1. Disjointness and $\perpp$-isomorphisms

Throughout this section, $X$ will always denote a locally compact and Hausdorff space, $H$ will denote a Hausdorff space, and $\theta\colon X\to H$ is a fixed continuous function. We denote by $C(X,H)$ the set of continuous functions from $X$ to $H$. For two functions $f,g\colon X\to H$, we denote

\begin{equation*} [f\neq g]=\left\{x\in X:f(x)\neq g(x)\right\}\quad\text{and}\quad[f=g]=X\setminus[f\neq g]. \end{equation*}

1.1. Disjointness relations

The main idea is that any manageable algebraic structure on $H$ will naturally lead us to consider a specific function $\theta$, which behaves as a “neutral element”, and which will be used to separate points of $X$. See Example 1.3 for the classical setting. We first generalize the notion of support, in the obvious manner.

Definition 1.1.

Given $f\in C(X,H)$, we define the $\theta$-support of $f$ as

\begin{equation*} \supp^\theta(f)=\overline{[f\neq\theta]}. \end{equation*}
We define $\sigma^\theta(f)$ as the interior of $\supp^\theta(f)$, and $Z^\theta(f)$ as the complement of $\supp^\theta(f)$:
\begin{equation*} \sigma^\theta(f)=\operatorname{int}\supp^\theta(f)\qquad\text{and}\qquad Z^\theta(f)=X\setminus\supp^\theta(f). \end{equation*}

Let $C_c(X,\theta)$ be the set of continuous functions from $X$ to $H$ with compact $\theta$-support. Whenever there is no risk of confusion, we will drop $\theta$ from the notation and write simply $\supp(f)$, $\sigma(f)$, $Z(f)$ and $C_c(X)$.

Now we define the following relations: Given $f,g\in C(X,H)$,

  1. $f\perp g$: if $[f\neq\theta]\cap[g\neq\theta]=\varnothing$; we say that $f$ and $g$ are weakly disjoint;

  2. $f\perpp g$: if $\supp(f)\cap\supp(g)=\varnothing$; we say that $f$ and $g$ are strongly disjoint;

  3. $f\subseteq g$: if $\sigma(f)\subseteq\sigma(g)$;

  4. $f\Subset g$: if $\supp(f)\subseteq\sigma(g)$.

Note that $Z^\theta(f)$ is the complement of $\sigma^\theta(f)$ in the lattice of regular open sets of $X$ (see [MR2466574, Chapter 10]). Also, $\sigma^\theta(f)$ is the regularization of $[f\neq\theta]$, and thus it follows immediately that

\begin{equation*} f\perp g\iff \sigma(f)\cap\sigma(g)=\varnothing, \end{equation*}
even though $[f\neq \theta]$ and $\sigma^\theta(f)$ are not equal in general.

Example 1.2.

Suppose $X=H=[0,1]$, $\theta=0$ (the zero map $[0,1]\to[0,1]$) and $f=\id_{[0,1]}$, the identity map of $[0,1]$. Then $[f\neq\theta]=(0,1]$ but $\sigma^\theta(f)=[0,1]$.

As stated above, when $H$ comes with additional structure, a particular choice of $\theta$ generally yields a suitable notion of support, and the relations above may be described in terms of this structure. This is the general technique used in the applications in Section 3.

Example 1.3.

If $H=\mathbb{R}$ or $\mathbb{C}$, and $\theta=0$ is the constant zero function, we obtain the usual notion of support. We may describe $\perp$ in terms of the multiplicative structure of $C_c(X)=C_c(X,0)$: $f\perp g$ if and only if $fg=0$, which is the only absorbing element of $C_c(X)$.

Example 1.4 (Kania‒Rmoutil, [MR3813611]).

