Does the Borel-$\sigma$-Algebra define the topology of a Set?

Question

Or is is possible that two Different Topologys $\tau_1$, $\tau_2$ generate the Same Borel-$\sigma$-Algebra?

Answer 1

Let $τ$ be a topology. In general, for every Borel set $A$, the topology $τ'$ generated by $τ ∪ \{A\}$ induces the same $σ$-algebra, since every $τ'$-open set is of the form $(A ∩ U) ∪ V$ where $U$ and $V$ are $τ$-open. So unless every Borel set is already open, there is always a different topology with the same induced $σ$-algebra.

Added: If every Borel set is already open, open sets are stable under complements and arbitrary intersections, and so the topology is just a sum of indiscrete spaces, i.e. discrete if $T_0$.

Answer 2

Consider an interesting countable space like for example $\mathbb{Q}$ with the usual topology $\tau_{st}$. Then $\mathcal{B}(\mathbb{Q})$ is a family of all subsets of $\mathbb{Q}$. In particular, $\tau = \mathcal{B}(\mathbb{Q})$ is a topology. Thus $\tau$ and $\tau_{st}$ generate the same $\sigma$-algerbra of sets.

For a more complex example consider $\mathbb{R}$ with the topology $\tau_{dir}$ generated by sets $[a,b)$ for $a<b$. Pick also $\mathbb{R}$ with the usual topology $\tau_{st}$. Then $\tau_{st}\subsetneq \tau_{dir}$ (the latter topology is not first countable) and these two topologies generate the same $\sigma$-algebra of Borel sets.

Answer 3

Consider $X = \{a,b\}$ with $\tau_1 = \{\varnothing, \{a\}, \{a,b\} \}$ and $\tau_2 = \{\varnothing, \{a\}, \{b\}, \{a,b\} \}$. Then $\sigma(\tau_1)=\sigma(\tau_2) = \{\varnothing, \{a\}, \{b\}, \{a,b\} \}$.

Answer 4

The lower-limit topology on $\Bbb R$ (the Sorgenfrey line) and $\Bbb R$ in the usual topology are very non-homeomorphic topologies on the same set that have the same Borel $\sigma$-algebra. (The base elements of the Sorgenfrey line are Borel in the usual topology.)