Notes:Just what is in a generated sigma-algebra

Consider a topology on [ilmath]X[/ilmath], call it [ilmath]\mathcal{J} [/ilmath], what can we say about the [ilmath]\sigma[/ilmath]-algebra generated by [ilmath]\mathcal{J} [/ilmath], or [ilmath]\sigma(\mathcal{J})[/ilmath] as it is called?

Notes about generation

We know from [ilmath]\mathcal{J}\subseteq\sigma(\mathcal{J})[/ilmath] that automatically (by the implies-subset relation) every open set is in [ilmath]\sigma(\mathcal{J})[/ilmath]. We also know that [ilmath]\sigma(\mathcal{J})[/ilmath] is minimal, as small as it can be. This means:

• [ilmath]\sigma(\mathcal{J})[/ilmath] only contains things we can get to via the operations a [ilmath]\sigma[/ilmath]-algebra allows starting from the open sets, [ilmath]\mathcal{J} [/ilmath]

Example

Take the usual topology on [ilmath]\mathbb{R} [/ilmath] (which we will call [ilmath](\mathbb{R},\mathcal{J})[/ilmath]) generated by open sets of the form [ilmath](a,b)[/ilmath] (as the interval [ilmath]\{x\in\mathbb{R}\vert\ a< x< b\} [/ilmath]), and let us ask the question:

• Is [ilmath]\{5\}\in\sigma(\mathbb{R})[/ilmath]? The answer is "yes", there are several ways.
  1. First we can do the famous counter example as to why toplogies require closure under finite intersection only:

     Let [math]A=\bigcap_{n=1}^\infty\left(5-\tfrac{1}{n},5+\tfrac{1}{n}\right)[/math], it is easy to see that

     • We have used the [ilmath]\sigma[/ilmath]-[ilmath]\cap[/ilmath]-closed property, as we know that [ilmath]\forall n\in\mathbb{N}[(5-\tfrac{1}{n},5+\tfrac{1}{n})\in\mathcal{J}\implies(5-\tfrac{1}{n},5+\tfrac{1}{n})\in\sigma(\mathcal{J})][/ilmath]