Let $X$, $H$ and $\theta$ as in the beginning of this section. Define the compatibility ordering on $C_c(X,\theta)$ by

\begin{equation*} f\preceq g\iff g|_{\supp^\theta(f)}=f|_{\supp^\theta(f)}. \end{equation*}
Then $\theta$ is the minimum of $\preceq$ in $C_c(X,\theta)$. We can describe weak disjointness in $C_c(X,\theta)$ by
\begin{equation*} f\perp g\iff\inf_{\preceq}\left\{f,g\right\}=\theta\text{ and }\left\{f,g\right\}\text{ has a }\preceq\text{-upper bound.} \end{equation*}

We will, moreover, be interested in recovering $X$ not from the whole set $C_c(X)$, but instead from a subcollection $\mathcal{A}\subseteq C_c(X)$. We will need to assume, however, that there are enough functions in $\mathcal{A}$ in order to separate points of $X$, and this is attained by assuming that an appropriate version of Urysohn's Lemma is valid. (This is the same type of assumption as made in [MR0020715] and in [MR2324919].)

Definition 1.5.

Let $\mathcal{A}\subseteq C_c(X)$ be a subset containing $\theta$. Denote $\sigma(\mathcal{A})=\left\{\sigma(f):f\in\mathcal{A}\right\}$. We say that $(X,\theta,\mathcal{A})$ (or simply $\mathcal{A}$) is

  1. weakly regular if $\sigma(\mathcal{A})$ is a basis for the topology of $X$.

  2. regular if for every $x\in X$, every neighbourhood $U$ of $x$ and every $c\in H$ there is $f\in\mathcal{A}$ with $f(x)=c$ and $\supp(f)\subseteq U$.

We will need to analyze relations between $\subseteq,\Subset,\perp$ and $\perpp$. In order to deal with finitely many functions simultaneously, we will need to adapt the notion of open cover of a set to this language.

Definition 1.6.

Suppose $\mathcal{A}\subseteq C_c(X)$ is weakly regular. A family $A\subseteq\mathcal{A}$ is a cover of an element $b\in\mathcal{A}$ if the implication

\begin{equation*} h\perp a\text{ for all }a\in A\Longrightarrow h\perp b. \end{equation*}
is valid for all $h\in\mathcal{A}$.

Lemma 1.7.

Suppose $\mathcal{A}$ is weakly regular, and let $A\subseteq\mathcal{A}$ and $b\in\mathcal{A}$. The following are equivalent:

  1. $A$ is a cover of $b$;

  2. The closure of $\bigcup_{a\in A}[a\neq\theta]$ contains $\supp(b)$.

Proof. (1)$\Rightarrow$(2): Let $x\in\supp(b)$. Take an open neighbourhood of $x$ of the form $\sigma(h)$, $h\in\mathcal{A}$. Since $\supp(b)=\overline{\sigma(b)}$, the intersection $\sigma(h)\cap\sigma(b)$ is nonempty and thus $h$ and $b$ are not weakly disjoint. From $A$ being a cover, $h$ is not weakly disjoint to some $a\in A$, which means that $\sigma(h)\cap[a\neq\theta]$ is nonempty. Since $\mathcal{A}$ is weakly regular, then $x$ is in the closure of $\bigcup_{a\in A}[a\neq\theta]$. (2)$\Rightarrow$(1): Suppose $h\in\mathcal{A}$ is such that $h\perp a$ for all $a\in A$. This means that
\begin{equation*} (\bigcup_{a\in A}[a\neq\theta])\cap[h\neq\theta]=\varnothing. \end{equation*}
Taking the closure of the first term and using (2) we conclude that $[b\neq\theta]\cap[h\neq\theta]\subseteq\supp(b)\cap[h\neq\theta]=\varnothing$, so $h\perp b$.

If $\mathcal{A}\subseteq C_c(X)$ and $\theta\in\mathcal{A}$, note that $\subseteq$ is a preorder on $\mathcal{A}$, whose minimum is $\theta$. Alternatively, $\theta$ is the only element of $\mathcal{A}$ such that $\theta\perp\theta$. Thus the function $\theta$ is uniquely determined in terms of either $\perp$ or $\subseteq$. We now proceed to prove that $\subseteq$ and $\perp$ “carry the same information”, as do $\Subset$ and $\perpp$.

