Let $X$ be the Stone space of the free $\sigma$-algebra $A$ on $\omega_1$ free generators.
Is $X$ separable (i.e. does $X$ contain a countable dense set)?
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$\begingroup$ Given the free $\sigma$-algebra $A_E$ on $E$ with free generators $(x_t)_{t\in E}$, is it correct that mapping $A_E$ to $2^{2^E}$ by $x_t\mapsto \{Y\in 2^E:t\in Y\}$ induces an isomorphism from $A_E$ onto the $\sigma$-algebra of Borel subsets of the compact space $2^E$? Surjectivity is clear, I'm just not sure for injectivity. If it indeed holds, this is a pretty concrete description (from which the cellularity fact is obvious). $\endgroup$– YCorCommented Feb 7, 2019 at 22:03
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$\begingroup$ @YCor If you replace Borel by Baire, your statement is correct (and can be found in Halmos's book, for example). However, it is false with Borel if $E$ is uncountable - a singleton in $2^E$ is a Borel set that is not a Baire set. $\endgroup$– Robert FurberCommented Jun 27, 2019 at 16:14
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1 Answer
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No.
Just looking at countably many generators we can produce a continuum of pairwise disjoint clopen subsets of $X$. Moreover, since $|A|=2^{\aleph_0}$, we have that $2^{\aleph_0} \leq c(X) \leq d(X) \leq w(X) \leq 2^{\aleph_0}$, where $c$, $d$ and $w$ denote cellularity, density and weight respectively.
answered Feb 5, 2019 at 20:25
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$\begingroup$ Thank you @Ramiro de la Vega for the answer. Would $X$ satisfy at least the countable chain condition? (Equivalently, is the free $\sigma$-algebra on $\omega_1$ free generators ccc?) $\endgroup$– LJGCCommented Feb 6, 2019 at 10:08
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1$\begingroup$ @LJGC No, to say that $X$ is ccc is the same as saying that $c(X)=\aleph_0$. $\endgroup$ Commented Feb 6, 2019 at 13:40
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$\begingroup$ @LJGC of course not, the generators themselves are a counterexample. $\endgroup$ Commented Feb 7, 2019 at 19:15
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$\begingroup$ To get continuum pairwise disjoint clopen subsets requires a little argument. I guess one is the following: write the generators as $(x_q)_{q\in\mathbf{Q}}$. For $q\in\mathbf{Q}$, define $a_q=\sup_{r\le q}x_r$. For $r\in\mathbf{R}$, define $b_r^+=\inf_{q>r}a_q$ and $b_r^-=\sup_{q<r}a_q$, and $c_r=b_r^+-b_r^-$. Then the $c_r$ are all nonzero (as we see using a representation by intervals) and are pairwise disjoint. $\endgroup$– YCorCommented Feb 7, 2019 at 19:46
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$\begingroup$ @YCor Would not just countable meets like $\left(\bigwedge_{n\in S}x_n\right)\land\left(\bigwedge_{m\notin S}\neg x_m\right)$ for all possible $S$ work? $\endgroup$ Commented Feb 7, 2019 at 21:07