## Locally compact Hausdorff space that is not normal

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), but 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.

-

## 4 Answers

Spacebook, a database-driven version of Steen and Seebach's Counterexamples in Topology, lists the following locally compact Hausdorff (i.e. T2) spaces that are not normal. (Some of these appear already in other answers.)

Deleted Tychonoff Plank

Open Uncountable Ordinal Crossed with Uncountable Cartesian Product of Unit Interval

Rational Sequence Topology

Thomas’s Plank

-

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 countanly 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.

-

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.) we have that this space is not normal.

-
 This is an excellent example! Is there an explicit construction of two closed, disjoint sets that cannot be separated in this space? – mathahada Feb 27 2011 at 11:57 The Jones' lemma argument is a nice counting argument: in a separable space there are at most $\mathfrak{c} = 2^{\aleph_0}$ many real-valued functions (as each function is determined by its restriction to the countable dense subset), and for every subset $B$ of the closed discrete set $A$ we need a distinct function continuous real-valeud $f_B$ that is 1 on $B$ and 0 on $A \setminus B$, both of which are closed and disjoint in the whole space. This does need Urysohn functions and some set theory, but you cannot really do topology without these anyway, so it's useful in a course, I think. – Henno Brandsma Feb 27 2011 at 13:06 So to recap (comments can only be so long...), I don't know of any direct argument (2 closed sets that cannot be separated), but the counting argument is nice enough, and can also be used to show non-normality of $SxS$, where $S$ is the Sorgenfrey line, or for Mrowka's $\Psi$-space (I think the rational sequence topology is called $\Psi$-like, by some authors, as it's very similar, that first space is also pseudocompact and non-countably compact, and needs a MAD family on $\mathbb{N}$, see the
show 1 more comment

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.

-
I think this is also the standard example of a non-normal subspace of a normal space. – Andreas Blass Jan 26 2011 at 0:34
See also related mathoverflow.net/questions/30662/… – Joel David Hamkins Jan 26 2011 at 0:55
Also, Counterexamples in Topology by Steen and Seebach is good for this type of question. The Tychonoff plank is mentioned in there. – Todd Trimble Jan 26 2011 at 1:03
The example below is slightly "nicer", in that it is first countable as well, while points like $(\omega_1, 0)$ do not have countable local base. – Henno Brandsma Feb 27 2011 at 17:31