Proposition 1.8.

Suppose $\mathcal{A}$ is weakly regular. If $f,g\in\mathcal{A}$, then

  1. $f\subseteq g\iff\forall h(h\perp g\Rightarrow h\perp f)$;

  2. $f\perp g\iff$ The $\subseteq$-infimum of $\left\{f,g\right\}$ is $\theta$;

    1. yeah

  3. $f\subseteq g\iff\forall h(h\Subset f\Rightarrow h\Subset g)$;

  4. $f\subseteq g\iff\forall h(h\perpp g\Rightarrow h\perpp f)$;

  5. $f\perpp g\iff\exists h_1,k_1,\ldots,h_n,k_n\in\mathcal{A}$ such that $\left\{h_1,\ldots,h_n\right\}$ is a cover of $f$, $h_i\Subset k_i$ and $\phantom{f\perpp g\iff...}k_i\perp g$ for all $i$;

  6. $f\Subset g\iff\forall b\in\mathcal{A}$, $\exists h_1,\ldots,h_n\in\mathcal{A}$ such that $\left\{h_1,\ldots,h_n,g\right\}$ is a cover of $b$ and $h_i\perpp f$.

By items (a) and (b), $\perp$ and $\subseteq$ are equi-expressible (i.e., each one is completely determined by the other). By (c) and (d) one can recover $\subseteq$ (and hence $\perp$) from either $\Subset$ or $\perpp$, which in turn implies, from (e) and (f), that $\Subset$ and $\perpp$ are also equi-expressible.

Proof.

Items (a)-(d) are easy consequences of weak regularity of $\mathcal{A}$, and $X$ being a regular topological space for items (c)-(d).

  1. $\Rightarrow$: Suppose $f\perpp g$. Given $x\in \supp(f)$, weak regularity of $\mathcal{A}$ and regularity of the topological space $X$ give us $h_x,k_x\in\mathcal{A}$ such that $x\in\sigma(h_x)$, $h_x\Subset k_x$ and $k_x\perp g$. Compactness of $\supp(f)$ allows us to find the elements $h_i,k_i$ we need, by going to a subcover of $\left\{\sigma(h_x):x\in\supp(f)\right\}$.

    $\Leftarrow$: Suppose such $h_i,k_i$ exist. Then by Lemma 1.7,

    \begin{equation*} \supp(f)\subseteq\bigcup_{i=1}^n\supp(h_i)\subseteq\bigcup_{i=1}^n\sigma(k_i)\subseteq X\setminus\supp(g), \end{equation*}
    and so $f\perpp g$.

  2. $\Rightarrow$: Suppose $f\Subset g$ and take any $b\in\mathcal{A}$. Since $\supp(b)\setminus\sigma(g)$ is compact and does not intersect $\supp(f)$, we can take $h_1,\ldots,h_n\in\mathcal{A}$ such that $h_i\perpp f$ and $\supp(b)\setminus\sigma(g)\subseteq\bigcup_i\sigma(h_i)$, which implies that $\left\{h_1,\ldots,h_n,g\right\}$ is a cover of $b$.

    $\Leftarrow$: By compactness os $\supp(f)$ and $\supp(g)$, take $b_1,\ldots,b_M$ in $\mathcal{A}$ such that $\supp(f)\cup\supp(g)\subseteq\bigcup_{k=1}^M\sigma(b_k)$. For each $k$ take functions $h_i^k$ satisfying the right-hand side of (f), relative to $b_k$.

    Given $k$, we have $\sigma(b_k)\subseteq\bigcup_i\supp(h_i^k)\cup\supp(g)$, so by taking complements we obtain $\bigcap_iZ(h_i^k)\cap Z(g)\cap\sigma(b_k)=\varnothing$, or equivalently $\bigcap_iZ(h_i^k)\cap\sigma(b_k)\subseteq\sigma(b_k)\setminus Z(g)\subseteq\supp(g)$. Taking interiors on both sides yields $\bigcap_iZ(h_i^k)\cap\sigma(b_k)\subseteq\sigma(g)$.

