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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 http://en.wikipedia.org/wiki/Borel_set 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](http://en.wikipedia.org/wiki/Cantor_set).2012-05-19
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    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.2012-05-19
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    Oh, but you also state that R\Q is a Borel set. I guess I'm still confused.2012-05-19
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    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?2012-05-19
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    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 irrationals2012-05-19

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