Stone–Weierstrass in General

By [L4] and [L5], the metric topology makes $K$ a compact Hausdorff topological space.

By [L2] and the equivalence (a)$\iff$(b) in [L3], a function $K\to \mathbb{R}$ is continuous in the metric sense exactly when it is continuous for the metric topologies, so the two ambient sets $C(K,\mathbb{R})$ are equal.

The pointwise addition, scalar multiplication, and multiplication in [L1] and [L6] are the same operations on the common ambient set from step 1.2, and the constant-function and point-separation clauses have the same quantifiers; hence condition 1 implies condition 2 and condition 2 implies condition 1.

By [L1], choose $g\in L$ with $g(x)\mathrm{\ne }g(y)$ and put $a:=(\beta -\alpha )\mathrm{/}(g(y)-g(x))$ and $b:=\alpha -ag(x)$; the denominator is nonzero because $g$ separates $x$ and $y$.

The affine combination $h:=ag+b$ belongs to $L$ by [L1], and substitution gives $h(x)=ag(x)+\alpha -ag(x)=\alpha$ and $h(y)=\alpha +a(g(y)-g(x))=\beta$.

Fix $x\in X$ and let $H_x:=\{h\in L:h(x)=f(x)\}$; for $h\in H_x$ put $U_h:=\{z\in X:h(z)>f(z)-\epsilon \}$, an open set by continuity.

The family $(U_h{)}_{h\in H_x}$ covers $X$: for any $y\in X$, [L1] supplies $h\in L$ with $h(x)=f(x)$ and $h(y)=f(y)>f(y)-\epsilon$, so $h\in H_x$ and $y\in U_h$.

By compactness and [L2], finitely many $U_{h_0},\dots ,U_{h_n}$ cover $X$; their pointwise maximum $g:=h_0\vee \cdots \vee h_n$ belongs to $L$, satisfies $g(z)>f(z)-\epsilon$ for every $z\in X$, and satisfies $g(x)=f(x)$ because every $h_j(x)=f(x)$.

Let $G:=\{g\in L:g(z)>f(z)-\epsilon \text{ for every }z\in X\text{, and }g(x)=f(x)\text{ for some }x\in X\}$, a subset of $L$ formed by comprehension rather than by selecting one function per point, and for $g\in G$ put $V_g:=\{z\in X:g(z)<f(z)+\epsilon \}$; each $V_g$ is open. The family $(V_g{)}_{g\in G}$ covers $X$: given $x\in X$, step 3.1 produces a member of $L$ with both defining properties, so it lies in $G$, and it contains $x$ in its $V_g$ because $g(x)=f(x)<f(x)+\epsilon$.

By compactness and [L2], finitely many $V_{g_0},\dots ,V_{g_m}$ cover $X$; their pointwise minimum $u:=g_0\wedge \cdots \wedge g_m$ belongs to $L$.

Every $g_j$ is greater than $f-\epsilon$ everywhere by step 3.1, so $u>f-\epsilon$ everywhere; and at every $z$ some $V_{g_j}$ contains $z$, so $u(z)\le g_j(z)<f(z)+\epsilon$. Thus $\mathrm{\mid }u(z)-f(z)\mathrm{\mid }<\epsilon$ for every $z\in X$.

If $X=\mathrm{\varnothing }$, then $C(X,\mathbb{R})$ contains only the empty function, which is a constant function and hence belongs to the unital lattice $L$; the displayed approximation condition holds vacuously, there being no $x$ to test. The topological reading is not asserted here, because [L3] supplies the uniform metric only on a nonempty domain.

Assume $X\mathrm{\ne }\mathrm{\varnothing }$. For distinct $x,y$, apply [L1] with $\alpha =f(x)$ and $\beta =f(y)$; for $x=y$, the constant function with value $f(x)$ belongs to $L$. Thus $L$ has the two-point duplication property relative to $f$.

Apply [L2] with the positive error $\min \{\epsilon ,1\}\mathrm{/}2$ to obtain $g\in L$ satisfying $\mathrm{\mid }g(x)-f(x)\mathrm{\mid }<\min \{\epsilon ,1\}\mathrm{/}2<\epsilon$ for every $x\in X$.

Suppose further that $X\mathrm{\ne }\mathrm{\varnothing }$, which is where [L3] defines the uniform metric. The approximant of step 2.1 then satisfies $\bar{\rho }(f,g)\le \min \{\epsilon ,1\}\mathrm{/}2$, so every uniform-metric neighbourhood of every $f$ meets $L$; hence $L$ is dense in the topology of uniform convergence.

If $X=\mathrm{\varnothing }$, then $C(X,\mathbb{R})$ consists of the unique empty function, which is the zero element of the vector subspace $A$; hence $A=\bar{A}=C(X,\mathbb{R})$ and the claim is immediate.

Assume $X\mathrm{\ne }\mathrm{\varnothing }$. By the definition of uniform closure, every $f\in \bar{A}$ has, for every positive error, a continuous approximant from $A$, so [L2] confirms that all such uniform limits remain continuous.