    Now from $h_i^j\perpp f$ we obtain

    \begin{align*} \supp(f)&\subseteq\bigcap_{i,j}Z(h_i^j)\cap\bigcup_{k=1}^M\sigma(b_k)\subseteq\bigcup_{k=1}^M\left[\bigcap_i Z(h_i^k)\cap\sigma(b_k)\right]\subseteq \sigma(g), \end{align*}
    so $f\Subset g$.

Remark

One should be careful with the connections between the pairs of relations $(\perp,\perpp)$ and $(\subseteq,\Subset)$. For example, $\perp$ and $\perpp$ may coincide but $\subseteq$ and $\Subset$ may not and vice-versa. See the example below.

Example 1.9.

Let $X=H=\mathbb{R}$ and $\theta=0$, so that we are dealing with the usual notion of support. Let $\left\{(a_n,b_n):n\in\mathbb{N}\right\}$ (where $a_n\smallerthan b_n$) be a countable basis of open intervals for the usual topology of $\mathbb{R}$. Up to small modifications, we may assume that all the numbers $a_n$, $b_n$ and $b_n+1$ are distinct. In particular, the sets $U_n:=(a_n,b_n)$ have pairwise disjoint boundaries.

For each $n$, let $f_n\in C_c(\mathbb{R})$ with $\sigma(f_n)=U_n$, e.g. $f_n(x)=\max(0,(x-\widetilde{a_n})(\widetilde{b_n}-x))$, and let $\mathcal{A}=\left\{f_n:n\in\mathbb{N}\right\}$, which is weakly regular. Then $\perp$ and $\perpp$ coincide on $\mathcal{A}$, as do $\subseteq$ and $\Subset$, since the boundaries of all $U_n$ are pairwise disjoint.

Letting $V=(\widetilde{a_1},\widetilde{b_1}+1)$ and $g_V$ be a continuous function with $\sigma(g_V)=V$, then $\perp$ and $\perpp$ still coincide in $\mathcal{A}\cup\left\{g_V\right\}$, however $\subseteq$ and $\Subset$ do not, since $f_1\subseteq g_V$ but not $f_1\Subset g_V$.

Alternatively, set $W=(\widetilde{b_1},\widetilde{b_1}+1)$ and let $g_W$ be any continuous function with $\sigma(g_W)=W$. Then $\subseteq$ and $\Subset$ still coincide in $\mathcal{A}\cup\left\{g_W\right\}$, however $\perp$ and $\perpp$ do not, because $f_1\perp g$ but not $f_1\perpp g$.

1.2. $\perpp$-ideals

Recall that $X$, $H$ and $\theta\in C(X,H)$ are fixed, as in the beginning of the section. We fix also a weakly regular family $\mathcal{A}\subseteq C_c(X,\theta)$.

One technique that is commonly used in the proofs of Banach‒Stone type theorems is to describe an order-isomorphism between $X$ and “maximal ideals” of $\mathcal{A}$, where the notion of an “ideal” depends on whatever kinds of algebraic signature one is working. See for example [MR0029476, Lemma 2.2], [MR2324919, Proposition 2.7], [MR0020715, Lemma 3]. This idea also appears in some manner in the proofs of the main results of [MR3162258] and [MR1060366], and of [MR1357166, p. 170, Théorème 3].

We will follow this idea by considering $\perpp$-ideals. Although their definition (1.12) is given simply in terms of the relation $\perpp$, Theorem 1.14 provides a much more manageable description of them.

A strengthening of the notion of cover will be necessary (see Lemma 1.11 for the intuition).

Definition 1.10.

