MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer
I think this is also the standard example of a non-normal subspace of a normal space. – Andreas Blass Jan 26 '11 at 0:34
See also related… – Joel David Hamkins Jan 26 '11 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 '11 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 '11 at 17:31
@Andreas Blass: Indeed, any non-normal locally compact Hausdorff space is an example of a non-normal subspace of a normal space. The one-point compactification is compact Hausdorff, hence normal. – Michael Jul 22 '14 at 0:35

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.

share|cite|improve this answer
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 '11 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 '11 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 <a href="">original paper</a>. – Henno Brandsma Feb 27 '11 at 13:13
[continuing] Such spaces are non-normal already by the theorem that normal pseudocompact spaces are countably compact. So if a course would cover these notions, I'd include $\Psi$-space as well. – Henno Brandsma Feb 27 '11 at 13:15
I don't think there is a constructive proof like for the Sogenfrey line, at least if the rational sequences are not given beforehand. If we merely consider two disjoint subsets of the irrationals we can arrange for sequences with even denominators to converge to the first set, and for sequences with odd denominators to converge to the other and then basic neighborhoods suffice to separate the sets. – mathahada Feb 27 '11 at 14:18

$\pi$-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

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.