The set $\bar{A}$ is a real vector subspace: approximants to $f$ and $g$ add to an approximant to $f+g$, scalar multiples approximate scalar multiples, and the zero function belongs to $A$.

The set $\bar{A}$ is closed under multiplication. Indeed, for $f,g\in \bar{A}$, [L3] gives finite bounds $M_f:={\max }_X\mathrm{\mid }f\mathrm{\mid }$ and $M_g:={\max }_X\mathrm{\mid }g\mathrm{\mid }$. Given $\eta >0$, choose $a,b\in A$ with $\parallel a-f{\parallel }_{\mathrm{\infty }}<\min \{1,\eta \mathrm{/}(2(M_g+1))\},\parallel b-g{\parallel }_{\mathrm{\infty }}<\eta \mathrm{/}(2(M_f+1))\mathrm{.}$ Then $\mathrm{\mid }a\mathrm{\mid }\le M_f+1$ and $\mathrm{\mid }ab-fg\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }b-g\mathrm{\mid }+\mathrm{\mid }g\mathrm{\mid }\mathrm{\mid }a-f\mathrm{\mid }<\eta$ pointwise. Thus products of members of $\bar{A}$ again lie in $\bar{A}$.

Fix $f\in \bar{A}$ and $\epsilon >0$. By [L3], $\mathrm{\mid }f\mathrm{\mid }$ has a maximum $M\ge 0$; if $M=0$ then $f=0$ and $\mathrm{\mid }f\mathrm{\mid }=0\in \bar{A}$.

If $M>0$, apply [L1] on $[-M,M]$ to choose a polynomial $q$ with $\mathrm{\mid }q(t)-\mathrm{\mid }t\mathrm{\mid }\mathrm{\mid }<\epsilon \mathrm{/}2$ there, and put $p(t):=q(t)-q(0)$. Then $p(0)=0$, $\mathrm{\mid }p(t)-\mathrm{\mid }t\mathrm{\mid }\mathrm{\mid }<\epsilon$ on $[-M,M]$, and steps 1.3 and 2.1 give $p(f)\in \bar{A}$. Hence $\mathrm{\mid }f\mathrm{\mid }$ is uniformly approximable by members of $\bar{A}$, and therefore belongs to the closed set $\bar{A}$. Together with the $M=0$ case in step 2.2, this proves $\mathrm{\mid }f\mathrm{\mid }\in \bar{A}$ for every $f\in \bar{A}$.

For $f,g\in \bar{A}$, the pointwise identities $f\vee g=(f+g+\mathrm{\mid }f-g\mathrm{\mid })\mathrm{/}2$ and $f\wedge g=(f+g-\mathrm{\mid }f-g\mathrm{\mid })\mathrm{/}2$, together with steps 1.3 and 3.1, put both functions in $\bar{A}$; hence $\bar{A}$ is a real vector sublattice.

If $X=\mathrm{\varnothing }$, then $C(X,\mathbb{R})$ contains only the empty function, which is a constant function and therefore belongs to the unital algebra $A$; thus $A=C(X,\mathbb{R})$.

Assume $X\mathrm{\ne }\mathrm{\varnothing }$ and let $B:=\bar{A}$ be the uniform closure. By [L1], $B$ is a real vector sublattice and a real function algebra; it is unital and point-separating because it contains $A$.

By [L2], the vector sublattice $B$ is dense in $C(X,\mathbb{R})$, while by definition $B$ is closed; hence $B=C(X,\mathbb{R})$, which says exactly that $A$ is uniformly dense.

If $X=\mathrm{\varnothing }$, 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.

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

By [L1], finitely many $U_{a_0},\dots ,U_{a_n}$ cover $X$. The function $h:=a_0^2+\cdots +a_n^2$ lies in $A$ and satisfies $h(x)>0$ for every $x\in X$.

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

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

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

In the case of step 4.2, every $x\in X$ satisfies $\mathrm{\mid }u(x)-1\mathrm{\mid }=h(x)\mathrm{\mid }p(h(x))-1\mathrm{/}h(x)\mathrm{\mid }<h(x)\epsilon \mathrm{/}M\le \epsilon$; together with step 4.1 this proves the claim in all cases.

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

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

Let $A^{+}:=A+\mathbb{R}1=\{a+c1:a\in A,c\in \mathbb{R}\}$, where $1$ is the constant-one function, itself continuous because the preimage of every open set under it is $\mathrm{\varnothing }$ or $X$. Sums and real multiples of such members again have this form, and $(a+c1)(b+d1)=(ab+da+cb)+cd1$, 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 $C(X,\mathbb{R})$.

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

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 $\epsilon >0$, use step 1.3 to choose $a+c1\in A^{+}$ within $\epsilon \mathrm{/}2$ of $f$. Evaluating at $x_0$, where $a(x_0)=0$ and $f(x_0)=0$, gives $\mathrm{\mid }c\mathrm{\mid }=\mathrm{\mid }a(x_0)+c-f(x_0)\mathrm{\mid }<\epsilon \mathrm{/}2$, so $\mathrm{\mid }a(x)-f(x)\mathrm{\mid }\le \mathrm{\mid }a(x)+c-f(x)\mathrm{\mid }+\mathrm{\mid }c\mathrm{\mid }<\epsilon$ for every $x$; hence $I_{x_0}\subseteq \bar{A}$.

