Stone Weierstrass, Lattice of Functions is Dense in c(S)

Approaching the Floor

Let L be a nonempty lattice of functions on a compact space S, using the ≤ relation, as described in the previous section.

For any x in S, let h(x) be the greatest lower bound of fi(x), for all fi in L.  Assume h(S) is well defined and continuous.  A sequence of functions from L approaches h uniformly.

Given ε, and x, find i such that fi(x) is between hx and hx+ε/3.  Then find an open set Ox that keeps h within ε/3 of h(x), and fi within ε/3 of fi(x).  Now h does not stray by more than ε across Ox.  Since finitely many open sets cover S, there are finitely many functions fi to consider, so set f to the minimum of these functions.  Since L is a lattice, f is in L.  Thus f is the function that is uniformly within ε of h.

Separating x from a closed Set

Assume there is a function f in L, for any pair of points x and y, such that f(x) ≠ f(y).  Thus S is hausdorff, and normal.  Further, assume f in L implies cf and c+f are in L, for any real constant c.

Let S have at least two points to separate, so that L is nonempty.  Hence 0×f+c implies the existence of every constant function c in L.

Given x and y in S, scale and translate, and L contains a function that maps x to a and y to b, for any real numbers a and b.

Now let's separate x from a closed set C.  Let a and b be real numbers, with a < b.  Find functions fy mapping x to a and y to b+1 for every y in C.  By continuity fy remains above b for an open set about y.  Since C is compact, finitely many functions will serve.  Let f be the maximum of all these functions, and the constant function a, which exists, since L is a lattice.  Now f(S) ≥ a, f(x) = a, and f(C) > b.

L is Dense in c(S)

Next consider an arbitrary continuous function h in c(S), and let L′ be the lattice inside L with functions ≥ h.  Note that min and max remain above h, so L′ is a lattice.  Since h has an upper bound b, and the constant function b is in L′, L′ is nonempty.  We wish to show h is pointwise the greatest lower bound of the functions in L′.

Remember that b is the upper bound on h.  Fix a point x in S.  For any ε, find an open set containing x such that h does not stray by more than ε, and let C be the complement.  Set a = h(x) + ε, and there is some function f in L with f ≥ a, f(x) = a, and f(C) > b.  Verify that f ≥ h, hence f is in L′.  This can be done for each ε, hence the functions of L′, applied to x, approach h(x).

The above holds for every x, so h(S) becomes the floor of L′.  Apply the theorem at the top of this page, and functions in L′ approach h uniformly.

This holds for any continuous function h, hence every function in c(S) is approximated uniformly from above in L.  (We could approach from below; it's just a matter of taste.)  The closure of such a lattice in the metric space c(S) is all of c(S), and L is dense in c(S).