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This is related to my last question.

Since a discrete $X$ is locally compact Hausdorff, its Stone-Čech compactification $i:X\rightarrow \beta X$ is a homeomorphism onto an open dense subspace of $\beta X.$ Thus $\beta X$ contains a dense discrete subspace $X$.

My question is: Does there exist dense discrete $Y\subseteq \beta X$ distinct from $X$?

Following up on Sassatelli Giulio’s, answer:

Does $\beta X\setminus X$ contain any discrete dense subspace?

I know there is a homeomorphism $\beta X\rightarrow \beta X\setminus X,$ but is the image a dense subspace?

Miles Gould
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No, if $X$ is discrete, then $i(X)$ is the one and only dense discrete subspace of $\beta X$.

If $Y$ is a topological space and $U\subseteq Y$ is dense, then $U$ contains all the isolated points. Since $X$ is discrete and it's embedded onto an open subset, $i(X)$ is made of isolated points, and therefore all dense subsets of $\beta X$ contain $i(X)$. Since $i(X)$ is dense, adding some $y\notin i(X)$ would result in a non-discrete set (because $y$ is not isolated in $i(X)\cup \{y\}$).

This has little to do with $\beta X$: it is true of any dense open embedding of a discrete space.

Sassatelli Giulio
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Does $\beta X\setminus X$ contain any discrete dense subspace?

Recall that sets of the form $\overline{A}$ for $A\subseteq X$ form a basis of $\beta X$, where $X$ is discrete.

Suppose $Y\subseteq \beta X\setminus X$ is a dense discrete subspace and for some $y\in Y$ take $A\subseteq X$ such that $\overline{A}\cap Y = \{y\}$. Since $Y$ is dense, if $B\subseteq X$ is infinite then $Y\cap \overline{B}\neq\emptyset$.

Write $A = A_1\cup A_2$ where $A_1\cap A_2 = \emptyset$ and $A_1, A_2$ are infinite. Then $\overline{A_1}\cap \overline{A_2} = \emptyset$, but since $\overline{A_i}\cap Y\neq\emptyset$ we need to have $y\in \overline{A_1}\cap\overline{A_2}$, contradiction.

So if $X$ is an infinite discrete space, then $\beta X\setminus X$ has no discrete dense subspace.

Alternatively you can show $\beta X\setminus X$ has no isolated points. For suppose that $y\in \beta X\setminus X$ is isolated, then there is $A\subseteq X$ such that $\overline{A} = A\cup \{y\}$. But since $A$ is $C^\ast$-embedded in $X$, $\overline{A}$ is the Stone-Cech compactification of $A$ and so $\overline{A}\setminus A$ must be infinite.

Jakobian
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  • Does the proof of the latter require $X$ discrete? Or does it holds for all locally compact Hausdorff $X$? I’m not familiar with the notion of $C^*$-embedding. – Miles Gould Mar 03 '25 at 15:57
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    @MilesGould It requires that $X$ is discrete. Moreover, for any compact Hausdorff space $Z$ there exists locally compact Hausdorff space $X$ such that $\beta X\setminus X\cong Z$. – Jakobian Mar 03 '25 at 16:01
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    @MilesGould You don't need to know that $\overline{A}$ is the Stone-Cech compactification of $A$, you just need to know there is at least two elements of $\overline{A}\setminus A$. And to see this, we can write $A = A_1\cup A_2$ where $A_1, A_2$ are infinite and consider free ultrafilters containing $A_1$ and $A_2$ respectively. They will be distinct and both in $\overline{A}\setminus A$. – Jakobian Mar 03 '25 at 16:42