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Since the total space of a cover is locally homeomorphic to the base space, local topological properties (like local (path) connectedness, T1 etc.) lift from the base space to the total space. The same holds for local compactness, if we assume the base space is Hausdorff.

My question is, given a non-Hausdorff locally compact space $X$, must every cover of $X$ be locally compact.

My intuition says no, because the compact neighbourhoods in $X$ might be too big to be "seen" by the covering structure. I can't think of any counterexamples, mainly because I don't remember any non-compact non-Hausdorff spaces right now and won't have access to Steen & Seebach for a couple of days. Although, when writing this, it strikes me that there could be conditions on $X$, which would ensure that the covering projection is a proper map, in which case we would be done.

NB: By a locally compact space I mean a space in which every point has a compact neighbourhood.

Edit: Thinking further, if the base space is T3/regular (the weaker of the two, whatever your convention might me) and locally compact every cover is locally compact, basically for the same reason as in the Hausdorff case.

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  • $\begingroup$ Have you tried making a "non-locally compact circle" and then mimicked the covering of the circle by the line? Here your "non-locally compact circle" might be something like the rational points on a circle, or some other bizarre creature. $\endgroup$
    – wckronholm
    May 27, 2011 at 22:23
  • $\begingroup$ @wckronholm I think you misunderstood me. I want the base space to be locally compact and not Hausdorff (looking at it, the phrasing might be a bit confusing). I've been thinking that the one-point compactification of the rationals might provide a counterexample. As far as I know, there are theorems characterizing the homeomorphic type of the rationals, from which it should be possible to construct an appropriate cover. $\endgroup$
    – Miha Habič
    May 29, 2011 at 9:31
  • $\begingroup$ @Miha you're right. I've had this question of yours running around the back of my head for a few days, and it got a bit jumbled up. Have you had any luck constructing a cover of $\mathbb{Q}^+$? $\endgroup$
    – wckronholm
    May 30, 2011 at 5:09
  • $\begingroup$ @wckronholm Maybe something like this could work: pick a proper neighbourhood $U$ of $\infty$ in $\mathbb{Q}^+$. Then take as the total space $\mathbb{Q}^*\times\{1/n;n\in\mathbb{N}\}\cup U\times 0$ with the obvious projection. I'm almost sure this is a covering space. I think it isn't locally compact on the 0-th level, but I'm not so sure about that. Any thoughts? $\endgroup$
    – Miha Habič
    Jun 3, 2011 at 13:32
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    $\begingroup$ See the answer on mathoverflow (mathoverflow.net/questions/67181/…) $\endgroup$
    – Taras Banakh
    Nov 28, 2017 at 18:13

1 Answer 1

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As mentioned in the comments, this question was subsequently asked on MathOverflow. It was answered there by Taras Banakh (see his answer there), so I am writing this answer to close this version as well.

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