The question arose while comparing the notions of compactness, countable compactness, local compactness, and "Lindelofness" in Hausdorff spaces. It is straightforward to show that compactness implies any of the other properties. I found ready counterexamples (I will be glad to provide them if asked) for all but one of the other possible implications, namely the question of whether countable compactness implies local compactness. The paper "On Countably Compact Nonlocally Compact Spaces" by T. B. Rushing shows that, in general, countable compactness does not imply local compactness. The examples he provides, however, are not Hausdorff.

  • $\begingroup$ What happens to the examples in the paper if they are made Hausdorff? $\endgroup$ – Loop Space Aug 30 '10 at 7:35
  • $\begingroup$ What does it mean to "make" a space Hausdorff if it is not so already? (Please forgive the question. I am still a novice in such matters.) $\endgroup$ – Austin Mohr Aug 30 '10 at 21:08
  • $\begingroup$ There is a "Hausdorffication": Every topological space has a biggest hausdorff quotient, that is a quotion $X\to HX$ such that all continouos maps $X\to Y$ into hausdorff spaces factor as $X\to HX\to Y$. $\endgroup$ – Johannes Hahn Mar 24 '13 at 0:41

Examples abound: take for instance a $\Sigma$-product of two-point spaces. To be specific let $X$ be the set of points in $\lbrace0,1\rbrace^{\omega_1}$ that have only countably many coordinates that are $1$. This set is dense but not open in the product, hence not locally compact but it is countably compact as each countably infinite subset sits in a compact subset of $X$.

  • $\begingroup$ Hi KP. It seems that I am too slow typing my answers. And of course, your example is more elegant than mine. $\endgroup$ – Stefan Geschke Aug 30 '10 at 8:40

Le $X$ be any Hausdorff, sequentially compact, not compact space (e.g. $\omega_1$ with the order topology). Then $X^\mathbb{N}$ is Hausdorff, sequentially (hence countably) compact, and not locally compact, because any set with non-empty interior is mapped surjectively onto $X$ by some projection, thus is not compact as $X$ itself is not.

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    $\begingroup$ Rmk: you need to start with $X$ sequentially, not just countably compact, because sequential compactness is preserved by countable products (via the diagonal argument), while countable compactness may be lost already in $X^2$. $\endgroup$ – Pietro Majer Aug 30 '10 at 8:30

Example 3.10.17 of Engelking's General Topology (Heldermann Verlag Berlin 1989) is the following dense subspace of $I^{\mathbb R}$, where $I=[0,1]$:

Let $X$ be the set of all $(x_t)_{t\in\mathbb R}\in I^{\mathbb R}$ such that $x_t$ is different from zero for at most countably many $t\in\mathbb R$.
Engelking shows that the space is countably compact but not compact.
Almost the same argument shows that the space is not locally compact. I sketch the proofs below.

Being a subspace of $I^{\mathbb R}$, the space is Hausdorff.
Countable compactness follows from the fact that every countable subset of $X$ is contained in a subspace of $X$ that is homeomorphic to a countable product of closed unit intervals.

The space is not locally compact:
Let $x\in X$ and let $U\subseteq X$ be an open neighborhood of $x$.
By shrinking $U$ if necessary we may assume that $U$ is the intersection of $X$ with a product of non-degenerate intervals. Now the closure of $U$ in $X$ is dense in a product of closed intervals, but not equal to this product. It follows that the closure of $U$ in $X$ is not compact.

Hence $X$ is not locally compact.


Even worse! ;) You might have a Hausdorff countably compact space which is not locally countably compact in the stronger sense that every point has a neighborhood base of countably compact sets (of course, for Hausdorff spaces, the strong and weak forms of "locally compact" are equivalent; but you need $T_3$ to get the corresponding equivalence for "locally countably compact").

Consider $ \omega_2$, the third infinite cardinal, with the usual order topology, and make the topology finer by adding $A= \{ \alpha \in \omega_2 | \alpha \text{ has cofinality } \omega_1 \}$ as an open set. So, a base for the topology is given by the open intervals together with the sets of the form $L \cap A$, for $L$ an open interval.

With this topology, $ \omega_2$ remains countably compact, since every infinite subset has a complete accumulation point. It obviously remains Hausdorff. Notice that, with this topology, any ordinal of cofinality $ \omega_1$ is an isolated point, provided it is not a limit of a set of ordinals of cofinality $ \omega_1$. If you consider instead an ordinal $\alpha$ of cofinality $ \omega_1$ such that $\alpha$ is also a limit of ordinals of cofinality $ \omega_1$, then $A$ is a neighborhood of $\alpha$, but no neighborhood $U$ of $\alpha$ contained in $A$ is countably compact, since any such $U$ contains a closed infinite discrete set, by the above remark (I mean, the infinite discrete set is closed in U).

I am sure that this example or something similar appears in the literature, but I do not recall where. Probably there are also simpler counterexamples (in the counterexample above, too, you can start with an ordinal much smaller than $ \omega_2$, of course). I just found that N. Noble (Two examples on preimages of metric spaces , Proc. Amer. Math. Soc. 36 (1972), 586-590) mentions that the existence of such counterexamples is relevant in constructing other counterexamples about $k$-spaces.

A lot of time has gone by, but I think it might be useful to point this out.


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