Locally compact Hausdorff space that is not normal

Question

What is a good example of a locally compact Hausdorff space that is not normal? It seems to be well-known that not all locally compact Hausdorff spaces are normal (and only a weaker version of Urysohn's Lemma holds in general in the locally compact Hausdorff case). However, I can't seem to think of any examples that demonstrate this, and I have tried all of the "standard" topological counterexamples such as the long line, etc.

Answer 1

I think the Tychonoff plank serves as an example. It is obtained by taking the product of the two ordinals $\omega_1+1$ and $\omega+1$, each with the order topology, and removing the corner point $(\omega_1,\omega)$. The product is a compact Hausdorff space, so the plank, as an open subspace, is locally compact. But it is not normal, because the "edges" $\omega_1\times\{\omega\}$ and $\{\omega_1\}\times\omega$ don't have disjoint neighborhoods.

Answer 2

Another nice elementary example is the rational sequence topology. For every irrational number $x$ we pick a sequence $q(x)_n$ of rational numbers, all different, that converge to $x$ (in the usual topology on $\mathbb{R}$). A topology on $\mathbb{R}$ is then defined by specifying basic neighbourhoods: a rational number $q$ has $ \{ q \}$ as a basic neighbourhood (it is isolated), while an irrational number $x$ has basic neighbourhoods of the form $\{ q(x)_n : n \ge k \}$, $k \in \mathbb{N}$. One checks that this defines a topology in which the irrationals are closed and discrete (in itself), $\mathbb{Q}$ is dense (and open), and $X$ is Hausdorff and zero-dimensional (basic open sets are clopen, this uses that 2 sequences that converge to 2 different irrationals only have at most finitely many terms in common, so no basic neighbourhood can have another irrational in its closure), so Tychonov, and locally compact, as all basic neighbourhoods are compact (finite or convergent sequences). But by Jones' lemma (in a normal space, with dense set $D$ and closed discrete set $A$ we have that $2^{|A|} \le 2^{|D|}$), a proof of which can be found here, e.g.; Wayback Machine) we have that this space is not normal.

Answer 3

$\pi$-Base, a database-driven version of Steen and Seebach's Counterexamples in Topology, lists the following locally compact, Hausdorff spaces that are not normal. You can view the search result to learn more about these spaces.

$[0,1) \times I^I$

Deleted Tychonoff Plank

Rational Sequence Topology

Thomas’s Plank

Answer 4

Let me give a construction that gives many locally compact spaces that are not normal. If $X$ is a completely regular space and $C$ is a compactification of $X$, then $X$ is paracompact if and only if $X\times C$ is normal. Therefore if $X$ is locally compact but not paracompact, then $X\times C$ is locally compact but not normal. Furthermore, the paracompact locally compact spaces are precisely the spaces that can be partitioned into a collection of $\sigma$-compact clopen sets (recall that a space is $\sigma$-compact if and only if it can be written as a countable union of compact sets). In particular, if $X$ is a connected locally compact space that is not $\sigma$-compact and $C$ is a compactification of $X$, then $X\times C$ is not normal.

Answer 5

My example of a locally compact space which is not normal ,is the Katetov space. this space is defined as follows:

$K$ =$\beta\mathbb{R}$-($cl_{\beta \mathbb{R}}$$\mathbb{N}$-$\mathbb{N}$). this space has The countable subset $\mathbb{N}$ as a closed subset with any accumulation point in $K$. that's why, this space is not countably compact. but this space is pseudocompact.(You can easily check this claim). then this space is not normal. because a normal pseudocompact space is countably compact.