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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 ...

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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 ...

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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 ...

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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 ...

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By Proposition 3.4 of D. Dikranjan; L. Stoyanov. An elementary approach to Haar integration and Pontryagin duality in locally compact abelian groups. Topology Appl. 158 (2011), no. 15, 1942--1961. if $G$ is an infinite abelian group and $\mathcal T$ is a Hausdorff precompact group topology on $G$, there are infinitely many continuous homomorphisms $f:G\to ... 0 There is a more elementary complete proof for this claim in: D. Dikranjan; L. Stoyanov. An elementary approach to Haar integration and Pontryagin duality in locally compact abelian groups. Topology Appl. 158 (2011), no. 15, 1942--1961. Precompact in this claim, which is called Comfort-Ross Theorem, is not going to include Hausdorffness. So if$PC(G)$is ... 1 Question 1: This is a standard Baire category argument. Let$U$be some small neighborhood of$c$. Since$G(X)$acts continuously on$X$, there is a neighborhood$V$of the identity in$G(X)$, such that$V^{-1} V c \subseteq U$. Choose a neighbourhood$V_1$of the identity, such that the closure$\overline{V_1}$is contained in$V\$, and is compact. Since ...

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