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