A finite family $B\subseteq\mathcal{A}$ is said to be a strong cover of an element $a\in\mathcal{A}$ if there is another finite family $\widetilde{B}\subseteq\mathcal{A}$ such that:

  1. For all $\widetilde{b}\in\widetilde{B}$, there is some $b\in B$ with $\widetilde{b}\Subset b$;

  2. $\widetilde{B}$ is a cover of $a$ (see Definition 1.6).

The following proposition shows that strong covers encode information about the closures of sets. It is a direct consequence of the definition of $\Subset$ and Lemma 1.7.

Lemma 1.11.

A finite family $B\subseteq\mathcal{A}$ is a strong cover of $a\in\mathcal{A}$ if and only if $\supp(a)\subseteq\bigcup_{b\in B}\sigma(b)$.

Definition 1.12.

A $\perpp$-ideal in $\mathcal{A}$ is a subset $I\subseteq\mathcal{A}$ such that, for all $a\in\mathcal{A}$,

\begin{equation*} a\in I\iff \text{there is a finite subset }B\subseteq I\text{ which is a strong cover of }a. \end{equation*}

Note that every $\perpp$-ideal of $\mathcal{A}$ contains $\theta$ (since the empty set is a strong cover of $\theta$).

We will now prove that the lattice of open subsets of a space $X$ is order-isomorphic to the lattice of $\perpp$-ideals of a weakly regular tuple $(X,\theta,\mathcal{A})$.

Definition 1.13.

Suppose $(X,\theta,\mathcal{A})$ is weakly regular. Given an open set $U\subseteq X$, denote $\mathbf{I}(U)=\left\{f\in\mathcal{A}:\supp(f)\subseteq U\right\}$, and , given a $\perpp$-ideal $I\subseteq\mathcal{A}$, denote $\mathbf{U}(I)=\bigcup_{f\in I}\sigma(f)$.

Lemma 1.11 and weak regularity of $\mathcal{A}$ imply that $\mathbf{I}(U)$ is a $\perpp$-ideal of $\mathcal{A}$ for any open $U\subseteq X$.

Theorem 1.14.

Suppose $(X,\theta,\mathcal{A})$ is weakly regular.

  1. For every $\perpp$-ideal $I$ of $\mathcal{A}$, $I=\mathbf{I}(\mathbf{U}(I))$;

  2. For every open subset $U\subseteq X$, $U=\mathbf{U}(\mathbf{I}(U))$;

  3. The map $U\mapsto\mathbf{I}(U)$ is an order isomorphism between the lattices of open sets of $X$ and $\perpp$-ideals of $\mathcal{A}$.

Proof.

  1. Let $I$ be a $\perpp$-ideal. The inclusion $I\subseteq \mathbf{I}(\mathbf{U}(I))$ follows easily from the definition of $\perpp$-ideals: if $f\in I$, then take a finite strong cover $B\subseteq I$ of $f$, so that $\supp(f)\subseteq\bigcup_{b\in B}\sigma(b)\subseteq\mathbf{U}(I)$.

    Conversely, if $f\in\mathbf{I}(\mathbf{U}(I))$ then $\supp(f)\subseteq\mathbf{U}(I)=\bigcup_{b\in I}\sigma(b)$. Using compactness of $\supp(f)$ we find a finite family $B\subseteq I$ with $\supp(f)\subseteq\bigcup_{b\in B}\sigma(b)$, so $B$ is a strong cover of $f$ (Lemma 1.11) and therefore $f\in I$.

  2. Suppose $U\subseteq X$ is open. By weak regularity of $(X,\theta,\mathcal{A})$ and since $X$ is regular, we have

    \begin{equation*} U=\bigcup_{\substack{f\in\mathcal{A}\\\supp(f)\subseteq U}}\sigma(f)=\bigcup_{f\in\mathbf{I}(U)}\sigma(f)=\mathbf{U}(\mathbf{I}(U)). \end{equation*}

  3. The previous items prove that $U\mapsto\mathbf{I}(U)$ is a bijection, with inverse $I\mapsto\mathbf{U}(I)$. It is clear that both maps are order-preserving.

