Stone-Cech Compactification of the real line

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.

• 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
• 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
• @MartinSleziak : I edited my definition of an F-space sorry for that. – user132068 Dec 4 at 9:11
• 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
• 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.