# Example of a locally compact + Hausdorff + ¬normal + connected space

Playing with $$\pi$$-base for a while, I found that In fact, only four spaces on $$\pi$$-base satisfy the first three properties. My question is:

Can we prove or disprove the existence of a locally compact + Hausdorff + ¬normal + connected space?

Many thanks!

Related:

• What happens if you take the Tychonoff-plank-minus-corner example and make it connected by filling in intervals between consecutive points in both $\omega_1+1$ and $\omega+1$. In other words, take the one-point compactification $\bar L=L\cup\{*\}$ of the long line $L$, form the product $\bar L\times[0,1]$, and delete the corner point $(*,1)$? This is locally compact and $T_2$ because it's an open subspace of the compact $T_2$ space $\bar L\times[0,1]$. It seems to not be normal as $L\times\{1\}$ and $\{*\}\times[0,1)$ can't be separated, and it looks connected. – Andreas Blass Mar 12 at 18:53
• @AndreasBlass a separation for those sets would imply one for the deleted Tychonoff plank and so cannot exist. The example works I think. – Henno Brandsma Mar 12 at 21:32

The fourth example is easily "fixable":

Let $$X= L(\omega_1) \times [0,1]^{[0,1]}$$ (we could also take $$[0,1]^{\omega_1}$$ as the second factor if we care about minimal weight, with the same result), and where $$L(\omega_1)$$ is the so-called long line, which is a "connectified" version of $$\omega_1$$ (or $$[0,\Omega)$$, as $$\pi$$-base calls it; the first uncountable ordinal (in the order topology)).

• $$X$$ is Hausdorff as the product of two Hausdorff spaces.
• $$X$$ is locally compact as the product of the locally compact $$L(\omega_1)$$ and the (locally) compact $$[0,1]^{[0,1]}$$. (local compactness is finitely productive.)
• $$X$$ is connected as the product of two connected spaces.
• $$X$$ is not normal, as it contains $$L(\omega_1) \times \beta L(\omega_1)$$ as a closed subspace, where $$\beta L(\omega_1)$$ is the Cech-Stone compactification of $$L(\omega_1)$$ and equals its one-point compactfication (which is homeomorphic to the extended long line ($$L^\ast$$ on the linked page)), because $$[0,1]^{[0,1]}$$ is a universal space for all spaces of weight $$\le \aleph_1$$ (even for all spaces of weight $$\le \mathfrak{c}$$) by Tychonoff's embedding theorem) and so contains (a homeomorphic copy of) $$\beta L(\omega_1)$$ as a (necessarily closed) subspace. Tamano's theorem that $$X$$ is paracompact Hausdorff iff $$X \times \beta X$$ is normal, plus the fact that $$L(\omega_1)$$ is not paracompact implies that $$X$$ is not normal. This is in essence the same argument (with $$\omega_1$$ replaced by its connected variant $$L(\omega_1)$$) that we also can hold to show non-normality of the fourth space from the $$\pi$$-base list (which has no special name that I know of).

So we could also use a space of the form $$Y \times \beta Y$$, where $$Y$$ is a locally compact, connected, Hausdorff space. This makes $$\beta Y$$ connected (as $$Y$$ is dense in it) and we have the same essential argument (minus the universal space argument, but we need Tamano's theorem) that such a product will be an example.

So $$\mathbb{R} \times \beta \mathbb{R}$$ works too, e.g. (but has a much larger weight). So maybe you can make more special examples (with extra properties) this way.

I would have liked to have been able to make a connected version of Mrówka's $$\Psi$$-space, which is also pseudocompact and not countably compact. But I did not see how to do that...

The hunt for interesting counterexample is always a fun one.

Note: This post is long and ultimately doesn't answer OP's question. I just wanted to suggest a line of attack.

One thing that you might try first is the following, every locally compact Hausdorff space is $$T_{3}$$. We then might first broaden the objective to finding a connected $$T_{3}$$ space that isn't normal. The $$\pi$$-base does has some examples of these.

