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 :)
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.
