Is there an example of a sigma algebra that is not a topology? If this is not the case, is it possible to prove that all sigma algebras are topologies?
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5$\begingroup$ This question might be better suited to math.stackexchange. Note: if your $\sigma$-algebra includes singletons, as many do, then if it were a topology, it would have to include all subsets of the space. $\endgroup$– Joel David HamkinsCommented Jul 12, 2011 at 16:13
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4$\begingroup$ @Joel: I'm probably being silly, but is it clear that in $\mathtt{ZF}$ this gives the requested example? More precisely, can you pin down a set $A$ such that the $\sigma$-algebra $\sigma(\{\{a\} : a \in A\})$ generated by singletons of elements of $A$ forms a proper subset of $\mathcal{P}(A)$? Things seem to get tricky in models where all uncountable cardinals are singular. $\endgroup$– Clinton ConleyCommented Jul 12, 2011 at 19:31
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2$\begingroup$ Clinton, I was working (habitually) in ZFC, and thinking about the usual $\sigma$-algebra of Borel sets, which are a counterexample once you know there is a non-Borel set. But you are right that this cannot be proved in ZF, and so your comment makes an interesting question! Namely, is it consistent with ZF that every $\sigma$-algebra is a topology? I'm not sure if François's answer in his linked question provides the answer, but it is surely very relevant. $\endgroup$– Joel David HamkinsCommented Jul 12, 2011 at 20:28
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2$\begingroup$ @Clinton: I just added an update to my old answer. The wellordered case is much easier to prove, but the fact holds for all sets. $\endgroup$– François G. DoraisCommented Jul 12, 2011 at 20:45
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3$\begingroup$ Clinton, why not ask the question explicitly as an MO question: Is it consistent with ZF that every $\sigma$-algebra is a topology? François's answer with Gitik's model nearly answers it, and may very well provide a full answer, if we dig a bit deeper into it. $\endgroup$– Joel David HamkinsCommented Jul 12, 2011 at 21:55
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1 Answer
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In order to elaborate on Joel Hamkins comment: The $\sigma$ algebra $A$ generated by the open sets of $\mathbb{R}$ is not the power set of $\mathbb{R}$, since there are subsets of $\mathbb{R}$, which are not contained in $A$ by the axiom of choice. Now suppose $A$ is a topology, then for every subset $X \subset \mathbb{R}$ as the union $\cup_{x \in X} \left\{ x \right\}$ would be in the topology, and hence measurable, which contradicts the observation that $ A \neq P(\mathbb{R})$.
answered Jul 12, 2011 at 17:25
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$\begingroup$ How to render the brackets correctly? $\endgroup$ Commented Jul 12, 2011 at 17:28
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2$\begingroup$ A small nitpick: your statement that "this [is] equivalent to the existence of non Lebesgue measurable sets," is not quite accurate, since it is consistent with ZF+DC (from suitable assumptions) that every set is Lebesgue measurable, but in these models, not every set is Borel. (For example, this situation arises under AD.) That is, knowing that there are non-Lebesgue measurable sets is not the same as knowing that there are non-Borel sets. Meanwhile, it is consistent with ZF that every set is Borel, such as when AC fails so badly that the reals are a countable union of countable sets. $\endgroup$ Commented Jul 12, 2011 at 17:35
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1$\begingroup$ pm: It seems the backslashes needed to be escaped by another backslash before it would be parsed. $\endgroup$ Commented Jul 12, 2011 at 17:53
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$\begingroup$ @Joel: Interesting, I edited the answer. $\endgroup$ Commented Jul 16, 2011 at 19:23