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On page 85 of Jech's Set Theory (3rd Edition), a complete Boolean algebra $B$ is defined to be $\kappa$-distributive if

\begin{equation}\label{a}\tag{1} \prod_{\alpha < \kappa}\, \sum_{i \in I_\alpha} u_{\alpha, i} = \sum_{f \in \prod_{\alpha < \kappa}} \prod_{\alpha < \kappa} u_{\alpha, f(\alpha)} \end{equation}

for any collection $\{u_{\alpha, i}\}_{\alpha < \kappa, i \in I_\alpha}$ of $B$. This property is easily seen to hold for a complete algebra of sets; for an arbitrary Boolean algebra, Jech presents two different characterizations of \eqref{a} in Lemma 7.16.

My question is why $\kappa$-distributivity is not just a consequence of Stone's Representation Theorem (Theorem 7.11): Suppose I wish to verify \eqref{a} for some Boolean algebra $B$. By Stone's Representation Theorem, there is an isomorphism $\varphi$ mapping $B$ to $S$, where $S$ is an algebra of sets. It suffices to verify that

\begin{equation} \varphi\left(\prod_{\alpha < \kappa}\, \sum_{i \in I_\alpha} u_{\alpha, i}\right) = \varphi\left(\sum_{f \in \prod_{\alpha < \kappa}} \prod_{\alpha < \kappa} u_{\alpha, f(\alpha)}\right). \end{equation}

To that end,

\begin{align} \varphi\left(\prod_{\alpha < \kappa}\, \sum_{i \in I_\alpha} u_{\alpha, i}\right) &= \bigcap_{\alpha < \kappa} \varphi\left(\sum_{i \in I_\alpha} u_{\alpha, i}\right)\tag{2}\\ &= \bigcap_{\alpha < \kappa}\,\bigcup_{i \in I_\alpha} \varphi\left(u_{\alpha, i}\right)\tag{3}\\ &= \bigcup_{f \in \prod_{\alpha < \kappa}} \bigcap_{\alpha < \kappa} \varphi\left(u_{\alpha, f(\alpha)}\right)\tag{4}\\ &\:\,\, \vdots\\ &= \varphi\left(\sum_{f \in \prod_{\alpha < \kappa}} \prod_{\alpha < \kappa} u_{\alpha, f(\alpha)}\right)\tag{5}, \end{align}

where (2), (3) and (5) hold because an isomorphism is a complete homomorphism; and where (4) is due to \eqref{a} and $S$ being an algebra of sets.

Certainly this argument is flawed, but I cannot pinpoint where I go wrong.

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The meet and join of infinitely many elements in an algebra of sets need not be the intersection and union (respectively).

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  • 1
    $\begingroup$ Thank you, this clarifies things. Is there an accessible example of the situation you mentioned? $\endgroup$ – user480881 Feb 16 at 2:07
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    $\begingroup$ Consider the algebra of clopen subsets of the Cantor set $C$. In this algebra, consider a sequence that shrinks down to one point, say $X_n=C\cap[0,3^{-n}]$. Then the intersection of these sets is $\{0\}$, but that's not clopen; it's not in the algebra under consideration. The meet in the algebra is $\varnothing$. $\endgroup$ – Andreas Blass Feb 16 at 2:10
  • $\begingroup$ Brilliant, thanks! $\endgroup$ – user480881 Feb 16 at 2:21

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