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Here is an argument that if $A$ is a countable subset of $\beta \omega \setminus \omega$, then every compactification of $X = \beta \omega \setminus A$ is zero-dimensional.

Suppose $K$ is a compactification of $X$ and $f \colon \beta \omega \to K$ is the Stone extension of the injection.

First note that, since every infinite compact subset of $\beta \omega$ is uncountable, every fiber $f^{-1}(y)$ is finite. The claim is that only finitely many fibers have size greater than one. For suppose not. Let $D$ be an infinite discrete subset of $K$ such that $f^{-1}(d)$ has at least two elements for each $d \in D$. Let $w_d$ and $z_d$ be distinct elements of $f^{-1}(d)$. Then $N = \cup_{d \in D}\{w_d,z_d\}$ is a copy of $\omega$ in $\beta \omega$ and the restriction of $f$ to $N$ is two-to-one. Therefore, by continuity, the restriction of $f$ to $Cl_{\beta \omega}N$ is also two-to-one, and, in particular, there is an element $y$ of $Cl_KA \setminus A$ such that $f^{-1}(y)$ has more than one element, a contradiction. This proves the claim.

The result now follows from the easy fact that if $X$ is any zero-dimensional space and $Y$ is obtained from $X$ by collapsing each of a finite number of finite sets to a point, then $Y$ is zero-dimensional.

It is not the case that the same argument works if $A$ is assumed only to be separable. In fact, if $A$ is any infinite compact subset of $\beta \omega \setminus \omega$ and $X = \beta \omega \setminus A$, then $X$ has a compactification which is not zero-dimensional. To get such a compactification, note that $A$ contains a copy of $\beta \omega \setminus \omega$, and let $K$ be the compactification obtained from $\beta \omega$ by mapping that copy to the interval $[0,1]$.

Here is an argument that if $A$ is a countable subset of $\beta \omega \setminus \omega$, then every compactification of $X = \beta \omega \setminus A$ is zero-dimensional.

Suppose $K$ is a compactification of $X$ and $f \colon \beta \omega \to K$ is the Stone extension of the injection.

First note that, since every infinite compact subset of $\beta \omega$ is uncountable, every fiber $f^{-1}(y)$ is finite. The claim is that only finitely many fibers have size greater than one. For suppose not. Let $D$ be an infinite discrete subset of $K$ such that $f^{-1}(d)$ has at least two elements for each $d \in D$. Let $w_d$ and $z_d$ be distinct elements of $f^{-1}(d)$. Then $N = \cup_{d \in D}\{w_d,z_d\}$ is a copy of $\omega$ in $\beta \omega$ and the restriction of $f$ to $N$ is two-to-one. Therefore, by continuity, the restriction of $f$ to $Cl_{\beta \omega}N$ is also (at least) two-to-one, and, in particular, there is an element $y$ of $Cl_KA \setminus A$ such that $f^{-1}(y)$ has more than one element, a contradiction. This proves the claim.

The result now follows from the easy fact that if $X$ is any zero-dimensional space and $Y$ is obtained from $X$ by collapsing each of a finite number of finite sets to a point, then $Y$ is zero-dimensional.

It is not the case that the same argument works if $A$ is assumed only to be separable. In fact, if $A$ is any infinite compact subset of $\beta \omega \setminus \omega$ and $X = \beta \omega \setminus A$, then $X$ has a compactification which is not zero-dimensional. To get such a compactification, note that $A$ contains a copy of $\beta \omega \setminus \omega$, and let $K$ be the compactification obtained from $\beta \omega$ by mapping that copy to the interval $[0,1]$.

Here is an argument that if $A$ is a countable subset of $\beta \omega \setminus \omega$, then every compactification of $X = \beta \omega \setminus A$ is zero-dimensional.

Suppose $K$ is a compactification of $X$ and $f \colon \beta \omega \to K$ is the Stone extension of the injection.

It is enough to show that if $p \in Cl_Kf(A)$, then every neighborhood of $p$ contains a clopen neighborhood $V$ of $p$ (since $K \setminus Cl_Kf(A)$ is an open subset of $\beta \omega$).

Suppose $U$ is a zero-set neighborhood of $p$. Let $S = \{ x \in A : f(x) \in Int_KU\}$. Then $S$ is countable and disjoint from the closure of the cozero-set $f^{-1}(K \setminus U)$, so, since $\beta \omega$ is an F-space, there are disjoint zero-sets $Z_1$ and $Z_2$ containing $S$ and $f^{-1}(K \setminus U)$, respectively. Since $\beta \omega$ is strongly zero-dimensional, there is a clopen subset $\widehat{V}$ of $\beta \omega$ such that $Z_1 \subseteq \widehat{V}$ and $\widehat{V} \cap Z_2 = \emptyset$. Then $V = f(\widehat{V})$ has the required property.

