$\sigma$-algebra generated by $\{ (a,b) : a,b \in \mathbb{Q} , a<b \}$

Question

Is $\sigma$-algebra generated by $\{ (a,b) : a,b \in \mathbb{Q} , a<b \}$ the borel sigma algebra of $\mathbb{R}$.

I wrote a proof and I am confused because I didn't find this result anywhere in the net.

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The proof that I suggest is as follows:

Given two reals $a$ and $b$ such that $a<b$ then by density of rationals in reals there are two rational sequences $(a_n)_{n \in \mathbb{N}}, (b_n)_{n \in \mathbb{N}} \subset (a,b) $ that converge respectively to $a$ and $b$, then $(a,b)= \cup_{n \in \mathbb{N}} (a_n,b_n)$ hence we deduce that $(a,b) \in \sigma (\{ (a,b) : a,b \in \mathbb{Q} , a<b \})$ and since $ \sigma (\{ (a,b) : a,b \in \mathbb{R} , a<b \}) = \mathbb{B} ( \mathbb{R} )$ we deduce that $\mathbb{B} ( \mathbb{R} ) \subset \sigma (\{ (a,b) : a,b \in \mathbb{Q} , a<b \})$ and the other inclusion is trivial hence we conclude the equality between the two sets.

Answer 1

Your proof is fine.

This is a specific example of a more general theorem: if $X$ is a topological space and if $X$ has a countable basis then the Borel $\sigma$-algebra of $X$ is generated by any basis for the topology of $X$. That, perhaps, explains why you have not found your specific example.