Borel sigma-algebra over [0,1] 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?
 A: 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 $\mathcal F$.
A: You should prove that if $A_{n} \in \mathcal{B}$ for each $n \in \Bbb N$, then $\bigcap \limits_{n = 1}^{\infty} A_{n} \in \mathcal{B}$.
Once you have the above fact, then since $[0, \frac{1}{n}) \in \mathcal{B}$ for each $n$, we have $\bigcap \limits_{n = 1}^{\infty} [0, \frac{1}{n}) = \{ 0 \} \in \mathcal{B}$.
Here is a proof for the first claim I made (try it yourself first, and when you want to see the proof, just move your mouse over the box below):
Claim: If $A_{n} \in \mathcal{B}$ for each $n \in \Bbb N$, then $\bigcap \limits_{n = 1}^{\infty} A_{n} \in \mathcal{B}$.
Proof:

 Since $A_{n} \in \mathcal{B}$ for each $n \in \Bbb N$, and $\mathcal{B}$ is a $\sigma$-algebra (which is closed under complements), then $A_{n}^{c} \in \mathcal{B}$ for each $n$.  Then since by definition $\sigma$-algebras are closed under countable unions, we have $\bigcup \limits_{n = 1}^{\infty} A^{c}_{n} \in \mathcal{B}$.  But then the complement of this set is in $\mathcal{B}$.  But $(\bigcup \limits_{n = 1}^{\infty} A^{c}_{n})^{c} = \bigcap \limits_{n = 1}^{\infty} A_{n}$, which proves $\bigcap \limits_{n = 1}^{\infty} A_{n} \in \mathcal{B}$, as desired.


UPDATE:
As requested in the comments, if $\mathcal{B}$ represents the smallest $\sigma$-algebra containing sets of the form $(a,b)$ with $a, b \in [0,1]$, then there is no way to get $\{ 0 \}$.  We can get $\{ 0, 1 \}$ by noticing this set is the complement of $(0,1)$.  But there is no way to "separate" $\{0 \}$ from the other elements as a measurable set.
A: Actually your definition is wrong: the Borel $\sigma$-algebra of $[0,1]$ is the smallest $\sigma$-algebra containing all open sets of $[0,1]$.
Clearly, every open set $U \subset (0,1)$ is a (at most) countable union of disjoint intervals of the form $(a_i, b_i)$. This means that $U \in \mathcal{B}$ (I denote by $\mathcal{B}$ your definition of Borel $\sigma$-algebra).
However, as you pointed in your question, there are some open subsets touching the boundary of $[0,1]$. As an example, $[0,1)$ is open in $[0,1]$, hence it is a Borel subset of $[0,1]$. Actually, $[0,1) \notin \mathcal{B}$: this shows that $\mathcal{B}$ is not the Borel algebra of $[0,1]$.
