Context: self-studying topology without tears, now at question 2.2.3.
Question: Let $\mathcal{B}$ be the collection of all open intervals $(a,b) \in \mathbb{R}$ with $a \lt b$ and $a, b$ rational numbers. Prove that $\mathcal{B}$ is a basis for the euclidean topology on $\mathbb{R}$.
I found this approach online and I have some questions:
We shall prove that if $S \subseteq \mathbb{R}$ is an open set then given any $x \in S$ we can find an open interval $(a_q,b_q) \subseteq S$ where $a_q,b_q \in \mathbb{Q}$.
Since $S$ is an open set, we know that there exists $a,b \in \mathbb{R}$ such that $x \in (a,b) \subseteq S$. Further, since the rationals are dense in $\mathbb{R}$, given any $r \in \mathbb{R}$, there exists a sequence of rationals monotonously converging to $r$. For instance, there exists a sequence of monotonously increasing $b_n \in \mathbb{Q}$ such that $\lim_{n \rightarrow \infty}b_n = b \in \mathbb R$.
Similarly, there exists a sequence of monotonously decreasing $a_n \in \mathbb{Q}$ such that $\lim_{n \rightarrow \infty}a_n = a$. This means that $(a,b) = \lim_{n \to \infty} (a_n,b_n)$. Hence, if $x \in (a,b)$, then $x \in (a_n,b_n)$ for some $n \in \mathbb N$ where $a_n,b_n \in \mathbb{Q}$. Hence, given any open set $S$, for every $x \in S$, there exists an open interval with rational endpoints containing $x$ contained within $S$.
This step here is a bit of a leap for me:
Hence, the set of open intervals with rational endpoints generate the same topology as the euclidian topology.
I do not get this last step. Okay, so we can construct a rational interval by using a density argument. If the interval is not big enough we can always make it bigger, eventually approaching some real number. This gives us the upper and lower bounds on the interval. Now we have constructed this interval, what is the reasoning that connects this basis, and the topology it generates to the Euclidean topology?
The book's definition of a basis for a topology is that any open set is a union of members from this basis.
Let $U$ be open. For each $x \in U$ there exist an interval $I_x=(a_x,b_x)$ such that $a_x, b_x \in \mathbb Q$ and $x \in I_x \subset U$. Now we claim that $U=\cup_{x \in U} I_x$. Since each $I_x$ is a subset of $U$ it follows that $\cup_{x \in U} I_x \subset U$. Now let $x_0 \in U $. Then $x_0 \in I_{x_0} \subset U$. Since $I_{x_0}$ is one of the sets in the union $\cup_{x \in U} I_x$ it follows that $x_0 $belongs to this union.