Q1. Is there a compact connected Hausdorff space (with at least two points) in which every non-empty $G_\delta$ set has non-empty interior? (Without the requirement for connectedness, every finite $T_1$ space is an example, and a more interesting example is the remainder of the Stone-Cech compactification of the integers, see this MSE question and also Parovicenko space.)

There are versions of the above question for LOTS (i.e. linearly ordered topological spaces with the order also called open-interval topology), and for topological groups.

Q2. Is there a compact connected LOTS in which every non-empty $G_\delta$ set has non-empty interior? (I believe the following is a restatement: Is there a compact connected LOTS which is not first-countable at any point?)

Q3. Is there a compact connected Hausdorff topological group in which every non-empty $G_\delta$ set has non-empty interior? (Such a group would necessarily have to be non-metrizable.)

Question 1 was inspired by the following MSE question (in particular by my answer there). Questions 2 and 3 are related and I feel the answers (to all questions) are probably known, in that case would you please provide reference(s).

Edit. Thanks to @ChristianRemling for pointing out that I am interested in examples with at least 2 points. (It follows that such examples, being connected and normal, will have at least continuum many points.)

Edit. The following answer was posted by @Alessandro Vignati (who perhaps could not post it is a comment): "You should have a look at the Stone-Cech reminder of $[0,1)$. I suspect this may be an example of what you're looking for in Q1".

First thank you for the suggested answer. While I agree that the Stone-Cech remainder of $[0,1)$ (or of $[0,\infty)$) is an example to look at, and I had already taken a (brief) look at related papers, I could not deduce the answer, so I posted the above question. Perhaps I am overlooking something obvious, but at any rate unless I see a more specific reference to the suggested result, or a (sketch of) proof, I would not consider the above question answered. Here are a couple of links to papers related to the Stone-Cech remainder $H^*$ of the half-line $H=[0,\infty)$ (indeed $H^*$ is compact and connected, and perhaps satisfies the condition about $G_\delta$ sets I am asking about, but I do not quite see this part, even if I agree it is not something unreasonable to suspect): http://arxiv.org/pdf/math/9805008.pdf and http://arxiv.org/pdf/0708.0838.pdf

  • $\begingroup$ @JosephVanName Thank you ... I either didn't know or had forgotten this terminology, though I spent some time searching for $P$-spaces, but $P$-space is a too strong condition (I think when every $G_\delta$ set is itself open). $\endgroup$ – Mirko Apr 5 '15 at 19:15
  • 1
    $\begingroup$ It was proved by Fine and Gillman that the remainder of a locally-compact realcompact space, e.g.[0,1), is an almost $P$-space. See N. Fine and L. Gillman, Extensions of continuous functions in $\beta\mathbb{N}$, Bull. Amer. Math. Soc. 66 (1960) 376–381. $\endgroup$ – Ramiro de la Vega Apr 6 '15 at 15:22

This is a proof of Alessandro Vignati's guess: Let $G_n$ be a decreasing sequence of nonempty open subsets of $X=\beta[0,1)\setminus[0,1)$. Choose $x\in\bigcap G_n$ - by assumption the intersection is nonempty. Then for each $n$ choose a pair of disjoint subsets of $\beta[0,1)$, namely $U_n\ni x$ and $V_n\supseteq X\setminus G_n$. They exist by normality of $\beta[0,1)$. Then we have $\bar U_n\cap X\subseteq G_n$. We may also assume that the sequence $U_n$ is decreasing. For every $n\in\mathbb{N}$ choose an open interval $I_n\subseteq [0,1)$ such that:

  • $I_n\subseteq U_n$
  • the supremum of $I_n$ is less than $1$
  • the infimum of $I_n$ is more than $1-{1\over n}$
  • to make things more elegant we may assume that $I_n$ sits above $I_{n-1}$

Such $I_n$ always exists since $U_n\cap X$ nonempty forces $U_n\cap[0,1)$ to be unbouded ($X$ has an empty interior in $\beta[0,1)$). Note that for $n<m$ we have $I_m\subseteq U_m\subseteq U_n$. Let $U=\bigcup I_n$ then for every $n$ there exists an $a<1$ such that $U\cap[a,1)\subseteq U_n$. Therefore $G=(\mathop{\rm int}\bar U)\cap X$ is contained in $\bigcap G_n$, and it is nonempty as $U$ is open and unbouded.


