I'm having some confussions about the differences between $\sigma$-algebras and topologies over sets. My main confussion is about wether or not one can be generalized by the other one. The definitions are evidently similar:
Definition 1:
A $\sigma$-algebra is a set $\Sigma\subseteq P(X)$, where $P(X)$ is the power set of a given set $X$ such that, for every $A_i\in\Sigma$, the following axioms hold:
- $X\in\Sigma$
- $A^\prime_i\in\Sigma$
- $\left(\bigcup_{i\in\mathbb{N}}A_i\right)\in\Sigma$
Also, it can be inferred from the axioms that $\Sigma$ is closed under countable (finite) intersections, and that $\emptyset\in\Sigma$ (wikipedia).
We also have that
Definition 2
A topology over a set $X$ is a class $\Sigma$ of subsets of $X$ that satisfies, for every $A_i,A_j\in\Sigma$:
- $A_i\cup A_j\in\Sigma$
- $A_i\cap A_j\in\Sigma$
Also, by definition, $\emptyset,X\in\Sigma$.
Given these two definitions, i claim:
$U$ is closed for $U\in T\Leftrightarrow T$ is a measurable space
Where $T$ is a topological space.
Proof: Let $T=(X,\Sigma)$ be a topological space, such that all open subsets $U\in T$ are closed sets. Since $U$ is open and closed (clopen), then $U^\prime$ is also clopen, so $U^\prime\in T$. By definition, $X\in T$ and $T$ is closed under arbitrary unions, so $\Sigma$ is a $\sigma$-algebra and the space $T$ is measurable.
To prove the other way of the implication, we start with measurable space $(X,\Sigma)$, where $\Sigma$ is a $\sigma$-algebra. We define an open subset $U$ to be open iff $U\in\Sigma$. By definition, $U^\prime\in\Sigma$, so $U^\prime$ is open. Suppose $A,B\in\Sigma$ are open. $\Sigma$ is closed under countable unions and intersections, so $A\cap B\in \Sigma$ and $A\cup B\in\Sigma$ are open. Finally, every $U^\prime,A\cup B,A\cap B\in\Sigma$ is open so $\Sigma$ is a topology, and the space $T=(X,\Sigma)$ is a topological space where $U\in T$ is closed. $\square$
This proof seems ok, and appears to correctly generalize measurables spaces in terms of topological ones. Is this proof ok?
