I have a question in $\beta\mathbb{R}$, the Stone-Cech compactification of the real line $\mathbb{R}$. My question is: is $\beta\mathbb{R}$ a $\mathrm{F}$-space, i.e., the closure of two disjoint open $F_{\sigma}$-sets are disjoint? I know that $\beta\mathbb{R}\setminus\mathbb{R}$ is a $\mathrm{F}$-space, but not if the whole space has this property.

Thank you for your help in advance :)

  • This is not a vector space. Could you provide a correct link for F-space? And Fréchet space only makes sense for vector space, so this sounds senseless. – YCor Dec 4 at 8:57
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    Or perhaps you mean the meaning of F-space as the one used in Alan Dow's paper Some set-theory, Stone–Čech, and F-spaces doi.org/10.1016/j.topol.2011.06.007 and some other related papers cited there? – Martin Sleziak Dec 4 at 9:10
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    @MartinSleziak : I edited my definition of an F-space sorry for that. – user132068 Dec 4 at 9:11
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    Perhaps you could add at least some reference to the fact that the Stone-Čech remainder is an F-space - I'd guess that the paper by Gillman and Henriksen: Rings of continuous functions in which every finitely generated ideal is principal doi.org/10.1090/S0002-9947-1956-0078980-4 seems like a reasonable candidate - and I suppose that after that we can delete all comments related to clarification of the question. – Martin Sleziak Dec 4 at 9:16
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    Yes Gillman and Hendriksen is the right reference for this. One can also prove that the closure of every open $F_{\sigma}$-set is open (this property is also known under $\sigma$-Stonean if I am right. This would imply that we have a F-space. – user132068 Dec 4 at 10:25

The answer is no, essentially because $\mathbb{R}$ embeds as a locally compact open subspace of $\beta\mathbb{R}$, and $\mathbb{R}$ is not an F-space.

In detail, for the purposes of this answer I will write $\mathbb{R} \subseteq \beta\mathbb{R}$. The facts we will use are that $\mathbb{R}$ is an open subspace of $\beta\mathbb{R}$ because it is locally compact, and that compact subsets of $\mathbb{R}$ are compact, and therefore closed, in $\beta\mathbb{R}$.

Consider $(0,1)$ and $(1,2)$ in $\mathbb{R}$. These are disjoint opens in $\mathbb{R}$, therefore in $\beta\mathbb{R}$ (because $\mathbb{R}$ is an open subset). The first set $(0,1) = \bigcup\limits_{i=1}^\infty [2^{-i},1-2^{-i}]$, so is $F_\sigma$ in $\beta\mathbb{R}$ (because closed bounded intervals are compact in $\mathbb{R}$, and therefore closed in $\beta\mathbb{R}$). A similar argument shows that $(1,2)$ is $F_\sigma$. Their closures are $[0,1]$ and $[1,2]$ in $\mathbb{R}$, and as these are compact, they are also their closures in $\beta\mathbb{R}$. These are not disjoint.

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