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A separating real function algebra is dense or its closure consists exactly of the functions vanishing at one point

Statement

Let $X$ be a compact Hausdorff space and let $A \subseteq C (X, R)$ be a point-separating real function algebra, not necessarily unital. Exactly one of the following descriptions applies when $X$ is nonempty:

$A$ has no common zero, and its uniform closure is $C (X, R)$ ;
there is a unique $x_{0} \in X$ at which every member of $A$ vanishes, and the uniform closure of $A$ is exactly $I_{x_{0}} : = {f \in C (X, R) : f (x_{0}) = 0} .$

If $X = \emptyset$ , the first conclusion holds: $A = C (X, R)$ .

Facts & Assumptions

Given: A compact Hausdorff space $X$ and a point-separating real function algebra $A \subseteq C (X, R)$ .

[L1]

A real function algebra is a real vector subspace closed under pointwise multiplication; it is point-separating when each distinct pair is distinguished by one member, and nowhere-vanishing when each point has some member nonzero there (Unital, point-separating, and nowhere-vanishing real function algebras on a compact Hausdorff space).

[L2]

A nowhere-vanishing real function algebra on a compact space uniformly approximates the constant-one function (A nowhere-vanishing real function algebra on a compact space approximates the constant one).

[L3]

Every unital point-separating real function algebra on a compact Hausdorff space is uniformly dense in $C (X, R)$ (Real Stone–Weierstrass theorem for compact Hausdorff spaces).

[L4]

On nonempty $X$ , the topology of uniform convergence on $C (X, R)$ is the metric topology of the restricted uniform metric (Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on $Y^{X}$ and on $C (X, Y)$ ).

Proof

technique · direct

1.1L1

If $X = \emptyset$ , then $C (X, R)$ contains only the empty function, which is the zero element of the vector subspace $A$ , so the first conclusion holds.

1.2L1

Assume $X \neq \emptyset$ and let $Z : = {x \in X : a (x) = 0 for every a \in A}$ . Point separation implies that $Z$ has at most one element, because two distinct members of $Z$ could not be distinguished by any $a \in A$ .

1.3L1L3algebra

Let $A^{+} : = A + R 1 = {a + c 1 : a \in A, c \in R}$ , where $1$ is the constant-one function, itself continuous because the preimage of every open set under it is $\emptyset$ or $X$ . Sums and real multiples of such members again have this form, and $(a + c 1) (b + d 1) = (a b + d a + c b) + c d 1$ , so $A^{+}$ is a real function algebra in the sense of [L1]; it is unital by construction and point-separating because it contains $A$ . Hence [L3] makes $A^{+}$ uniformly dense in $C (X, R)$ .

2.1step 1.2L1L2

If $Z = \emptyset$ , then $A$ is nowhere-vanishing, so [L2] says that it uniformly approximates the constant-one function.

2.2step 1.2step 1.3L4choosealgebra

Suppose instead $Z = {x_{0}}$ , so that step 1.2 makes $x_{0}$ the unique point at which every member of $A$ vanishes. For $f \in I_{x_{0}}$ and $ε > 0$ , use step 1.3 to choose $a + c 1 \in A^{+}$ within $ε / 2$ of $f$ . Evaluating at $x_{0}$ , where $a (x_{0}) = 0$ and $f (x_{0}) = 0$ , gives $∣ c ∣ = ∣ a (x_{0}) + c - f (x_{0}) ∣ < ε / 2$ , so $∣ a (x) - f (x) ∣ \leq ∣ a (x) + c - f (x) ∣ + ∣ c ∣ < ε$ for every $x$ ; hence $I_{x_{0}} \subseteq \overline{A}$ .

2.3step 1.2L4

Conversely, still in the case $Z = {x_{0}}$ , if $f \in \overline{A}$ then for every $ε > 0$ some $a \in A$ satisfies $∣ f (x_{0}) - a (x_{0}) ∣ < ε$ ; since $a (x_{0}) = 0$ , this forces $f (x_{0}) = 0$ , so $\overline{A} \subseteq I_{x_{0}}$ .

3.1step 2.1step 1.3L1L4choosealgebra

Suppose $Z = \emptyset$ . Given $f \in C (X, R)$ and $ε > 0$ , use the density in step 1.3 to choose $a + c 1 \in A^{+}$ within $ε / 2$ of $f$ , then use step 2.1 to choose $u \in A$ within $ε / (2 (∣ c ∣ + 1))$ of $1$ ; the member $a + c u \in A$ satisfies $∣ a + c u - f ∣ \leq ∣ a + c 1 - f ∣ + ∣ c ∣ ∣ u - 1 ∣ < ε$ everywhere, so the uniform closure of $A$ is all of $C (X, R)$ .

4.1step 1.2step 3.1step 2.2step 2.3∎

The alternatives $Z = \emptyset$ and $Z = {x_{0}}$ exhaust step 1.2; step 3.1 gives the full closure in the first case, while steps 2.2 and 2.3 give exactly $I_{x_{0}}$ in the second.

Depends on

Used by

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

E. Carlen, Notes on Topology and the Stone-Weierstrass Theorem, Theorem 1.26 (standard reference, not scraped)