If $A$ generates the topology $\mathcal{T}$ and the $\sigma$-field $\mathcal{F}$ then $\mathrm{Borel} (\mathcal{T}) =\mathcal{F}$? Let $\mathcal{A}$ a collection of subsets of $\mathbb{X}$. Let $ \mathcal{T}$ the topology generated by colection $\mathcal{A}$ and $\mathcal{F}$ the  $\sigma$-field generated by $ \mathcal{A}$.
Denote by , $\mathrm{Borel}(\mathcal{T})$  the $\sigma$-field of Borel sets of $\mathbb{X}$ whit respect the topology $\mathcal{T}$. 

Question1:  Is true that $\mathrm{Borel} (\mathcal{T}) =\mathcal{F}$?

Thank's.

Edit:

Question 2: Suppose now that $\mathbb{X}$ is countable and discrete with respect to some metric $d$  that generates the topology $\mathcal{T}$. Is true that $\mathrm{Borel} (\mathcal{T}) =\mathcal{F}$? 

If answer is not, I have a more question.

Question 3 If the answer to question 2 is still there some condition (topological, metric or condition of measurability) on $\mathbb{X}$ or $\mathcal{A}$ it is enough that the answer is yes?


Let $X$ be uncountable and $\mathcal{A}$ be the set of all singletons. The topology generated contains all subsets of $X$, but the $\sigma$-algebra generated contains only those subsets that are countable or have a countable complement. So the answer is no.
Edit: To the second and third question: It is enough that $\mathcal{A}$ is countable for the two $\sigma$-algebras to coincide. Since $\mathcal{A}\subseteq\mathcal{T}$, we have always $\sigma(\mathcal{A})\subseteq\sigma(\mathcal{T})=\mathrm{Borel} (\mathcal{T})$. Now, if $\mathcal{A}$ is countable, then the set of finite intersections of elements of $\mathcal{A}$ is countable too and forms a basis for the topological space $\mathcal{T}$. So every open set is a countable union of these finite intersections and therefore in $\sigma(\mathcal{A})$.  
