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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).

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.

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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 $0\notin A$.

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  • $\begingroup$ Thanks @KP Hart. In fact, $\overline{Y\cap ]0,+\infty[}^{Y}=Y\cap \overline{Y\cap ]0,+\infty[}^{\mathcal{T}}=Y\cap \overline{]0,+\infty[}^{\mathcal{T}}=Y\cap [0,+\infty[ $, because $Y$ is dense, $\endgroup$
    – Gabriel Medina
    Commented Mar 14, 2021 at 0:07
  • $\begingroup$ Now, my question is why $Y\cap [0,+\infty[$ is not open in $Y$?. Otherwise, there is $A\in\mathcal{T}$, such that $Y\cap [0,+\infty[=A\cap Y$. As $Y$ is dense, then $\mbox{int}_{\mathcal{T}}(A\setminus Y)=\emptyset$. $\endgroup$
    – Gabriel Medina
    Commented Mar 14, 2021 at 0:10
  • $\begingroup$ Then $0\in A\cap Y$, so $1=\lim_{h\to 0}\frac{m(A\cap ]-h,h[)}{2h}$. $\endgroup$
    – Gabriel Medina
    Commented Mar 14, 2021 at 0:11
  • $\begingroup$ $1=\lim_{h\to 0}\frac{m( (A\cap Y\cap]-h,h[) \cup (A\setminus Y\cap]-h,h[))}{2h}=\lim_{h\to 0}\frac{m( (Y\cap [0,h[) \cup ((A\setminus Y)\cap]-h,h[))}{2h} $. $\endgroup$
    – Gabriel Medina
    Commented Mar 14, 2021 at 0:17
  • $\begingroup$ At first I don't know if $Y$ is measurable, we only know that it has a positive outer measure and is dense. So I don't know how to get to a contradiction. $\endgroup$
    – Gabriel Medina
    Commented Mar 14, 2021 at 0:19

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