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Let $X$ be a totally disconnected compact Hausdorff space. Does $X$ always have a discrete dense subspace $Y$?

My motivation for asking this is that the Stone-Čech compactification of a discrete space is always totally disconnected.

If the answer is yes, is it also true if we replace compact with locally compact? What about for general totally disconnected $X$?

UPDATE: As @user14111 pointed out, the answer to all the above is no. In retrospect, so the following question is potentially more interesting:

If a topological space $X$ has a dense discrete subspace, what properties must it satisfy? For example, is $X$ totally disconnected, or does this only apply for compact Hausdorff spaces? What about local compact Hausdorff spaces?

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    $\begingroup$ The Cantor set is a totally disconnected compact Hausdorff space with no isolated points. $\endgroup$ Commented Mar 2 at 1:27
  • $\begingroup$ @user14111 Duh, my apologies. $\endgroup$ Commented Mar 2 at 1:43
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    $\begingroup$ Your new question seems too open-ended. Are there particular properties you are interested in? $\endgroup$ Commented Mar 2 at 1:45
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    $\begingroup$ Consider a non-$T_0$ space that is a discrete space, except with every point duplicated. Pick one point from each pair and you have a dense discrete subspace. However the space is not totally disconnected, as there is no way of separating a point from its pair. $\endgroup$ Commented Mar 2 at 1:56
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    $\begingroup$ The ray topology on $\Bbb{R}$ haz $\Bbb{Z}$ as a dense discrete subspace, and is $T_1$. The ray topology is connected. $\endgroup$ Commented Mar 2 at 2:00

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The first question was already answered in the comments, take $X$ to be the Cantor set.

If a topological space $X$ has a dense discrete subspace, what properties must it satisfy? For example, is $X$ totally disconnected, or does this only apply for compact Hausdorff spaces? What about local compact Hausdorff spaces?

Let $Z$ be any non-empty separable compact Hausdorff space, take a continuous surjection $f:\beta\mathbb{N}\setminus\mathbb{N}\to Z$ (this is possible since $\beta\mathbb{N}\setminus\mathbb{N}$ contains a copy of $\beta\mathbb{N}$ and $Z$ is separable), then there exists a compactification $X$ of $\mathbb{N}$ induced by $f$ such that $X\setminus \mathbb{N}\cong Z$ (since $\mathbb{N}$ is locally compact; theorem 3.5.13 in Engelking). So a compactification of $\mathbb{N}$ can contain any separable Tychonoff space you want.

In fact if $Z$ is any non-empty compact Hausdorff space, take an infinite discrete space $Y$ and a map $Y\to Z$ with dense image. Then since $\beta Y\setminus Y$ contains a copy of $\beta Y$, there is a continuous surjection $f:\beta Y\setminus Y\to Z$ and so a compactification $X$ of $Y$ such that $X\setminus Y\cong Z$. So a compactification of discrete space can contain any Tychonoff space you want, and any non-empty compact Hausdorff space arises as remainder of such compactification (here compactness is essential since a discrete space is locally compact).

So as you see, it doesn't have to be totally disconnected even if compact (e.g. $Z = [0, 1]$), and there is not much hope for its "properties". It can basically be anything.

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