1.3. The main theorems

The main theorem (1.17) now follows easily from the previous subsection. Fix two locally compact Hausdorff spaces $X$ and $Y$, and for $Z\in\left\{X,Y\right\}$ a Hausdorff space $H_Z$, a continuous map $\theta_Z\colon Z\to H_Z$, and a subset $\mathcal{A}(Z)\subseteq C_c(Z,\theta_Z)$.
Definition 1.15.

We call a map $T\colon\mathcal{A}(X)\to\mathcal{A}(Y)$ a $\perpp$-morphism if $f\perpp g$ implies $Tf\perpp Tg$; $T$ is a $\perpp$-isomorphism if it is bijective and both $T$ and $T^{-1}$ are $\perpp$-morphisms. $\perp$, $\subseteq$ and $\Subset$-isomorphisms are define analogously.

By Proposition 1.8(a), $\perp$-morphisms coincide with $\subseteq$ morphisms. We obtain:

Theorem 1.16.

Suppose $(X,\theta_X,\mathcal{A}(X))$ and $(Y,\theta_Y,\mathcal{A}(Y))$ are weakly regular and $T\colon\mathcal{A}(X)\to\mathcal{A}(Y)$ is a $\perp$-isomorphism. Let $f,g\in\mathcal{A}(X)$. Then $\sigma(f)\subseteq \sigma(g)$ if and only if $\sigma(Tf)\subseteq \sigma(Tg)$. In particular, $Z(f)=\varnothing$ if and only if $Z(Tf)=\varnothing$.

Assume $(X,\theta,\mathcal{A}(X))$ is weakly regular. Let $\widehat{\mathcal{A}(X)}$ be the collection of maximal $\perpp$-ideals of $\mathcal{A}$, and endow it with the topology generated by the sets

\begin{equation*} U(f)=\left\{I\in\widehat{\mathcal{A}(X)}:\exists g\Subset f\text{ such that }g\not\in I\right\},\qquad f\in\mathcal{A}(X). \end{equation*}
By Theorem 1.14, we obtain a bijection $\kappa_X\colon X\to\widehat{\mathcal{A}(X)}$, $\kappa_X(x)=\mathbf{I}(X\setminus\left\{x\right\})$. Since for all $x\in X$ and $f\in\mathcal{A}(X)$,
\begin{equation*} x\in\sigma(f)\iff\exists g\Subset f\text{ such that } x\in\supp(g)\iff\kappa_X(x)\in U(f), \end{equation*}
then $\kappa_X(\sigma(f))=U(f)$, which proves that $\kappa_X$ is a homeomorphism. Performing a similar procedure with $Y$ and using standard duality arguments, we obtain our main theorem:

Theorem 1.17.

If $\mathcal{A}(X)$ and $\mathcal{A}(Y)$ are weakly regular and $T\colon\mathcal{A}(X)\to\mathcal{A}(Y)$ is a $\perpp$-isomorphism then there is a unique homeomorphism $\phi\colon Y\to X$ such that $\phi(\supp(Tf))=\supp(f)$ for all $f\in \mathcal{A}(X)$ (equivalently, $\phi(\sigma(Tf))=\sigma(f)$, or $\phi(Z(Tf))=Z(f)$, for all $f\in\mathcal{A}(X)$).

Definition 1.18.

The unique homeomorphism $\phi$ associated with $T$ as in 1.17 will be called the $T$-homeomorphism.

We finish this section by proving that Theorem 1.17 is sharp, in the sense that the analogous result for $\perp$-isomorphisms is not true in general. We can even be more bold and find counter-examples in settings which are usually regarded as “well-behaved”; namely, we will consider only real-valued functions and the usual notion of support (i.e., $C_c(X)=C_c(X,0)$ for a space $X$), and compact spaces.

Moreover, we will provide two examples: one where the underlying topological spaces do not have the same small inductive dimension (Corollary 1.23), and one where they do (Corollary 1.25).

