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Does there exist a compact Hausdorff space such that each singleton is not in $\cal{G_\delta}$? Maybe not difficult example.

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A nonempty family $\cal{A}$ of subsets of a fixed set $X$ is called a $\sigma$-ring if for each $A,B\in \cal{A}$ we have $A\setminus B \in \cal{A}$ and for each sequence $A_n, n \in \mathbb{N},$ elements of $\cal{A}$ we have $\cup_{n=1}^\infty A_n \in \cal{A}$. – L.T Dec 20 '11 at 9:31
This may help: – David Mitra Dec 20 '11 at 9:43
up vote 7 down vote accepted

Yes, there are compact Hausdorff spaces in which no point is a $G_\delta$. (I’ll have to think more about the second question.)

Let $X=\beta\omega\setminus\omega$; then $X$ is a compact Hausdorff space in which no singleton is a $G_\delta$. To see this, let $p\in X$, and think of $p$ as a free ultrafilter on $\omega$. Let $\{V_n:n\in\omega\}$ be a countable family of open nbhds of $p$. Then for each $n\in\omega$ there is a set $A_n\subseteq\omega$ such that $p\in\hat A_n\subseteq V_n$, where $$\hat A_n=\{q\in X:A_n\in q\}$$ is a basic open set in $X$. Without loss of generality we may assume that $\bigcap_{n\in\omega}A_n=\varnothing$, that $A_{n+1}\subseteq A_n$, and further that $|A_n\setminus A_{n+1}|\ge 2$ for each $n\in\omega$. For each $n\in\omega$ choose an integer $b_n\in A_n\setminus A_{n+1}$, and let $B=\{b_n:n\in\omega\}$. Then the families $$\mathscr{F}=\{B\}\cup\{A_n:n\in\omega\}$$ and $$\mathscr{G}=\{\omega\setminus B\}\cup\{A_n:n\in\omega\}$$ are both centred, so they can be extended to free ultrafilters $q(\mathscr{F})$ and $q(\mathscr{G})$, respectively. Exactly one of $B$ and $\omega\setminus B$ belongs to $p$. If $B\in p$, then $$q(\mathscr{G})\in\bigcap_{n\in\omega}\hat A_n\setminus\{p\}\;,$$ and if $\omega\setminus B\in p$, then $$q(\mathscr{F})\in\bigcap_{n\in\omega}\hat A_n\setminus\{p\}\;,$$ so $$\{p\}\ne\bigcap_{n\in\omega}\hat A_n \subseteq\bigcap_{n\in\omega}G_n\;,$$ and $\{p\}$ is not a $G_\delta$.

Added: An even simpler example is $X=\{0,1\}^{\omega_1}$. If $x\in X$ belongs to some $G_\delta$-set $H$, there is an $\alpha<\omega_1$ such that $G=\{y\in X:y\upharpoonright\alpha= x\upharpoonright\alpha\}\subseteq H$, and clearly $\{x\}\subsetneqq G$.

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Very thanks for examples. – L.T Dec 23 '11 at 12:26

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