Only two though

With one of these spaces in hand, you can attempt to "locally compactify" the space. That is you can embed such a space into a locally compact Hausdorff space in such a way that it is a dense subspace. As connectedness is closed under closure this may be do the job so long as the local compactification doesn't introduce normality. There are then two questions, one of which there is an answer for:

1. How you locally compactify a ($$T_{3}$$) space?
2. How do you make sure that a local compactification is not normal?

I will admit that I do not know how to answer the second question. The first question however has a reasonably well studied answer. Namely, every separated local proximity space admits a local compactification. All of the following information was pulled from Naimpally and Warrack's book "Proximity Spaces".

Definition: A nonempty collection $$\mathcal{B}$$ of subsets of a topological space $$X$$ is called a boundedness (this might be old terminology) iff

(i) $$A\in\mathcal{B}$$ and $$B\subseteq A$$ implies $$B\in\mathcal{B}$$ (ii) $$\mathcal{B}$$ is closed under finite unions.

Basically a boundedness is a bornology without the requirement that the collection covers the space.

Definition: A local proximity psace is a triple $$(X,\alpha,\mathcal{B})$$ where $$X$$ is a set, $$\mathcal{B}$$ is a boundedness on $$X$$ and $$\alpha$$ is a binary relation on the power set of t$$X$$ satisfing:

(i) $$A\alpha B\implies B\alpha A$$

(ii) $$(A\cup B)\alpha C\iff$$ $$A\alpha C$$ or $$B\alpha C$$

(iii) $$A\alpha B\implies A,B\neq\emptyset$$

(iv) $$A\cap B\neq\emptyset\implies A\alpha B$$

(v) Let $$A\subseteq X$$ and $$B\in\mathcal{B}$$. If for every $$C\in\mathcal{B}$$ either $$A\alpha C$$ or $$(X\setminus C)\alpha B$$, then $$A\alpha B$$.

(vi) If $$A\alpha B$$, then there is a $$D\in\mathcal{B}$$ such that $$D\subseteq B$$ and $$A\alpha D$$.

If in addition to these axioms the relation $$\alpha$$ satisfies

$$\{x\}\alpha\{y\}\iff x=y$$

then the local proximity space $$(X,\alpha,\mathcal{B})$$ is called separated.

The topology on a local proximity space $$(X,\alpha,\mathcal{B})$$ is defined by means of the closure operator

$$\overline{A}:=\{x\in X\mid \{x\}\alpha A\}$$

Some good facts to have about local proximity spaces are the following:

(i) The closure of bounded sets are bounded.

(ii) Every compact subset of a local proximity space is bounded.

Finally, how do we locally compactify? Effectively we define a means of getting a one-point compactification.

Definition: Let $$(X,\alpha,\mathcal{B})$$ be a local proximity space an define a binary relation $$\delta$$ on the power set of $$X$$ by

$$A\delta B\iff A\alpha B\,\text{ or }A,B\notin\mathcal{B}$$

Then $$\delta$$ is called the Alexandroff extension of $$\alpha$$, and is a proximity on $$X$$. Moreover, if $$\alpha$$ was a separated localy proximity, then $$\delta$$ is separated.

Every separated proximity space has a compactification called its Smirnov compactification. There is a certain point in the Smirnov compactification that you can delete, leaving you with a locally compact Hausdorff space. If your original local proximity space is connected, then this locally compact hausdorff space will be connected as well. The parts of the theorem that would matter for you are the following:

Theorem: Given a separated local proximity space $$(X,\alpha,\mathcal{B})$$, there is a locally compact Hausdorff space $$L$$ and a one-to-one map $$f:X\rightarrow L$$ such that:

(i) $$A\alpha B\iff\overline{f(A)}\cap\overline{f(B)}=\emptyset$$ in $$L$$

(ii) $$B\in\mathcal{B}\iff\overline{f(B)}$$ is compact in $$L$$

(iii) $$\overline{f(X)}=L$$

This is probably where I would start the search for the space you want if I didn't have an obvious example. I am not at all sure how to make sure that normality isn't gained in this process.

EDIT: I forgot to mention, every completely regular space admits a proximity relation whose induced topology is the original topology.