Let us fix some notation, and recall some basic facts about Stone duality. We refer to [MR1507106], dml124080] for details (see also [MR861951, II.4.4])

Notation

Given a Hausdorff space $X$, denote by $\operatorname{RO}_K(X)$ the generalized Boolean algebra of regular open subsets of $X$ with compact closure, and by $\operatorname{KO}(X)$ the generalized Boolean algebra of compact-open subsets of $X$. Given $A\in \operatorname{RO}_K(X)$, we define $\Sigma_X(A)=\left\{f\in C_c(X):\sigma(f)=A\right\}$.

Given a generalized Boolean algebra $B$, let $\operatorname{Spec}(B)$ be the spectrum of $B$, i.e., the space of non-trivial lattice homomorphisms from $B$ to the two-element lattice $\left\{0,1\right\}$, with the topology of pointwise convergence. (Equivalently, it may be regarded as the space of ultrafilters of $B$.)

Stone duality

The usual form of Stone duality states that the category of Stone (i.e., zero-dimensional, compact Hausdorff) spaces is dual to that of Boolean algebras. This extends to the locally compact spaces and generalized Boolean algebras, and in particular we obtain: Every zero-dimensional, locally compact Hausdorff space $X$ is (naturally) homeomorphic to $\operatorname{Spec}(\operatorname{KO}(X))$, and every generalized Boolean algebra $B$ is (naturally) isomorphic to $\operatorname{KO}(\operatorname{Spec}(B))$. For a more general version, see [bicestarlinglcsd].

In order to find non-homeomorphic spaces $X$ and $Y$ such that $C_c(X)$ and $C_c(Y)$ are $\perp$-isomorphic, we need the following result:

Proposition 1.19.

Suppose that:

  1. $X$ and $Y$ are separable, locally compact Hausdorff spaces;

  2. For all nonempty $A\in\operatorname{RO}_K(X)$ and $B\in\operatorname{RO}_K(Y)$, $\Sigma_X(A)$ and $\Sigma_Y(B)$ are nonempty;

  3. $\varphi\colon\operatorname{RO}_K(X)\to\operatorname{RO}_K(Y)$ is an order isomorphism (with respect to set inclusion).

Then $C_c(X)$ and $C_c(Y)$ are $\perp$-isomorphic.

Proof.

Given $A\in\operatorname{RO}_K(X)$, the sets $\Sigma_X(A)$ and $\Sigma_Y(\varphi(A))$ have the same cardinality: They are either singletons if $A=\varnothing$, or have cardinality $2^{\aleph_0}$ otherwise, by (ii) and since $X$ and $Y$ are separable. Consider any bijection $T_A\colon\Sigma_X(A)\to\Sigma_Y(\varphi(A))$. Then the map

\begin{equation*} T\colon C_c(X)\to C_c(Y),\qquad T(f)=T_{\sigma(f)}(f) \end{equation*}
is a $\perp$-isomorphism.

The following are two technical lemmas which will allow us to construct spaces $X$ and $Y$ satisfying the hypotheses of the theorem above.

Lemma 1.20.

Suppose that $\mathfrak{C}$ is a zero-dimensional, locally compact Hausdorff space and $\operatorname{KO}(\mathfrak{C})$ is conditionally complete (i.e., every bounded family has a supremum). Then $\operatorname{RO}_K(\mathfrak{C})=\operatorname{KO}(\mathfrak{C})$.

Proof.

The only non-trivial part is proving $\operatorname{RO}_K(\mathfrak{C})\subseteq\operatorname{KO}(\mathfrak{C})$. Given $A\in\operatorname{RO}_K(\mathfrak{C})$, the family $\left\{V\in\operatorname{KO}(\mathfrak{C}):V\subseteq A\right\}$ is bounded in $\operatorname{KO}(\mathfrak{C})$, so let $U$ be its supremum in $\operatorname{KO}(\mathfrak{C})$. As $\mathfrak{C}$ is zero-dimensional we have $A\subseteq U$. To prove the reverse inclusion, we first show that $U\setminus\overline{A}=\varnothing$.

