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Recall that a topological space is extremally disconnected if the closure of any open set is open.

A continuous map is quasi-open if it maps nonempty open sets onto sets with nonempty interior. For some reason this class of maps shows up in the literature under different names, the most common of which is semi-open.

If $K$ is an extremally disconnected compact Hausdorff space without isolated points does there exist an infinite cardinal $\alpha$ and a quasi-open continuous map $\varphi:K\to\{0,1\}^\alpha$?

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    $\begingroup$ Since the projection $\{0,1\}^{\alpha}\rightarrow\{0,1\}^{\omega}$ is quasi-open and since the composition of two quasi-open mappings is quasi-open, you only need to assume that $\alpha=\omega$. $\endgroup$ Aug 17 at 1:10
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Let $B$ be a complete Boolean algebra such that if $p_{n}$ is a finite partition of $B$ for all $n$, then there are $b_{n}\in p_{n}$ for all $n$ where $\bigwedge_{n\in\omega}b_{n}\neq 0$. Let $X=S(B)$ be it's Stone space. Then let $f:X\rightarrow\{0,1\}^{\alpha}$ be a continuous function. Then let $P_{n}$ be the partition of $\{0,1\}^{\alpha}$ where we set $(x_{\alpha})_{\alpha}=(y_{\alpha})_{\alpha}(P_{n})$ if and only if $x_{i}=y_{i}$ for each $i<n$. Then $P_{n}$ is a partition of $\{0,1\}^{n}$ into clopen sets, so $Q_{n}=\{f^{-1}[U]\mid U\in P_{n}\}^{+}$ is a partition of $X$ into clopen sets. Therefore, by our hypothesis, there exists some non-empty open subset $U\subseteq X$ and open sets $U_{n}\in Q_{n}$ such that $U\subseteq\bigcap_{n}U_{n}$. Therefore, if we set $O_{n}\in Q_{n},U_{n}=f^{-1}[O_{n}]$, then $U\subseteq\bigcap_{n}f^{-1}[O_{n}]=f^{-1}[\bigcap_{n}O_{n}]$. Therefore, $f[U]\subseteq\bigcap_{n}O_{n}$, but clearly $\bigcap_{n}O_{n}$ has empty interior. Thus, $f[U]$ also has empty interior. In other words, there is no quasi-open $f:X\rightarrow\{0,1\}^{\alpha}$.

Let $B$ be a complete Boolean algebra such that if $p_{n}$ is a finite partition of $B$ for all $n$, then there are $b_{n}\in p_{n}$ for all $n$ where $\bigwedge_{n\in\omega}b_{n}\neq 0$. Let $X=S(B)$ be it's Stone space. Then let $f:X\rightarrow\{0,1\}^{\alpha}$ be a continuous function. Then let $P_{n}$ be the partition of $\{0,1\}^{\alpha}$ where we set $(x_{\alpha})_{\alpha}=(y_{\alpha})_{\alpha}(P_{n})$ if and only if $x_{i}=y_{i}$ for each $i<n$. Then $P_{n}$ is a partition of $\{0,1\}^{n}$ into clopen sets, so $Q_{n}=\{f^{-1}[U]\mid U\in P_{n}\}^{+}$ is a partition of $X$ into clopen sets. Therefore, by our hypothesis, there exists some non-empty open subset $U\subseteq X$ and open sets $U_{n}\in Q_{n}$ such that $U\subseteq\bigcap_{n}U_{n}$. Therefore, if we set $O_{n}\in Q_{n},U_{n}=f^{-1}[O_{n}]$, then $U\subseteq\bigcap_{n}f^{-1}[O_{n}]=f^{-1}[\bigcap_{n}O_{n}]$. Therefore, $f[U]\subseteq\bigcap_{n}O_{n}$, but clearly $\bigcap_{n}O_{n}$ has empty interior. Thus, $f[U]$ also has empty interior. In other words, there is no quasi-open $f:X\rightarrow\{0,1\}^{\alpha}$.

For problems like this, it is helpful to use Stone duality to turn properties of compact totally disconnected spaces into properties of Boolean algebras.

Suppose that $X$ be a compact totally disconnected space. Then a mapping $f:X\rightarrow\{0,1\}^{\omega}$ is dual to a mapping $\phi:F_{\omega}\rightarrow B$ where $F_{\omega}$ is the free Boolean algebra on countably many generators $(x_{n})_{n}$. By freeness, the Boolean algebra homomorphisms $\phi:F_{\omega}\rightarrow B$ are in a one to one correspondence with the functions $s:\{x_{n}|n\in\omega\}\rightarrow B$. In other words, the Boolean algebra homomorphisms $\phi:F_{\omega}\rightarrow B$ are in a one-to-one correspondence with the sequences $(b=a_{n})_{n}$ of elements in $B$. Therefore, let $b_{n}=\phi(x_{n})$ for all $n$.

Now, $f$ is quasi-open if whenever $U$ is clopen in $X$, we have $\bigcap_{n\leq N}\pi_{n}^{-1}[\{c_{n}\}]\subseteq f[U]$. Said, differently, $f$ is quasi-open iff whenever $U$ is a clopen subset of $X$, there is some sequence $(c_{n})_{n\leq N}$ of bits where if $(d_{n})_{n\in\omega}$ is an extension of the sequence $(c_{n})_{n\leq N}$, then $(d_{n})_{n\in\omega}=f(x)$ for some $x\in X$.

