In the article **AN EXAMPLE INVOLVING BAIRE SPACES** (<https://www.ams.org/journals/proc/1975-048-01/S0002-9939-1975-0362249-1/S0002-9939-1975-0362249-1.pdf>) of H. E. White Jr. it is shown that, assuming CH, there exists a Baire space $Y$ such that $Y\times Y$ is not Baire.


For the construction of this space, is used the density topology on $\mathbb{R}$ (for details of this topology you can see <https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-62/issue-1/The-density-topology/pjm/1102867878.full>).




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Denote by $\mathcal{T}$ the density topology on $\mathbb{R}$ and by $\mathcal{E}$ the Euclidean topology on $\mathbb{R}$.


The construction of the space $Y$ begins with an enumeration $(F\_{\alpha})\_{\alpha<\omega\_{1}}$ of all $\mathcal{E}$-Borel sets of measure zero (Lebesgue measure on $\mathbb{R}$). Then, by transfinite recursion on $\omega\_{1}$ is construct a sequence $(Y\_{\alpha})\_{\alpha<\omega\_{1}}$ of countable rational vector subspaces of $\mathbb{R}$ and finally our space is $Y=\bigcup\_{\alpha<\omega\_{1}}Y\_{\alpha}$.


My question is the following :


In the article it is mentioned that $Y$ is not extremally disconnected, does anyone have any idea how to prove that fact?


Remember that a topological space $X$ is **extremally disconnected** if the closure of every open subset of $X$ is open.


Thanks a lot.



CH and the density topology on $\mathbb{R}$
According to your first reference your space is dense with respect to the density topology; this implies that, in $Y$, the closure of $Y\cap(0,\infty)$ is $Y\cap[0,\infty)$; the latter set is not open in $Y$ because its upper density at $0$ is less than or equal to $\frac12$.


That last sentence needs some amplification.To complete the argument hinted at in the last sentence: because $Y$ is a vector space over $\mathbb{Q}$ it is closed under shifting and scaling by rational numbers. From this it follows that there is a constant $c$ such that $m^\*(Y\cap(a,b))=c\cdot(b-a)$ for every interval $(a,b)$, and $c>0$ because $Y$ has positive outer measure. Now let $A$ be open in the density topology such that $A\cap Y\subseteq[0,\infty)$. Then $A\cap(-\infty,0)$ is disjoint from $Y$ and so $m(A\cap(a,b))\le(1-c)(b-a)$ whenever $a<b\le0$; but this implies $m(A\cap(-\infty,0))=0$. Therefore the density of $A$ at $0$ is at most $\frac12$, and so $0\notin A$.


