Let A" be a set, A a family of real-valued functions on X which contains the constants, pA the weak uniformity generated by A, and U{nAX) the collection of uniformly continuous functions to the real line R.The problem is how to construct U(\iAX) from A. The main result here is: For A a vector lattice, the collection of suprema of countable, finitely Aequiuniform, order-one subsets of A + is uniformly dense in U(pAX).Two less technical corollaries: If A is a vector lattice (resp., vector space), then the collection of functions which are finitely A-uniform and uniformly localfy-A (resp., uniformly locally piecewise-A) is uniformly dense in U(pAX).Further, for any A, a finitely A-uniform function is fust a composition F (a,,... ,af) for some a{.^ e A and F uniformly continuous on the range of (fli.ap) in Rf.Thus, such compositions are dense in U(iiAX).For BU(iiAX), the compositions with F e BU(Rr) are dense (B denoting bounded functions).So, in a sense, to know U(pAX) it suffices to know A and subspaces of the spaces Rp, and to know BU(pAX), A and the spaces Rr suffice.In case A is a vector lattice and A ** BA (i.e., A consists of bounded functions), the problem of describing U(pAX) in terms of A has an elegant solution: A is uniformly dense in U(pAX) (i.e., if / G U(pAX) and t > 0 there is a E A with \f(x) -a(x)\ < e for all x E A").This can be seen to be essentially equivalent to the usual Stone-Weierstrass Theorem.The result appears to have been first published in 1955 by Maak [M] and Nbeling and Bauer [NB]; we give a short proof in the course of proving our main theorem.If A contains unbounded functions, the situation is more complicated: A = U(R)\N is a "uniformly closed" vector lattice, while pA is (uniformly) discrete.With X = R, let A be the piecewise linear functions (finitely many pieces); pA is the usual uniformity on R and the closure of A consists only of continuous eventually linear functions.
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