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If you're working on $\mathsf {ZF}$ and you assume the compactness theorem for propositional logic, then you have the prime ideal theorem, and thus you can show that the dual of the category of Boolean algebras is equivalent to the category of Hausdorff $0$-dimensional spaces.

There is a natural functor $F$ yielding the equivalence, namely the covariant functor $F$ given by $F(\mathbb B)=S(\mathbb B)$; the stone space of $\mathbb B$, and for a homomorphism of Boolean Algebras $f:\mathbb A\rightarrow \mathbb B$, we consider the continuous function $F(f):S(\mathbb B)\rightarrow S(\mathbb A)$ given by $F(f)(u)=\{a\in \mathbb A: f(a)\in u\}$ for all ultrafilters $u$ of $\mathbb B$.

So, my question is, if we're only assuming $\mathsf {ZF}$, and you have that the dual of the category of Boolean algebras is equivalent to the category of Hausdorff $0$-dimensional spaces, does the compactness theorem for propositional logic hold?; this is only idle curiosity.

Thanks

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  • $\begingroup$ What's $S$ in the second paragraph? $\endgroup$ – Asaf Karagila Apr 7 '13 at 18:33
  • $\begingroup$ @Asaf: Likely the Stone space of $\mathbb{B}$. $\endgroup$ – user642796 Apr 7 '13 at 18:36
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    $\begingroup$ Asking whether two categories are equivalent, rather than asking whether a particular functor is an equivalence, seems to be an unnatural thing to do. $\endgroup$ – Zhen Lin Apr 7 '13 at 18:43
  • $\begingroup$ @Zhen Lin: If am I not mistaken, the duality theorem states that this functor gives the equivalence, no? So it seems like a reasonable (read: natural?) question after all. $\endgroup$ – Asaf Karagila Apr 7 '13 at 18:44
  • $\begingroup$ Yes, that is a reasonable question. But the last one says, suppose that the two categories are equivalent for unknown reasons; is the obvious functor an equivalence? This is unnatural. $\endgroup$ – Zhen Lin Apr 7 '13 at 18:57
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Let $\textbf{Bool}$ be the category of boolean algebras, and let $\textbf{Stone}$ be the category of Stone spaces (however you define it). Suppose $F : \textbf{Bool}^\textrm{op} \to \textbf{Stone}$ is a weak equivalence of categories (i.e. fully faithful and essentially surjective on objects); we will deduce the boolean prime ideal theorem.

First, note that $\textbf{Bool}$ has an initial object, namely the boolean algebra $2 = \{ 0, 1 \} = \mathscr{P}(1)$, and it has a terminal object, namely the boolean algebra $1 = \{ 0 \} = \mathscr{P}(\emptyset)$. Since $F$ is a weak equivalence, $F(2)$ must be a terminal object (so a one-point space) and $F(1)$ must be an initial object (the empty space). [It is very easy to check these preservation properties even in the absence of a quasi-inverse for $F$.]

Let $B$ be a boolean algebra. Clearly, prime ideals of $B$ correspond to boolean algebra homomorphisms $B \to 2$, and hence, to continuous maps $F(2) \to F(B)$, which are the same thing as points of $F (B)$. But if $F (B)$ is empty, then the canonical map $F(1) = \emptyset \to F(B)$ is a homeomorphism, and so the canonical homomorphism $B \to 1$ is an isomorphism. [Again, this is easy to check even in the absence of a quasi-inverse for $F$.] Thus, $B$ has a prime ideal if and only if it is non-trivial.

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  • $\begingroup$ What about principal ultrafilters? (Read: Isn't $F(B)$ never empty?) $\endgroup$ – Asaf Karagila Apr 7 '13 at 20:02
  • $\begingroup$ Principal ultrafilters only make sense in the context of powersets, no? $B$ could be an atomless boolean algebra. $\endgroup$ – Zhen Lin Apr 7 '13 at 23:45
  • $\begingroup$ D'oh! And I was struggling with this when I was trying to figure out an answer... Why didn't I think of that? :-) $\endgroup$ – Asaf Karagila Apr 8 '13 at 6:08

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