# Does there exist a Hausdorff group which is not locally compact?

A topological space is countably compact if every countable open cover has a finite subcover.

A topological space $X$ is locally compact if any point has a neighbourhood which is compact.

A topological space $X$ is locally countably compact if any point has a neighbourhood which is countably compact. That is, for any point $x\in X$, we can find a neighbourhood $U$ of $x$, such that $U$ is a countably compact space.

I need two specific examples:

1. $G$ is a Hausdorff topological group, but $G$ is not locally compact.

2. $G$ is a Hausdorff topological group, and $G$ is locally countably compact, but $G$ is not locally compact.

Or can we give a wealth of examples about them?

Thanks a lot.

• The first part is easy because a Hausdorff topological vector space is locally compact if and only if finite dimensional, a result given early on in Rudin's Functional Analysis. So for part (1) take any infinite dimensional normed vector space (considered as an additive group). – hardmath Jan 14 '16 at 5:44
• Isn't there a "countably" too much in the second paragraph? – Hagen von Eitzen Jan 15 '16 at 7:17
• @HagenvonEitzen edited it... – Henno Brandsma Jan 15 '16 at 8:28

The first is easy: $\mathbb{Q}$ is a Hausdorff (metrisable) topological group, that is not locally compact. Or we can use $\mathbb{Z}^\mathbb{N}$ in the product topology (which is a topological group that is (as a topological space) homeomorphic to the irrationals, another non-locally compact space).

Or, as the comment suggested, any infinite-dimensional topological vector space (which are never locally compact, as this implies finite-dimensionality in the context of a topological vector space).

The second is somewhat more complicated. I think the following works: consider the $\Sigma$-product of $\omega_1$ many copies of $\{0,1\}$, so $G = \{(x_\alpha)_{\alpha < \omega_1} \in \{0,1\}^{\omega_1}: |\{\beta < \omega_1: x_\beta = 1\}| \le \aleph_0 \}$, all sequences of $0$ and $1$ that have at most countably many $1$'s.

It is well-known that this $G$ is countably compact (a sequence in total has only countably many non-$0$ coordinates as well, and so essentially lives in a countable product of copies $\{0,1\}$ and has a limit point there), but not compact (being dense in $\{0,1\}^{\omega_1}$), and a Hausdorff topological group. And a basic neighbourhood of $0$ is (I'm pretty sure) homeomorphic to $G$ again, so also countably compact and non-compact. So it fits your description.

• I'm kind of confused. Does $\{0,1\}$ carry the discrete topology? Does $G$ carry the final topology induced by the uncountably many injections $\{0,1\} \to G$? – rwols Jan 15 '16 at 11:21
• $\{0,1\}$ is discrete and $G$ has the subspace topology from the product topology on $\{ 0,1 \}^{\omega_1}$ (it's a subgroup of the latter as well). – Henno Brandsma Jan 15 '16 at 11:42