Finite Boolean combinations of rational intervals, as a subspace of $\{0,1\}^{\mathbb{R}}$: is it $C^*$-embedded?

Question

Setup: Consider $X = \lbrace 0,1\rbrace ^{\mathbb{R}}$ with the product topology, which we might freely identify with the powerset of $\mathbb{R}$ through characteristic functions.

Let $D \subseteq X$ be the subset consisting of characteristic functions of finite Boolean combinations of intervals with rational endpoints. (I don't expect that the exact definition matters much, but in case it does: an “interval” here is allowed to be open, closed or half-open, and can even be a singleton or a half-line or the full real line. Equivalently, $D$ consists of functions $\mathbb{R} \to \lbrace 0,1\rbrace$ which are continuous — i.e., constant — outside of at most finitely many rationals.)

We give $D$ the topology it inherits as a subspace of $X$. Note that $D$ is countable, and that it is dense in $X$ (it is dense because any finite set of conditions of the form $h(x)=0$ or $h(x)=1$ for finitely many $x\in\mathbb{R}$ is satisfied by an element of $D$). For what it's worth, one might also note that $D$ embeds in $\lbrace 0,1\rbrace ^{\mathbb{Q}}$, and, as such, it is metrizable and, in fact, ultrametrizable. [Correction (2024-09-07): while it is true that $D$ can be seen as a subset of $\lbrace 0,1\rbrace ^{\mathbb{Q}}$, the induced topology is not the same.]

THE QUESTION: which of the following conditions, in which we have (1)⇔(2)⇔(3) ⇒ (4)⇔(5), are in fact true?

1. Any bounded continuous function $D\to\mathbb{R}$ extends (obviously uniquely) to all of $X$.

2. The Stone-Čech map $\beta D \to X$ associated to the subspace inclusion $D \to X$ is a homeomorphism.

3. The Stone-Čech map $\beta D \to X$ associated to the subspace inclusion $D \to X$ is injective.

4. The Boolean algebra $\operatorname{Clop}(D)$ of clopen subsets of $D$ is the free Boolean algebra generated by $(e_x)_{x\in\mathbb{R}}$, where $e_x$ is associated with the clopen subset $\lbrace h\in D : h(x)=1\rbrace$ of $D$.

5. Every clopen subset of $D$ is a finite Boolean combination of those of the form $\lbrace h\in D : h(x)=1\rbrace$ for $x\in\mathbb{R}$.

[Correction (2024-09-07): I had initially written the question thinking they were all equivalent (and indeed I would be surprised if it turned out that (4)/(5) are true but (1)/(2)/(3) are not), but my proof of (4)⇒(2) rested on the assertion that $D$ is strongly zero-dimensional, for which my proof was flawed, so I must correct this.]

Proof of the equivalences: The equivalence (1)⇔(2) follows from Gillman & Jerison, Rings of Continuous Functions, theorem 6.5(II). That (2)⇒(3) is trivial, and that (3)⇒(2) follows from the fact that $\beta D \to X$ is surjective (its image is closed because $\beta D$ is compact, and dense because $D$ is dense) and that a continuous bijection between compact spaces is a homeomorphism.

To see that (2)⇒(4), note that $X$ is the Stone space of the free Boolean algebra over $(e_x)_{x\in\mathbb{R}}$; and $\beta D$ is a compact subset of $X$, so it is a Stone space as well; so now we can apply Stone duality for Boolean algebras. Also note that $\operatorname{Clop}(D) = \operatorname{Clop}(\beta D)$ (this is the set of idempotents of the ring $C ^ *(D) = C ^ *(\beta D)$ of bounded continuous functions on $D$). Stone duality tells us that $\beta D \to X$ is an isomorphism iff $\operatorname{Clop}(X) \to \operatorname{Clop}(D)$ is one, which gives the implication (2)⇒(4). Finally, (5) tells us that $\operatorname{Clop}(X) \to \operatorname{Clop}(D)$ is surjective, and it is injective since $D$ is dense in $X$, so in fact (5)⇔(4). ∎

Motivation of the question: a standard proof of the fact that $\operatorname{card}(\beta\mathbb{N}) = 2^{2^{\aleph_0}}$ goes as follows. That $\operatorname{card}(\beta\mathbb{N}) \leq 2^{2^{\aleph_0}}$ is easy, so the point is to prove that $\operatorname{card}(\beta\mathbb{N}) \geq 2^{2^{\aleph_0}}$. For this, it is enough to construct a surjection $\beta\mathbb{N} \to X$ (where $X = \lbrace 0,1\rbrace ^{\mathbb{R}}$ is as above); but since $D$ is countable, we have a surjection $\mathbb{N} \to D$ which composed with the inclusion $D \to X$ gives a map $\mathbb{N} \to X$ whose associated Stone-Čech map $\beta\mathbb{N} \to X$ is surjective (for the same reason as in the proof above: the image is closed and dense). ∎ ❧ Now this proof involves the maps $\beta\mathbb{N} \to \beta D \to X$, both of which are surjective, and it is a natural question in this context to wonder whether the second is in fact a homeomorphism, which is what I ask here.