It's not hard to construct a compact connected LOTS with this property. Let $X_0$ be any countably saturated dense linear order, and let $X$ be its bounded Dedekind completion ("bounded" meaning also add points at $\pm\infty$). Explicitly, such an $X_0$ can be constructed by a transfinite induction of length $\omega_1$, where at each stage you add a new point into every cut (even cuts for which a sup or inf exists). Suppose $U_n$ is a sequence of open sets in $X$ and let $x\in\bigcap U_n$. Assume $x\neq\pm\infty$; if $x=\pm\infty$ the argument is similar. We can find a sequence of intervals $(a_n,b_n)\subseteq U_n$ with $a_n,b_n\in X_0$ and $$a_0<a_1<a_2<\dots<x<\dots<b_2<b_1<b_0.$$

Since $X_0$ is countably saturated, there exist $a,b\in X$ such that $a_n<a<b<b_n$ for all $n$. Thus $\bigcap U_n$ contains the $(a,b)$ and has nonempty interior (note that $x$ itself might not be in the interior; it could be $\sup a_n$ or $\inf b_n$, but it can't be both).

  • $\begingroup$ I couldn't do the "but it can't be both" part on my own yesterday (using somewhat simpler but related ideas), thank you for being specific with the construction. I wasn't sure of the meaning of "countably saturated", wikipedia suggests there are two versions of the definition, I assume you meant the version with "countable sets of parameters" rather than "finite parameter sets". $\endgroup$ – Mirko Apr 5 '15 at 19:40

The completely regular spaces $X$ such that every $G_{\delta}$ set has a non-empty interior are called almost $P$-spaces. The following facts will help you construct numerous examples of almost $P$-spaces.

Recall that a $P$-space is a completely regular space where every $G_{\delta}$-set is open. Therefore, the notion of an almost $P$-space is a weakening of the notion of a $P$-space. If $X$ is locally compact and realcompact, then $\beta X\setminus X$ is an almost $P$-space (this fact generalizes the answer of Adam Przeździecki). A completely regular space $X$ is an almost $P$-space if and only if its Hewitt Realcompactification $\upsilon X$ is an almost $P$-space.

A product of spaces $X\times Y$ is an almost $P$-space if and only if both $X$ and $Y$ are almost $P$-spaces. It is easy to show that a dense subspace of an almost $P$-space is an almost $P$-space. However, a closed subspace of an almost $P$-space is not necessarily an almost $P$-space. In fact, every completely regular space can be embedded as a closed subspace of an almost $P$-space. If $B$ is a Boolean algebra, then the Stone space of $B$ is an almost $P$-space if and only if $B$ has no countable partitions. In particular, if $B$ is $\aleph_{1}$-saturated, then the Stone space $S(B)$ is an almost $P$-space. The papers 1,2 give more information on almost $P$-spaces (with the link to 2 provided by Mirko).

For question 2, an example of a totally ordered almost $P$-space is given in Proposition 2.3 of 2.

  1. Almost P-spaces. Chang Il Kim. Commun. Korean Math. Soc. 18 (2003), No. 4, pp. 695–701

  2. Ronnie Levy. Almost-P-Spaces. Can. J. Math., Vol. XXIX, No. 2, 1977, pp. 284-288

  • $\begingroup$ Thank you, I also found another paper by Ronnie Levi about almost $P$-spaces $\endgroup$ – Mirko Apr 5 '15 at 20:33
  • $\begingroup$ Mirko. Thanks for providing the link to that paper. $\endgroup$ – Joseph Van Name Apr 5 '15 at 20:47
  • $\begingroup$ though I misspelled his name, should be Ronnie Levy. His definition is that in an almost P-space, $G_\delta$ sets have dense interiors (clearly he means dense in the $G_\delta$ set, and his version of the definition is equivalent to requiring that the interior be non-empty when the $G_\delta$ set is non-empty). $\endgroup$ – Mirko Apr 5 '15 at 20:54

I claim that no infinite compact group can be an almost $P$-space. In other words, in every compact group there is a nonempty $G_{\delta}$ that does not contain any nonempty open set.

Suppose that $G$ is an infinite compact group. Let $\mu$ be the Haar probability measure on $G$. I claim that every non-empty open subset of $G$ has positive measure. Suppose that $U\subseteq G$ is a non-empty open set. Then $G=\bigcup_{g\in G}Ug$, so by compactness, there are $g_{1},...,g_{n}$ with $G=Ug_{1}\cup... U g_{n}$. However, since $\mu(U)=\mu(Ug_{1})=...=\mu(Ug_{n})$, we conclude that $1=\mu(G)\leq\mu(Ug_{1})+...+\mu(Ug_{n})=n\mu(U)$, so $\mu(U)\geq\frac{1}{n}$. However, it is easy to construct a non-empty $G_{\delta}$ subset of $G$ with measure zero. Therefore $G$ is a non-empty $G_{\delta}$-set that does not contain any non-empty open set. A similar argument shows that every non-discrete locally compact abelian group is not an almost $P$-space.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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