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Since Bd (x, 54 rx ) ⊂ Bd (y, 14 ry + 54 rx ), we must have that 14 ry + 54 rx ≥ ry . It follows that ry ≤ 53 rx . Since x ∈ Bd (y, 14 ry ), we have that Bd (y, 14 ry ) ⊂ Bd (x, 24 ry ) ⊂ Bd (x, 56 rx ) ⊂ Ux . We have shown that St(x, V) ⊂ Ux . The next result indicates a close connection between full normality and the existence of continuous pseudometrics. Let d be a pseudometric of X. A cover L of X is d-uniform if there exists r > 0 such that the family {Bd (x, r) : x ∈ X} is a refinement of L.

Then we have that fV (x) = 1 and, for each y ∈ X V , we have that fV (y) = 0. As a consequence, we have that Bd (x, 1) ⊂ BdV (x, 1) ⊂ V . By the foregoing, the family {Bd (z, 1) : z ∈ X} refines U. 2. Applications. In this section, we shall give some applications of our first method of construction of pseudometrics. The second method will be used in the next chapter, with paracompact spaces. Our first result gives a topological characterization of pseudometrizability of a space. 41 1 Theorem A space X is pseudometrizable iff X has a sequence Un 2 such that, for every x ∈ X, the sequence St (x, Un ) n∈N n∈N of open covers is a nbhd base at x.

Then the formula d(x, y) = |f (x) − f (y)| defines a (continuous) pseudometric d of X. Proof. Exercise. We are going to define certain pseudometrics as supremums of pseudometrics of the type appearing in Lemma 9. The problem here is that a supremum of continuous pseudometrics is not necessarily continuous, if the supremum has infinitely many factors. To overcome this difficulty, we need to introduce some new concepts and results. 10 Definition Let L be a family of subsets of X and let p be a point of X.

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