Stone Weierstrass, Properties of c(S)

c(S) forms a Ring

Let c(S) be the set of continuous functions from a space S into the reals. We will prove c(S) is an R vector space and a ring.

Scale a function f by 0 and all of S maps to 0, which is continuous. Otherwise let c be nonzero. If f(x) = m, keep cf(x) to within ε of cm by restricting f to a neighborhood about x whose image lies in m±ε/c. Thus cf is continuous, and scaling by R is valid.

If f(x) = m and g(x) = n, an open set about x keeps f from straying by more than ε/2, and an open set about x keeps g from straying by more than ε/2. Intersect these open sets, and f+g is within ε of m+n. Thus f+g is continuous, and c(S) is an R vector space.

Let f(x) = m and g(x) = n, where m and n are nonzero. Choose open sets so that f does not stray by more than ε/3n and g does not stray by more than ε/3m. Intersect these open sets, and fg strays from mn by no more than 2ε/3+ε²/9mn. Keep ε less than 3mn and we're all right.

If m = 0, keep f within ε/2n of 0 and keep g within ε of n. The product strays by no more than ε/2+ε²/2n. Keep ε below n and we're all right.

If m = n = 0, keep f within ε of 0 and keep g below 1.

Those are all the cases, hence the product function is continuous. The 0 function and the 1 function act as identities, and c(S) forms a ring.

If f is nonzero everywhere, its inverse is in c(S). Let f(x) = m, and don't let f stray by more than ε. The inverse doesn't stray from 1/m by more than ε over m×(m+ε). With ε less than half of m, the difference is bounded by 2ε/m². This goes to 0 as ε approaches 0.

c(S) forms a Lattice

Define f ≤ g when f(x) ≤ g(x) for all x in S. This is a partial ordering on c(S).

The min and max of f and g are continuous functions. Keep f and g within ε of m and n respectively, and their max will remain within ε of max(m,n). With max as join and min as meet, ≤ turns c(S) into a lattice.

Normal

If S is normal, any pair of points x and y implies a function f in c(S) with f(x) = 0 and f(y) = 1. In other words, x and y can be separated. This is a consequence of urysohn's lemma.

Compact implies Complete Metric Space

Assume S is compact. If S is hausdorff it is also normal, whence the previous theorem applies.

The continuous image of a compact set is compact, hence each f(S) is closed and bounded.

Let |f| be the upper bound of the absolute value of f(S). Since f(S) is closed there is some x in S with f(x) = |f|. Note that f = 0 iff |f| = 0.

Let the distance from f to g be |f-g|. This is a symmetric operator. Suppose the triangular inequality fails. Thus |f(S)|+|g(S)|-|f(S)+g(S)| becomes negative, and is negative for a specific x. However, the images of x would violate the triangular inequality in R¹. Therefore |f,g| is a distance metric, and c(S) is a metric space.

If a sequence of functions is cauchy, then that sequence, applied to x in S, produces a cauchy sequence in R¹, which has a limit. The limit of a cauchy sequence of functions attains the limit at each x. We need to prove this is a continuous function. If all functions beyond f_n are within ε of the limit function g, and an open set keeps f_n within ε of f_n(x), then g is within 3ε of g(x) on that same open set, assuring continuity. The limit function is part of c(S), and c(S) forms a complete metric space.