TITLE: Borel sigma-algebra over [0,1]
QUESTION [8 upvotes]: I just started studying this, so forgive me if I get something wrong.
I have been given the following definition of a Borel $\sigma$-algebra over $\Omega=[0,1]$: It is the smallest $\sigma$-algebra that contains all intervals $(a,b)$ with $0\leq a<b\leq1$. Lets call this algebra $\mathcal{B}$.
Now apparently, every subinterval of $[0,1]$, including a half-open one like $[0,1)$, should be in $\mathcal{B}$. Even simpler, $\{0\}\in\mathcal{B}$ should be correct. Right?
I can not figure out how this would work - I know that complements, intersections and unions of any elements of $\mathcal{B}$ are again elements of $\mathcal{B}$.
With the given definition, I don't know how to obtain a subset that contains either $0$ or $1$ and not both: $0\leq a<b\leq 1, M = (a,b) => 0,1 \notin M => 0,1\in \overline{M}$.
I have searched using google and stackexchange, but I seem to have been given an uncommon definition. Is the definition wrong or am I missing something?

REPLY [5 votes]: You're right; the definition you quoted is wrong.  The family $\mathcal F$ of all subsets of $[0,1]$ that contain both or neither of $0$ and $1$ is a $\sigma$-algebra that contains the intervals $(a,b)$.  So all Borel sets according to the quoted definition are in $\mathcal F$.