Now, by Stone duality, $f$ is quasi-open if and only if for all $b\in B^{+}$, there is a sequence $(c_{n})_{n\leq N}$ where $c_{n}\in\{b_{n},b_{n}'\}$ for all $n$ such that if $(d_{n})_{n\in\omega}$ extends $(c_{n})_{n\leq N}$ and $d_{n}\in\{b_{n},b_{n}'\}$ for all $n$, then there is an ultrafilter $U\subseteq B$ such that $b\in U$ and $d_{n}\in U$ for all $n$.

We can remove the ultrafilters from the characterization of quasi-openness.

The mapping $f:X\rightarrow\{0,1\}^{\omega}$ is quasi-open if and only if for all $b\in B^{+}$, there exists a sequence $(c_{n})_{n\leq N}$ such that $c_{n}\in\{b_{n},b_{n}'\}$ for all $n\leq N$ and where if $(d_{n})_{n\in\omega}$ is a sequence that extends $(c_{n})_{n\leq N}$ with $d_{n}\in\{b_{n},b_{n}'\}$ for all $n\in\omega$, then $b\wedge\bigwedge_{n\leq M}d_{n}\neq 0$ for all $n$.

Therefore, if $X$ is a compact zero-dimensional space with $X=S(B)$ for some Boolean algebra $B$, then there exists a quasi-open map $f:X\rightarrow\{0,1\}^{\omega}$ if and only if there exists a sequence $(b_{n})_{n\in\omega}$ of elements in $B$ where for all $b\in B^{+}$, there exists an $N$ and a sequence $(c_{n})_{n\leq N}$ with $c_{n}\in\{b_{n},b_{n}^{+}\}$ for all $n\leq N$ such that whenever $(d_{n})_{n\in\omega}$ is a sequence with $c_{n}\in\{b_{n},b_{n}^{+}\}$ for all $n\in\omega$ that extends $(c_{n})_{n\leq N}$, then for all $M$, we have $b\wedge\bigwedge_{n\leq M}d_{n}\neq 0$.

Finally, there are examples of compact extremally disconnected spaces $X$ and quasi-open maps $f:X\rightarrow\{0,1\}^{\omega}$. Let $B=Ro(\{0,1\}^{\omega})$, and let $X=S(B)$. Let $C(\{0,1\}^{\omega})$ be the algebra of clopen subsets of $\{0,1\}^{\omega}$. Then clearly $F_{\omega}=C(\{0,1\}^{\omega})\subseteq\text{Ro}(\{0,1\}^{\omega})$, so let $(b_{n})_{n}$ be the canonical generators of $F_{\omega}$. Then whenever $b\in B^{+}$, there is some $N$ and $c_{n}\in\{b_{n},b_{n}'\}$ for $n\leq N$ where $0<\bigwedge_{n\leq N}c_{n}\leq b$. Therefore, there is a quasi-open mapping $f:X\rightarrow\{0,1\}^{\omega}$.

Now, recall that Sikorski's extension theorem states that if $A,B,C$ are Boolean algebras where $A\subseteq B$ and $C$ is complete, then every Boolean algebra homomorphism $f:A\rightarrow C$ can be extended to a Boolean algebra homomorphism $g:B\rightarrow C$. Therefore, by using Sikorski's extension theorem, we can obtain another characterization of the compact extremally disconnected spaces that map quasi-openly onto $\{0,1\}^{\omega}$.

Observe that if $B$ is a Boolean algebra, and $\phi:F_{\omega}\rightarrow B$, then the dual map $S(\phi):S(B)\rightarrow S(F_{\omega})$ is quasi-open if and only if for all $b\in B^{+}$, there exists a $c\in F_{\omega}^{+}$ where for all $d\in F_{\omega}^{+}$ with $d\leq c$ we have $\phi(d)\wedge b\neq 0$.

Therefore, if $B$ is a complete Boolean algebra, and $G_{\omega}$ is the completion of $F_{\omega}$, then by the Sikorski extension theorem, there exists some quasi-open continous map $f:S(B)\rightarrow\{0,1\}^{\alpha}$ if and only if there exists a Boolean algebra homomorphism $\phi:G_{\omega}\rightarrow B$ such that for all $b\in B^{+}$, there exists a $c\in F_{\omega}^{+}$ where for all $d\in G_{\omega}^{+}$ with $d\leq c$, we have $\phi(d)\wedge b\neq 0$.

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    $\begingroup$ The second paragraph seems to be the same as the first one. $\endgroup$
    – YCor
    Aug 17 at 7:54
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    $\begingroup$ Also it would be useful to summarize at the beginning the contents of your post (e.g., that you want to answer OP's question in the negative). $\endgroup$
    – YCor
    Aug 17 at 7:54
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    $\begingroup$ @JosephVanName There seems to be a lot of repetition/redundancy in the answer; also I don't see an explicit example of a Boolean algebra as you use in the beginning; maybe the complete algebra that forces CH, or the completion of $\mathcal{P}(\omega)/\mathit{fin}$? $\endgroup$
    – KP Hart
    Aug 17 at 16:32

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