# Sigma Algebra that is not a topology [closed]

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?

• 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. – Joel David Hamkins Jul 12 '11 at 16:13
• @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. – Clinton Conley Jul 12 '11 at 19:31
• 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. – Joel David Hamkins Jul 12 '11 at 20:28
• @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. – François G. Dorais Jul 12 '11 at 20:45
• 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. – Joel David Hamkins Jul 12 '11 at 21:55

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 $A \neq P(\mathbb{R})$.