Question

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.

Answer 1

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 $X$.

Answer 2

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.

Answer 3

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.