Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Reference to any book with a theorem that solves this issue are welcome. –  Elias Nov 12 '12 at 18:11

1 Answer 1

up vote 3 down vote accepted

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 $\sigma(\mathcal{A})$.

share|improve this answer
    
I edit the question. –  Elias Nov 12 '12 at 18:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.