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Claim: $\displaystyle \bigcup_{a,b \in \mathbb{Q}}[a,b)$ is a Borel set.
Solution: I arrive at $[a,b)=\displaystyle \bigcap_{a,b \in \mathbb{Q}}(a-\frac{1}{n},b)$. This means that $[a,b)$ is a countable intersection of open sets. Also, $\displaystyle \bigcup_{a,b \in \mathbb{Q}}[a,b)=\displaystyle \bigcup_{a,b \in \mathbb{Q}} \left(\displaystyle \bigcap_{a,b \in \mathbb{Q}}(a-\frac{1}{n},b)\right)$.

My question is: out of these arguments, can we say that $\displaystyle \bigcup_{a,b \in \mathbb{Q}}[a,b)$ is a Borel set? How? Any help is greatly appreciated.

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Well, $\bigcup_{a,b \in \mathbb{Q}}[a,b)$ is the whole real line. – Byron Schmuland Jan 23 '13 at 17:00
up vote 6 down vote accepted

Your first argument is a little flawed: I take it that you meant $$ [a,b)=\bigcap_{n\in\mathbb{N}}\left(a-\frac{1}{n},b\right) $$ which shows that $[a,b)$ is a countable intersection of open sets and hence it is an element of the Borel $\sigma$-field (i.e. is a Borel set). Is this clear?

Now to show that $$ \bigcup_{a,b\in\mathbb Q}[a,b) $$ you just have to note that it is a countable union of Borel sets and hence it is itself a Borel set. This is by the definition of a $\sigma$-field.

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Thank you sir Stefan. – Philip Benj Marcoby Eragon Jan 23 '13 at 9:54
You are welcome. – Stefan Hansen Jan 23 '13 at 9:55

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