If $W\in\operatorname{KO}(\mathfrak{C})$ and $W\subseteq U\setminus\overline{A}$, then $A\subseteq U\setminus W$, from which it follows that $U\subseteq U\setminus W$, so $W=\varnothing$. This proves that $U\setminus\overline{A}=\varnothing$, because $\mathfrak{C}$ is zero-dimensional, and so $U\subseteq\overline{A}$. However, $U$ is clopen and $A$ is regular open, which implies $A=U$.

Lemma 1.21.

If $X$ is a separable locally compact Hausdorff space, then $\mathfrak{C}=\operatorname{Spec}(\operatorname{RO}_K(X))$ is separable as well.

Proof.

Let $\left\{x_n:n\in\mathbb{N}\right\}$ be a countable dense subset of $X$. For each $n$, Zorn's Lemma implies that exists $\phi_n\in\mathfrak{C}$ such that $\phi_n(U)=1$ whenever $x_n\in U$. Then $\left\{\phi_n\right\}_n$ is easily seen to be dense in $\mathfrak{C}$.

Lemma 1.22.

If $X$ is a second-countable locally compact Hausdorff space and $A\in\operatorname{RO}_K(X)$, then there is $f\in C_c(X)$ such that $\sigma(f)=A$.

Proof.

First choose a countable family of compact subsets $K_n\subseteq A$ such that $\bigcup_nK_n=A$. For each $n$ we can, by Urysohn's Lemma and regularity of $X$, find a continuous function $f_n\colon X\to[0,1]$ such that $f_n(k)=1$ for all $k\in K_n$ and $\supp(f_n)\subseteq A$. Letting $f=\sum_{n=1}^\infty 2^{-n}f_n$ we obtain $[f\neq 0]=\sigma(f)=A$, because $A$ is regular.

Given locally compact Hausdorff $X$, let $\mathfrak{C}=\operatorname{Spec}(\operatorname{RO}_K(X))$. The generalized Boolean algebra $\operatorname{RO}_K(X)$ is conditionally complete, so by Stone duality, $\operatorname{KO}(\mathfrak{C})$ is also conditionally complete, and hence coincides with $\operatorname{RO}_K(\mathfrak{C})$. As a consequence of Lemmas 1.20, 1.21 and 1.22 and Proposition 1.19 when $X=[0,1]$, we conclude:

Corollary 1.23.

There exists a zero-dimensional, compact Hausdorff topological space $\mathfrak{C}$ (namely, $\mathfrak{C}=\operatorname{Spec}(\operatorname{RO}_K([0,1]))$) ‒ which is, in particular, not homeomorphic to $[0,1]$ ‒ such that $C(\mathfrak{C})$ and $C([0,1])$ are $\perp$-isomorphic.

For our second example, we will consider only second-countable spaces. The next lemma is again a technical lemma which can be proven by elementary topological considerations, so we omit its proof.

Lemma 1.24.

Let $X$ be a topological space and $U$ a dense open subset of $X$. Then the map

\begin{equation*} \varphi_U\colon \operatorname{RO}(X)\to\operatorname{RO}(U),\qquad \varphi_U(A)=A\cap U \end{equation*}
is an order isomorphism.

Let $\mathbb{S}^1=\left\{z\in\mathbb{C}:|z|=1\right\}$ be the complex unit circle.

Corollary 1.25.

$C([0,1])$ and $C(\mathbb{S}^1)$ are $\perp$-isomorphic.

Proof.

Let $X=(0,1)$ and $Y=\mathbb{S}^1\setminus\left\{1\right\}$. Then $X$ and $Y$ are homeomorphic, and two applications of Lemma 1.24 imply that $\operatorname{RO}([0,1])$ and $\operatorname{RO}(\mathbb{S}^1)$ are order-isomorphic. Lemmas 1.21 and 1.22, and Proposition 1.19, imply that $C([0,1])$ and $C(\mathbb{S}^1)$ are $\perp$-isomorphic.