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](https://math.stackexchange.com/questions/92905/an-example-of-topological-space-in-which-each-singleton-is-not-in-g-delta) and also [Parovicenko space](http://en.wikipedia.org/wiki/Parovicenko_space).)


There are versions of the above question for LOTS 
(i.e. [linearly ordered topological spaces](http://en.wikipedia.org/wiki/Order_topology) 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](https://math.stackexchange.com/questions/1215866/separation-in-compact-spaces) (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>



Is there a compact connected Hausdorff space in which every non-empty $G_\delta$ set has non-empty interior?
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.


