1},i}, x_{2} ∈ V_{s_{2},i} and s_{1} ≠ s_{2}, then ρ(x_{1}, x_{2}) > 1/2^{i},

and this will show that the families V_{i} are discrete, because every 1/2^{i+1}-ball meets at most one member of V_{i}.

Let us assume that s_{1} < s_{2}. By the definition of V_{s_{1},i} and V_{s_{2},i} there exist points c_{1}, c_{2} satisfying (1)-(3) such that x_{k} ∈ B(c_{k}, 1/2^{i}) ⊂ V_{s_{k},i} for k = 1, 2. From (3) it follows that B(c_{1}, 3/2^{i}) ⊂ U_{s_{1}} and from (1) we see that c_{2} ∉ U_{s_{1}}, so that ρ(c_{1}, c_{2}) ≥ 3/2^{i}. Hence,

ρ(x_{1}, x_{2}) ≥ ρ(c_{1}, c_{2}) − ρ(c_{1}, x_{1}) − ρ(c_{2}, x_{2}) > 1/2^{i},

which proves (4).

To conclude the proof of the theorem it suffices to show that for every t ∈ S and any pair k, j of natural numbers

(5)  if B(x, 1/2^{k}) ⊂ V_{t,j}, then B(x, 1/2^{i+k}) ∪ V_{s,i} = ∅ for i ≥ j + k and s ∈ S,

because for every x ∈ X there exist k, j and t such that B(x, 1/2^{k}) ⊂ V_{t,j} and thus the ball B(x, 1/2^{j+k}) meets at most j + k − 1 members of V.

It follows from (2) that the points c in the definition of V_{s,i} do not belong to V_{t}, j whenever i ≥ j + k; since B(x, 1/2^{k}) ⊂ V_{t,j}, we have ρ(x, c) ≥ 1/2^{k} for any such c. The inequalities j + k ≥ k + 1 and i ≥ k + 1 imply that B(x, 1/2^{j+k}) ∪ B(c, 1/2^{i}) = ∅, and this yields (5). ♢