It is not the case that the same argument works if $A$ is assumed only to be separable. In fact, if $A$ is any infinite compact subset of $\beta \omega \setminus \omega$ and $X = \beta \omega \setminus A$, then $X$ has a compactification which is not zero-dimensional. To get such a compactification, note that $A$ contains a copy of $\beta \omega \setminus \omega$, and let $K$ be the compactification obtained from $\beta \omega$ by mapping that copy to the interval $[0,1]$.

Here is an argument that if $A$ is a countable subset of $\beta \omega \setminus \omega$, then every compactification of $X = \beta \omega \setminus A$ is zero-dimensional.

Suppose $K$ is a compactification of $X$ and $f \colon \beta \omega \to K$ is the Stone extension of the injection.

First note that, since every infinite compact subset of $\beta \omega$ is uncountable, every fiber $f^{-1}(y)$ is finite. The claim is that only finitely many fibers have size greater than one. For suppose not. Let $D$ be an infinite discrete subset of $K$ such that $f^{-1}(d)$ has at least two elements for each $d \in D$. Let $w_d$ and $z_d$ be distinct elements of $f^{-1}(d)$. Then $N = \cup_{d \in D}\{w_d,z_d\}$ is a copy of $\omega$ in $\beta \omega$ and the restriction of $f$ to $N$ is two-to-one. Therefore, by continuity, the restriction of $f$ to $Cl_{\beta \omega}N$ is also two-to-one, and, in particular, there is an element $y$ of $Cl_KA \setminus A$ such that $f^{-1}(y)$ has more than one element, a contradiction. This proves the claim.

The result now follows from the easy fact that if $X$ is any zero-dimensional space and $Y$ is obtained from $X$ by collapsing each of a finite number of finite sets to a point, then $Y$ is zero-dimensional.

It is not the case that the same argument works if $A$ is assumed only to be separable. In fact, if $A$ is any infinite compact subset of $\beta \omega \setminus \omega$ and $X = \beta \omega \setminus A$, then $X$ has a compactification which is not zero-dimensional. To get such a compactification, note that $A$ contains a copy of $\beta \omega \setminus \omega$, and let $K$ be the compactification obtained from $\beta \omega$ by mapping that copy to the interval $[0,1]$.

Here is an argument that if $A$ is a separable subset of $\beta \omega \setminus \omega$, then every compactification of $X = \beta \omega \setminus A$ is zero-dimensional.

Suppose $K$ is a compactification of $X$ and $f \colon \beta \omega \to K$ is the Stone extension of the injection. Let $D$ be a countable dense subset of $A$.

It is enough to show that if $p \in Cl_Kf(A)$, then every neighborhood of $p$ contains a clopen neighborhood $V$ of $p$ (since $K \setminus Cl_Kf(A)$ is an open subset of $\beta \omega$).

Suppose $U$ is a zero-set neighborhood of $p$. Let $S = \{ x \in D : f(x) \in Int_KU\}$. Then $S$ is countable and disjoint from the closure of the cozero-set $f^{-1}(K \setminus U)$, so, since $\beta \omega$ is an F-space, there are disjoint zero-sets $Z_1$ and $Z_2$ containing $S$ and $f^{-1}(K \setminus U)$, respectively. Since $\beta \omega$ is strongly zero-dimensional, there is a clopen subset $\widehat{V}$ of $\beta \omega$ such that $Z_1 \subseteq \widehat{V}$ and $\widehat{V} \cap Z_2 = \emptyset$. Then $V = f(\widehat{V})$ has the required property.

Here is an argument that if $A$ is a countable subset of $\beta \omega \setminus \omega$, then every compactification of $X = \beta \omega \setminus A$ is zero-dimensional.

Suppose $K$ is a compactification of $X$ and $f \colon \beta \omega \to K$ is the Stone extension of the injection.

It is enough to show that if $p \in Cl_Kf(A)$, then every neighborhood of $p$ contains a clopen neighborhood $V$ of $p$ (since $K \setminus Cl_Kf(A)$ is an open subset of $\beta \omega$).

Suppose $U$ is a zero-set neighborhood of $p$. Let $S = \{ x \in A : f(x) \in Int_KU\}$. Then $S$ is countable and disjoint from the closure of the cozero-set $f^{-1}(K \setminus U)$, so, since $\beta \omega$ is an F-space, there are disjoint zero-sets $Z_1$ and $Z_2$ containing $S$ and $f^{-1}(K \setminus U)$, respectively. Since $\beta \omega$ is strongly zero-dimensional, there is a clopen subset $\widehat{V}$ of $\beta \omega$ such that $Z_1 \subseteq \widehat{V}$ and $\widehat{V} \cap Z_2 = \emptyset$. Then $V = f(\widehat{V})$ has the required property.

It is not the case that the same argument works if $A$ is assumed only to be separable. In fact, if $A$ is any infinite compact subset of $\beta \omega \setminus \omega$ and $X = \beta \omega \setminus A$, then $X$ has a compactification which is not zero-dimensional. To get such a compactification, note that $A$ contains a copy of $\beta \omega \setminus \omega$, and let $K$ be the compactification obtained from $\beta \omega$ by mapping that copy to the interval $[0,1]$.