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Let $X$ be a compact $T_0$ topological space such that every closed subset of $X$ is locally connected. Is there any characterization for such a space? I guess $X$ is Noetherian, but I cannot prove that. Any hint is appreciated.

Recall that a topological space $X$ is called Noetherian if any ascending chain of open subsets stabilizes after finitely many steps, equivalently, any non-empty set of closed subsets of $X$, ordered by inclusion, has a minimal element.

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    $\begingroup$ I think this would be more appropriate at mse. Incidentally, such spaces need not be Noetherian: consider for example the topology on $\mathbb{R}_{\ge 0}$ consisting of all upwards-closed sets. Every closed set in this topology is connected, and every open cover has a singleton subcover. (Even better, consider the cofinite topology on any infinite set - this is additionally $T_1$.) $\endgroup$
    – Noah Schweber
    Commented Apr 22, 2020 at 18:59
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    $\begingroup$ And trivially there are no $T_2$ examples with more than one point. $\endgroup$
    – Noah Schweber
    Commented Apr 22, 2020 at 19:05
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    $\begingroup$ @NoahSchweber, cofinite is Noeterian. $\endgroup$
    – Wlod AA
    Commented Apr 23, 2020 at 3:11
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    $\begingroup$ @NoahSchweber, cocountable is not compact (sorry to bother you about trivialities). $\endgroup$
    – Wlod AA
    Commented Apr 23, 2020 at 3:22
  • $\begingroup$ @WlodAA Welp, not my day. $\endgroup$
    – Noah Schweber
    Commented Apr 23, 2020 at 3:23

1 Answer 1

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Let $\ (Y\ S)\ $ be an arbitrary locally connected space. Let $\ Y\subset X\ $ -- sharp inclusion (not just $\, \subseteq).\ $ Define:

$$ T\ :=\ S\cup \{X\} $$

Then $\ (X\ T)\ $ is compact and every closed subset of $\ X\ $ is connected and locally connected.

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    $\begingroup$ You can use $\subsetneq$ \subsetneq for proper inclusion, to avoid needing a separate qualifier afterwards. $\endgroup$
    – LSpice
    Commented Apr 23, 2020 at 3:58
  • $\begingroup$ @LSpice, very nice, thank you. However, $\ \subsetneq\ $ is not aesthetic and it LOOKS ambiguous, it may be visually interpreted as "not contained": $\ Y\setminus X\ne\emptyset.\ $ (True, there is a special notation for this too but nevertheless...). $\endgroup$
    – Wlod AA
    Commented Apr 23, 2020 at 4:06

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