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Let $X$ be a topological space. $Y$ is a discrete subset of $X$ if it has a discrete topology induced by the topology of $X$. This is equivalent to the fact that for every $y\in Y$ there is an open $U\subset X$ such that $U\cap Y=\{y\}$. Consider a stronger condition: there is a collection $\{U_y\}_{y\in Y}$ of mutualy disjoint open sets, such that $y\in U_y$. I wonder if the two are equivalent for nice spaces, but my question today is following:

Is it true that if $X$ is locally compact and not compact, there is a discrete (in the stronger sense) set $Y\subset X$, such that $\overline{Y}$ is not compact?

In my specific case $X$ is locally compact, but this condition is not necessary in order to pose the question. If $X$ is metrizable and separable, it has a metrizable compactification, and $Y$ can be chosen as a sequence that converges in a conrolled way to an infinite element. If $X$ is locally compact and $\sigma$-compact, then there is a sequence of compacts $\{K_n\}$ such that $K_n\subset int~K_{n+1}$, and taking $Y$ to be the sequence of arbitrary $y_n\in int~K_{n+1}\backslash K_{n}$ will work.

I also have a reference request about existence of an infinite discrete subset in any infinite Hausdorff space. This follows from the result in this article, but I was not able to find a more direct reference. I found some research on this topic, but not this explicit statement.

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    $\begingroup$ To last your question there exists a direct argument: if a Hausdorff space $X$ is not discrete, then take a non-isolated point $x\in X$ and by induction construct a decreasing sequence $(U_n)_{n\in\omega}$ of neighborhoods of $x$ such that for every $n$ the set $U_{n+1}$ is disjoint with a neighborhood $V_n\subset U_n$ of some point $x_n\in U_n$. Then $\{x_n\}_{n\in\omega}$ will be a requred strongly discrete set in $X$. $\endgroup$
    – Taras Banakh
    Commented Apr 7, 2018 at 18:21
  • $\begingroup$ If your space is locally compact and $T_2$ then it's completely regular; furthermore assuming it's not countably compact, you should be able to thin-out, a countable set without accumulation point, to get your desired discrete set (in the strong sense.) If the space is countably compact, I suspect the answer does not easily follow. $\endgroup$
    – Not Mike
    Commented Apr 7, 2018 at 19:31

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The answer to this question is negative and can be obtained with the help of weak P-points in compact spaces with countable cellularity. A non-isolated point $p$ of a topological space $X$ is a weak $P$-point if $p$ is not an accumulation point of a countable set $C\subset X\setminus\{p\}$.

If $p$ is a weak P-point in a compact space $X$ with countable cellularity, then $X\setminus\{p\}$ is a locally compact space such that any strongly discrete subset is countable and hence has compact closure in $X\setminus\{p\}$.

It seems that examples of countably cellular spaces with weak P-points can be found using this result of Dow.

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  • $\begingroup$ Thank you! I have to admit that I did not know the definitions of neither countable celluarity, nor P-points. I guess if I knew, I wouldn't have asked the question. Yet another evidence of how numb the intuition is sometimes, and how broad the mathematics is. $\endgroup$
    – erz
    Commented Apr 7, 2018 at 23:18
  • $\begingroup$ @erz In fact, your intuition was not very far from the truth: by an old (1988) result of Tkachuk a Tychonoff space $X$ is compact if each discrete subspace of has compact closure. $\endgroup$
    – Taras Banakh
    Commented Apr 9, 2018 at 7:34
  • $\begingroup$ So these two notions of discreteness are different after all, right? $\endgroup$
    – erz
    Commented Apr 9, 2018 at 7:55

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