$\sigma$-algebra generated by a finite set $S$

Question

How the Borel $\sigma$-algebra generated by a finite set $S$ may look like?

Answer 1

The Borel-$\sigma$-Algebra contains any subset that can be formed by intersecting or taking countable unions of open sets. I assume you want your finite set to carry the discrete topology, i.e. every set is open. Then you're Borel-$\sigma$-Algebra will just be the entire power set.

Answer 2

I think we have to answer without the assumption of taking the discrete topology, i.e. the identification of our topology with the powerset of our space.

We can consider exempli gratia the space $Ω=\{a,b\}$ and the topology $\mathcal T=\{\varnothing, Ω, \{a\}\} \neq \mathcal P (Ω)$. We then construct $\mathcal B (Ω)$ (with the reminder that $\mathcal B (Ω)$ is the minimum $σ$-algebra containing $\mathcal T$) as follows:

It must be $\mathcal T \subseteq \mathcal B(Ω)$ and $\varnothing \in \mathcal B(Ω)$, so we get $\varnothing, Ω, \{a\} \in \mathcal B(Ω)$. We also need $\{a\}^c=\{b\}\in \mathcal B(Ω)$ (closure of complements). Finally the closure of countable unions is satisfied by above, thus $\mathcal B(Ω)=\mathcal P (Ω)$.

Of course this result is not universal for finite spaces: If $Ω=\{a,b,c\}$ and topology is as above, then obviously $\mathcal B(Ω)\neq\mathcal P (Ω)$. Consequently Borel sets depend on the topology.