Conversely, still in the case $Z=\{x_0\}$, if $f\in \bar{A}$ then for every $\epsilon >0$ some $a\in A$ satisfies $\mathrm{\mid }f(x_0)-a(x_0)\mathrm{\mid }<\epsilon$; since $a(x_0)=0$, this forces $f(x_0)=0$, so $\bar{A}\subseteq I_{x_0}$.

Suppose $Z=\mathrm{\varnothing }$. Given $f\in C(X,\mathbb{R})$ and $\epsilon >0$, use the density in step 1.3 to choose $a+c1\in A^{+}$ within $\epsilon \mathrm{/}2$ of $f$, then use step 2.1 to choose $u\in A$ within $\epsilon \mathrm{/}(2(\mathrm{\mid }c\mathrm{\mid }+1))$ of $1$; the member $a+cu\in A$ satisfies $\mathrm{\mid }a+cu-f\mathrm{\mid }\le \mathrm{\mid }a+c1-f\mathrm{\mid }+\mathrm{\mid }c\mathrm{\mid }\mathrm{\mid }u-1\mathrm{\mid }<\epsilon$ everywhere, so the uniform closure of $A$ is all of $C(X,\mathbb{R})$.

The alternatives $Z=\mathrm{\varnothing }$ 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.

If $X=\mathrm{\varnothing }$, [L1] gives the full-closure conclusion directly.

If $X\mathrm{\ne }\mathrm{\varnothing }$, the proper alternative in [L1] would give a point $x_0$ at which every member of $A$ vanishes, contradicting nowhere-vanishing at $x_0$; therefore the full-closure alternative holds and $A$ is uniformly dense.

For $f\in A$, self-adjointness and complex linearity put $u:=(f+\bar{f})\mathrm{/}2$ and $v:=(f-\bar{f})\mathrm{/}(2i)$ in $A$.

Sums, real scalar multiples, and products of real-valued members of $A$ are again real-valued by the displayed coordinate formulas in [L2] and [L3], so $A_{\mathbb{R}}$ is a real function algebra by [L1] and [L6]; if $A$ is unital, its real constant functions lie in $A_{\mathbb{R}}$.

The coordinate formulas in [L2], [L3], and [L7] give $u(x)=\mathrm{Re}f(x)$ and $v(x)=\mathrm{Im}f(x)$ for every $x$, so $u$ and $v$ are real-valued. They are continuous as maps into $\mathbb{R}$: each is continuous into $\mathbb{C}$ as a member of $A$, and by [L5] the distance $d_{\mathbb{C}}$ restricted to the real values agrees with $\mathrm{\mid }s-t\mathrm{\mid }$, so the corestriction of a real-valued continuous map to $\mathbb{R}$ is again continuous. Hence $u,v\in C(X,\mathbb{R})$, and [L5] also gives $\mathrm{\mid }u(x)-u(y)\mathrm{\mid }\le d_{\mathbb{C}}(f(x),f(y))$ and $\mathrm{\mid }v(x)-v(y)\mathrm{\mid }\le d_{\mathbb{C}}(f(x),f(y))$.

If $x\mathrm{\ne }y$, choose $f\in A$ with $f(x)\mathrm{\ne }f(y)$. Since $\mathrm{\Phi }$ in [L3] is injective, either $\mathrm{Re}f(x)\mathrm{\ne }\mathrm{Re}f(y)$ or $\mathrm{Im}f(x)\mathrm{\ne }\mathrm{Im}f(y)$, and step 2.1 places the corresponding separator in $A_{\mathbb{R}}$.

If every member of $A$ vanishes at $x$, then every member of $A_{\mathbb{R}}$ does. Conversely, if every member of $A_{\mathbb{R}}$ vanishes at $x$, then step 2.1 makes both real and imaginary parts of every $f(x)$ zero, so coordinate uniqueness in [L2] gives $f(x)=0$; hence the two common-zero sets are equal.

If $X=\mathrm{\varnothing }$, then $C(X,\mathbb{C})$ has only the empty function, which is the zero element of $A$, so the full-closure conclusion holds.

Assume $X\mathrm{\ne }\mathrm{\varnothing }$. By [L1] and [L2], the real-valued part $A_{\mathbb{R}}$ either is dense in $C(X,\mathbb{R})$ or has a unique common zero $x_0$ and closure equal to the real functions vanishing there.

In the dense alternative, let $F\in C(X,\mathbb{C})$ and $\epsilon >0$; the coordinate functions $\mathrm{Re}F$ and $\mathrm{Im}F$ are continuous by [L5] and [L6], so choose $u,v\in A_{\mathbb{R}}$ within $\epsilon \mathrm{/}2$ of them and put $h:=u+iv\in A$.