What can we say about a locally compact Hausdorff space whose every open subset is sigma compact? Can we say that it is metrizable or second countable?



We can say neither. A classical counterexample (due to Alexandroff and Urysohn) is the so-called Double Arrow space (description here; I believe that in Counterexamples in Topology it is called the Weak Parallel Line topology), which is compact Hausdorff (it's ordered even) and perfectly normal space. The latter implies all open subsets are $F_\sigma$ and thus $\sigma$-compact in particular. But it is not second countable (and thus not metrisable) as e.g. the upper half contains (a homeomorphic copy of) the Sorgenfrey line.


I investigated this question as follows. Suppose that $X$ is a locally compact Hausdorff space whose every open subset is $\sigma$-compact. Then Alexandroff compactification $\alpha X$ should be a compact Hausdorff space whose every open subset is $\sigma$-compact. A space whose every open subset is $F_\sigma$ is called perfect. It is known that a compact Hausdorff space with a perfect square (even with a $G_\delta$-diagonal) is metrizable (see, for instance, [Gru, 2.13]). I tried to prove or to find a theorem claimed that each Hausdorff compact perfect space is metrizable and at last I found a paper [John], which may contain an example different from Double Arrow space.

And even we cannot say that $X$ is metrizable, we can say some things about it knowing that $X$ is a perfect compact space without a point.

Suppose that $X$ has a $G_\delta$-diagonal. It seems the following. There is a countable family $\mathcal F$ of closed subsets of $\alpha X$ such that $\bigcup\mathcal F\cap (X\times X)=\Delta\backslash\{(\alpha,\alpha)\}$, where $\Delta$ is a diagonal of the space $\alpha X$. Also there is a countable family $\mathcal F_X$ of closed subsets of the space $\alpha X$ such that $\bigcup\mathcal F_X=\alpha X\backslash\{\alpha\}$. Put $\mathcal F'=\{F\cap F_1\times F_2:F\in\mathcal F, F_1,F_2\in\mathcal F_X\}\cup \{F\times\{\alpha\}\cup \{\alpha\}\times F:F\in\mathcal F_X\}$. Then $\bigcup\mathcal F'=\Delta\backslash \{(\alpha,\alpha)\}$, and therefore the space $\alpha X$ is metrizable.


[Gru] Gary Gruenhage, Generalized Metric Spaces, in K. Kunen and J. E. Vaughan, Handbook of set theoretic topology, Elsevier, 1984.

[John] Roy A. Johnson. A Compact Non-Metrizable Space Such That Every Closed Subset is a G-Delta. The American Mathematical Monthly, Vol. 77, No. 2, pp. 172–176.


No. The counterexamples can be seen in the above answers.

However we can conclude that $X$ is hereditarily Lindelöf.

Lemma 1: $X$ is a hereditarily Lindelöf space if and only if all open subspaces of $X$ have the Lindelöf property.)

Proof: Firstly note that $X$ is $\sigma$-compact and hence it is Lindelöf. Notice every open set of $X$ is also $\sigma$-compact, and hence every open set is also Lindelöf. Therefore $X$ is a hereditarily Lindelöf space.

If $X$ has some conditions, for example, $X$ has a $G_\delta$-diagonal, such that $X$ is locally metrizable. A locally metrizable Lindelöf space is metrizable, see Arhangel’skii A V, Buzyakova R Z. On some properties of linearly Lindelöf spaces[C]//Topology Proc. 1998, 23: 1-11.. So $X$ is a Lindelöf metrizable, and hence $X$ is also second countable.

Some notes:

  1. In fact, there is a nicer result from the paper.

    A locally metrizable $\omega_1$-Lindelöf is metrizable. $\omega_1$-Lindelöf is a property which is weaker than Lindelöf.

  2. Lindelöf and second countable are equivalent in the Metrizable spaces.

  3. $X$ is a hereditarily Lindelöf space if and only if all open subspaces of $X$ have the Lindelöf property. It can be seen the general topology by Engelking Page 194.

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    $\begingroup$ You don’t need the result from Arhangel’skiĭ & Buzyakova. $X$ is $T_3$ and Lindelöf, so it’s paracompact. The Smirnov metrization theorem says that a space $Y$ is metrizable iff $Y$ is locally metrizable and paracompact, so $X$ is metrizable. $\endgroup$ – Brian M. Scott Jun 5 '13 at 9:39
  • $\begingroup$ @Brain M.Scott: Thanks for the comment. I really appreciated it. $\endgroup$ – Paul Jun 5 '13 at 10:01
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    $\begingroup$ It is not locally metrisable, as the Double Arrow shows. This is even compact. If locally compact hereditarily Lindelöf spaces were locally metrizable, we could prove that ZFC has no compact $L$-spaces (and the Suslin hypothesis would be proved etc...). See my counterexample to the original question. $\endgroup$ – Henno Brandsma Jun 5 '13 at 11:11

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