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I'm trying to get my head around Borel sets (like so many before me!) IF I've understood them correctly, the definition on seems unnecessarily complicated. Isn't the following equivalent?

If for all x in A all values in the neighbourhood of x are also in A (in one direction, if x is the end of a closed interval) then A is a Borel set.

If this is not correct, what have I misunderstood?

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I'm not sure I follow your description of Borel sets, but note that e.g. $\mathbb{Q} = \bigcup_{q \in \mathbb{Q}} \{q\}$ is a countable union of points, hence a Borel set, and so is its complement $\mathbb{R} \smallsetminus \mathbb{Q}$. For a more "pathological" example see the Cantor set. – t.b. May 19 '12 at 16:19
Thanks. That helps. I was thinking about Borel sets in the context of probability, and therefore measure. This led me to wondering if isolated points could be in Borel sets, or if they had to be intervals. I have seen the set of irrational numbers (in [0, 1]) given as an example of a non-Borel set, but I now understand that it is the fact that this is not a countable union etc that prevents it from being Borel, and not because there are numbers missing from the neighbourhood of any x in that set. Of course, isolated points can be generated by the intersection of closed sets, anyway. – Call it a spade May 19 '12 at 17:06
Oh, but you also state that R\Q is a Borel set. I guess I'm still confused. – Call it a spade May 19 '12 at 17:08
OK, I can see that R\Q is the countable intersection of sets (the complements of all R\q). Is it then correct that the irrational numbers in [0,1] form a non-Borel set? – Call it a spade May 19 '12 at 17:26
The Borel sets form a $\sigma$-algebra so that if something is a Borel set, then so is it's complement. If you've shown the rationals are a Borel set, then so are the irrationals – Alex R. May 19 '12 at 17:52

Sticking to the real line, the simplest definition of Borel sets is from the Borel $\sigma$-algebra which is defined to be the smallest $\sigma$-algebra which contains all open sets. To be clear, if $\sigma_\alpha$ is a sigma algebra containing the open sets, then you can take the intersection of all $\sigma_\alpha$ to form the Borel sigma-algebra (convince yourself that sigma-algebras are stable under intersection).

If you want a constructive definition, start with the whole space and the empty set and then throw in every single open set and their complements. Then start taking unions and intersections and their complements. It sometimes doesn't pay to think about $\sigma$-algebras in this constructive way so sometimes it's more convenient to use the definition that they form the smallest sigma algebra.

The definition you've stated seems to be for an open set, that it contains a neighborhood around every point. Be careful because open sets are NOT stable under intersection. In the comments the rationals were constructed with open sets but even more simply, think of $\{0\}=\bigcap\limits_{n=1}^\infty (-1/n,1/n)$. In other words $\{0\}$ is a Borel set even though it's not open. There are non-Borel sets out there (there's an example of one in your wikipedia link) but luckily they are pretty obscure.

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