TITLE: Is there a set that is both a sigma algebra and a topology but not a powerset?
QUESTION [12 upvotes]: Is there a set that is both a sigma algebra, $\Sigma$, and a topology, $\tau$, but not a powerset, $\mathcal{P}$?

REPLY [12 votes]: If you add the Hausdorff condition, then you can achieve the full power set result, as I explain below. Meanwhile, in general, here is a sense in which every such algebra/topology is in fact a power set. 
Theorem. Every family $\cal{A}$ of sets that is both a $\sigma$-algebra and a topology on a set $X$ (or even just an algebra and a topology) is isomorphic, as a partial order under $\subseteq$, to a power set order. Specifically, $\cal{A}$ is isomorphic to the power set of the collection of minimal nonempty elements of $\cal{A}$, which form a partition of $X$. 
Proof. Suppose that $\cal{A}$ is closed under complement and is a topology on $X$. Thus, $\cal{A}$ is closed under complement and arbitrary unions, and hence by de Morgan it is closed under arbitrary intersection. For each $x\in X$, let $[x]$ be the intersection of all $A\in\cal{A}$ with $x\in A$. Thus, $[x]\in \cal{A}$, and these are simply the minimal nonempty elements of $\cal{A}$. These partition $X$, and form the equivalence classes of the relation $x\sim y$ iff $x$ and $y$ are in all the same elements of $\cal{A}$. Every element of $\cal{A}$ is a union of the various classes $[x]$, and furthermore, every union of classes $[x]$ is in $\cal{A}$. So the partial order $\langle \cal{A},{\subseteq}\rangle$ is isomorphic to the power set of the classes $\langle P(X/\sim),\subseteq\rangle$, since every element $A\in \cal{A}$ breaks into the union of pieces of the partition $A=\bigcup_{a\in A}[a]$. QED
Let us now add a separation requirement to the topology to achieve the full power set result. 
Theorem. Every family $\cal{A}$ of sets that is both an algebra and a Hausdorff topology on $X$ is the full power set of $X$.
Proof. If the topology is Hausdorff, or merely even $T_1$, then the relation $x\sim y$ I mention above is just $x=y$, and so $[x]=\{x\}$. So all the singletons are in $\cal{A}$, which is closed under arbitrary unions.  So ${\cal A}=P(X)$. QED