Given the above equivalences I feel like this should be a fairly straightforward matter to decide, but I only managed to find myself turning in circles, so I suspect I must have missed something obvious.

Answer 1

(I asked a related question on MathOverflow, to which I recieved an answer by KP Hart with references to a paper by Engelking and Pełczyński which is used as inspiration in the following direct answer to my own question.)

I claim that all statements (1)–(5) in the question are false: by the implications proved in the question itself, it is enough to show that (5) is false. I will do this by constructing a (fairly explicit) continuous function $F: D \to \{0,1\}$ which does not extend to a continuous function $X \to \{0,1\}$ (recall that clopen subsets of a topological space and continuous functions to $\{0,1\}$ can be identified by means of the characteristic function).

To do this, we will only use the fact that $D$ is countable and dense in $\{0,1\}^{\mathbb{R}}$ (and $\mathbb{R}$ itself can be replaced by an arbitrary infinite set).

We first construct a partition of $D$ into countably many nonempty clopen sets as follows:

Since $D$ is countable, let us enumerate its elements into a sequence $(h_n)_{n\in\mathbb{N}}$. Let $(x_n)_{n\in\mathbb{N}}$ be an arbitrary sequence of pairwise distinct real numbers, and let $v_n = h_n(x_n) \in \{0,1\}$. Inductively define a sequence of pairwise disjoint nonempty clopen subsets $C_n$ of $D$ by: $$ C_n \;:=\; \{h \in D : h(x_n) = v_n\} \,\setminus\, \bigcup_{i=0}^{n-1} C_i $$ Note that $\{h \in D : h(x_n) = v_n\} \subseteq D$ is clopen by definition of the product topology, so $C_n$ is clopen. That the $C_n$ are pairwise disjoint is obvious by construction. Furthermore, $C_n$ is nonempty because, as pointed out in the question, $D$ is dense in the sense that for any $n$ we can find $h$ such that $h(x_i) = 1-v_i$ for $i<n$ and $h(x_n) = v_n$, which ensures $h \in C_n$.

Finally, since $h_n \in \bigcup_{i=0}^n C_i$ by definition of $v_n$, we have $\bigcup_{i=0}^{+\infty} C_i = D$. So the $C_n$ are indeed a partition of $D$ into countably many nonempty clopen sets.

Now define the function $F\colon D \to \{0,1\}$ as follows: if $h \in C_i$, we let $F(h) = 0$ when $i$ is even or $F(h) = 1$ when $i$ is odd. This function is defined on all of $D$ by what has just been said, and it is continuous (viꝫ. locally constant) because the $C_i$ are open.

I now claim that $F$ does not extend to a continuous function on $X$ with values in $\{0,1\}$ (or indeed even not even in $\mathbb{R}$, as $F^2 = F$ on $D$ implies that any real-valued extension to $X$ would satisfy the same and thus take values in $\{0,1\}$).

To show this, assume by contradiction that $\widehat{F} \colon X \to \{0,1\}$ is such an extension.

As pointed out in the question, $X$ is the Stone space of the free Boolean algebra on $e_x$ (identified with $h \mapsto h(x)$) for $x \in \mathbb{R}$ so by Stone duality any clopen subset of $X$ is a finite Boolean combination of the $\{h\in X : h(x)=1\}$. This means that $\widehat{F}(h)$ can depend only on finitely many values $h(y_1),\ldots,h(y_m)$ of $h$ (in the sense that if $h,h'$ take the same values at $y_1,\ldots,y_m$, then $\widehat{F}(h)=\widehat{F}(h')$).

Let $n$ be such that $x_i \not\in \{y_1,\ldots,y_m\}$ if $i\geq n$. As explained above, by density of $D$ we can find $h$ such that $h(x_i) = 1-v_i$ for $i<n$ and $h(x_n) = v_n$ (where $(x_n)$ and $(v_n)$ are as constructed above). Let $w_j = h(y_j)$ for $1\leq j\leq m$. Again by density of $D$ we can find $h'$ such that $h'(x_i) = 1-v_i$ for $i<n$ and $h'(y_j) = w_j$ (note that if there is a value common to the $x_i$ and the $y_j$, these conditions are compatible because it is the value of $h$ there) and, this time, $h'(x_n) = 1-v_n$ and $h'(x_{n+1}) = v_{n+1}$.

The conditions we have given imply that $h \in C_n$ and $h' \in C_{n+1}$, so $F(h) \neq F(h')$. But since $h$ and $h'$ take the same values $w_1,\ldots,w_m$ on $y_1,\ldots,y_m$, we have $\widehat{F}(h)=\widehat{F}(h')$, contradicting the assumption that $\widehat{F}$ extends $F$. ∎