Proving the equality of two $\sigma$-algebras given in terms of generators My lecturer says as follows; 
$C_0 = \{(a,b) : -\infty \le a\le b< \infty\}$
$$C_\mathrm{open} = \{ A \in \mathbb{R} : A\text{ open} \}$$ 
He goes on to show that $\sigma(C_0) = \sigma(C_{\mathrm{open}})$; 
Clearly, $\sigma(C_0)$ is in $\sigma(C_{\mathrm{open}})$
So now I need to show the other way round, 
$\sigma(C_{\mathrm{open}})$ is in $\sigma (C_0)$
I do this by showing that $C_{\mathrm{open}}$ is in $\sigma (C_0)$
He says; take $A$ as subset of $\mathbb{R}$ which is open, then 
$$A= \bigcup_{x \in X} (x-\varepsilon_x, x + \varepsilon_x)$$ 
Then he says $A \cap\mathbb{Q}$ is a subset of 
$$A= \bigcup_{x \in X} (x-\varepsilon_x, X + \varepsilon_x)$$ 
and $A\cap\mathbb{Q} = \{y_1,\ldots\}$ such that there exists $x_n$ s.t $y_n$ is in $(x_n-\varepsilon_{x_n}, x_n+\varepsilon_{x_n})$ for all $n$
of course  $$\bigcup_{n=1}^{\infty} (x_n-\varepsilon_{x_n}, x_n+\varepsilon_{x_n})$$
is a subset of $A$ 
Then he says let $$\varepsilon_{n-} = \sup\{\varepsilon>0\mid (x_n-\varepsilon, x_n] \subseteq A\}$$
and let 
$$\varepsilon_{n+} = \sup\{\varepsilon>0\mid [x_n, x_n+\varepsilon)\subseteq A\}$$
Now we need to show that $A$ is a subset of 
$$\bigcup_{n=1}^{\infty}(x_n-\varepsilon_{n-},x_n+\varepsilon_{n+})$$
Take $x$ in $A$ then as before $x$ is in $(x-\varepsilon_x, x + \varepsilon_x)$. 
$\mathbb{Q}$ is dense, so there exists $n$ s.t $y_n$ is in $(x-\varepsilon_x, x + \varepsilon_x)$ which is a subset of $(x_n-\varepsilon_{n-},x_n+\varepsilon_{n+})$ by definition of $\varepsilon_{n+}$ and $\varepsilon_{n-}$.
This then implies that $A$ is a subset of 
$$\bigcup_{n=1}^{\infty}(x_n-\varepsilon_{n-} , x_n+\varepsilon_{n+})$$
So our arbitrary $A$ is in $\sigma(C_0)$ 
Can anybody explain this to me, I don't understand where the steps come from and lead to. All I know is you need to use the rationals because they're countable, unlike the reals. 
Thanks
 A: What your instructor does is the following: He tries to write $A$ as a countable union of open intervals. He first writes $A$ as a uncountable union of intervals $(x-\varepsilon_x, x + \varepsilon_x)$. He then picks a rational number $y_n$ from each interval
and picks then a point $x_n$ such that $y_n\in(x_n-\varepsilon_{x_n}, x_n+\varepsilon_{x_n})$. This step is actually superflous, one can just take $x_n=y_n$. He then puts a maximally large open intervall around $x_n$. This is where the $\varepsilon_{n-}$ and $\varepsilon_{n+}$ come in. They
are constructed so that $(x_n-\varepsilon_{n-},x_n+\varepsilon_{n+})$ is the largest possible open interval around $x_n$ that lies still in $A$. Then he verifies that the union of these sets coincides with $A$. The motivation that this is a useful approach comes from the
fact that every open subset of $\mathbb{R}$ is the countable union of open intervals. These open intervals are actually taken to be as large as possible:
Let $A\subseteq\mathbb{R}$ be open. Define an equivalence relation $\equiv$ on $A$ such that $x\equiv y$ if and only if, there is an open interval $(a,b)\subseteq A$ that contains both $x$ and $y$. That $\equiv$ is reflexive and symmetric is obvious. 
It is transitive since the union of two open intervals that are not disjoint is again an open interval. This also shows that the equivalence classes are all open intervals. Since equivalence classes are disjoint, this shows that $A$ is the disjoint union of open intervals.
Now every open interval contains some rational number and disjoint intervals contain different rational numbers. It follows that $A$ is the disjoint union of countably many open intervals.
The only caveat is that the open intervals are not required to be finite. But a moments thought shows that at most two of the intervals are not finite, and we can write them as a countable union of finite non-disjoint intervals.  
Note: The intervals $(x_n-\varepsilon_{n-} , x_n+\varepsilon_{n+})$ are actually the equivalence classes.
A: The steps to show that $\sigma(C_{\mathrm{open}}) \subset \sigma(C_0)$ are as follows:
Show that any open set $A$ can be written as a countable union of elements of $C_0$. From this it follows that $A \in \sigma(C_0)$. Since $A$ was 'any' set, this means that $C_{\mathrm{open}} \subset \sigma(C_0)$. From this it follows that $\sigma(C_{\mathrm{open}}) \subset \sigma(C_0)$.
To show that any open set $A$ can be written as a countable union of elements of $C_0$, I follow the basic construction above, but hopefully a little more clearly:
I will construct a countable union of intervals that are contained in $A$. Select $q \in A \cap \mathbb{Q}$, and define the interval $I_q$ as the largest open interval containing $q$ that is contained in $A$. The explicit construction is exactly the same as above:
Let $\overline{x}_q = \sup \{ x \; | \;  [q, x) \subset A \}$, and $\underline{x}_q = \inf \{ x \; | \;  (x,q] \subset A \}$. Then you can show that $I_q = (\underline{x}_, \overline{x}_q)$. It should be clear that $I_q \subset A$.
Let $B = \cup_{q \in \mathbb{Q} \cap A} I_q$, from above, it should be clear that $B \subset A$ (since each $I_q \subset A$). To finish, I just need to show that $A \subset B$.
Suppose $x\in A$, then since $A$ is open, we have $(x-\epsilon, x+\epsilon) \subset A$ for some $\epsilon > 0$. At least one $q \in \mathbb{Q}$ lies in this interval, and by construction $(x-\epsilon, x+\epsilon) \subset I_q$, hence we have $x \in B$. So we conclude that $A$ can be written as the countable union of intervals.
