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?
closed as off topic by Joel David Hamkins, Gerald Edgar, quid, Qiaochu Yuan, Qfwfq Jul 12 '11 at 18:18Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question. 


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})$. 

