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A nowhere-vanishing real function algebra on a compact space approximates the constant one

Statement

Let $X$ be a compact Hausdorff space and let $A \subseteq C (X, R)$ be a nowhere-vanishing real function algebra, not necessarily unital. Then the constant-one function belongs to the uniform closure of $A$ : for every $ε > 0$ there is $u \in A$ such that $∣ u (x) - 1 ∣ < ε for every x \in X .$

Facts & Assumptions

Given: A compact Hausdorff space $X$ , a nowhere-vanishing real function algebra $A \subseteq C (X, R)$ , and a real $ε > 0$ .

[L1]

If an indexed family of open subsets of an ambient space covers a compact subset, finitely many indexed members cover it, with the empty-set case stated separately (A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, clause 2).

[L2]

If $X$ is compact and nonempty and $g : X \to R$ is continuous, then $g [X]$ has a maximum and a minimum (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, clause 2).

[L3]

For $a \leq b$ , every continuous real function on $[a, b]$ is a uniform limit of polynomials (Polynomials are uniformly dense in $C ([a, b], R)$ for every closed interval).

[L4]

A nowhere-vanishing real function algebra has, for every $x \in X$ , some $a \in A$ with $a (x) \neq 0$ , and it is closed under real linear combinations and pointwise products (Unital, point-separating, and nowhere-vanishing real function algebras on a compact Hausdorff space).

Proof

technique · direct

1.1L4

If $X = \emptyset$ , then the unique empty function is simultaneously the zero and constant-one function and belongs to the vector subspace $A$ , so the conclusion is immediate.

1.2L4

Assume $X \neq \emptyset$ . For each $a \in A$ let $U_{a} : = {x \in X : a (x) \neq 0}$ ; these sets are open by continuity, and they cover $X$ by the nowhere-vanishing clause in [L4].

2.1step 1.2L1L4algebra

By [L1], finitely many $U_{a_{0}}, \dots, U_{a_{n}}$ cover $X$ . The function $h : = a_{0}^{2} + \dots + a_{n}^{2}$ lies in $A$ and satisfies $h (x) > 0$ for every $x \in X$ .

3.1step 2.1L2

By [L2], $h$ has a minimum $m$ and maximum $M$ ; step 2.1 gives $0 < m \leq M$ .

4.1step 3.1L4algebra

If $m = M$ , then $h$ is the positive constant $m$ and $u : = m^{- 1} h \in A$ is exactly the constant-one function.

4.2step 3.1L3L4choose

If $m < M$ , use [L3] to choose a polynomial $p$ satisfying $∣ p (t) - 1 / t ∣ < ε / M$ on $[m, M]$ ; then $u : = h p (h)$ lies in $A$ , because the polynomial $t \mapsto t p (t)$ has zero constant term.

5.1step 4.1step 4.2step 3.1algebra∎

In the case of step 4.2, every $x \in X$ satisfies $∣ u (x) - 1 ∣ = h (x) ∣ p (h (x)) - 1 / h (x) ∣ < h (x) ε / M \leq ε$ ; together with step 4.1 this proves the claim in all cases.

Depends on

Used by

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

Dependency tree · two levels

28 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, proof of Theorem 1.26 (standard reference, not scraped)