Let $q_1, q_2, ...,$ be the rational numbers enumerated. Consider the countable collection $$\mathcal{B} = \{ B_{\frac{1}{n}}(q_i) \ | \ i,n \in \mathbb{N} \}$$ of open balls centered at rational numbers. Show that $\mathcal{B}$ is a basis for the Euclidean topology on $\mathbb{R}$.
First we need to show that $\mathcal{B}$ is a basis of a topology, so we need to check the two conditions:
(i): $x \in \mathbb{R}$ belongs to some $B \in \mathcal{B}$.
(ii): For $x \in B_1 \cap B_2$, there exists $B_3 \in \mathcal{B}$ such that $x \in B_3 \subset B_1 \cap B_2$.
For (i): This is true because there exists $q \in \mathbb{Q}$ such that $B_{\frac{1}{n}}(q) \ni x$. i.e., $(q-x) < \frac{1}{n}$.
For (ii): Suppose $B_1$ has a radius of $\frac{1}{n}$ centered at $q_1$, and $B_2$ has a radius of $\frac{1}{m}$ centered at $q_2$. Then, $|(q_1 - q_2)| < \frac{1}{n} + \frac{1}{m}$ We can find $B_3$ with radius $\frac{1}{k}$ with $\frac{1}{k} < min \{ |x-q_1|, |x-q_2|, \}$, this $B_3$ will be in $B_1 \cap B_2$ and will include $x$.
Thus, $\mathcal{B}$ is a basis that constructs a topology, say $\mathcal{T}'$.
Now I'm stuck at how to prove that this topology is the standard topology on $\mathbb{R}$, say $\mathcal{T}$.
If I want to show that the topology induced by $\mathcal{B}$, $\mathcal{T}'$ is equal to the standard topology on $\mathbb{R}$, $\mathcal{T}$.
We need to show that $\mathcal{B} \subset \mathcal{T} \implies \mathcal{T}' \subset \mathcal{T}$, and also $\mathcal{T} \subset \mathcal{T}'$. This is the part I don't know how to. Any help will be greatly appreciated.
Thanks in advance!
Here's a hint. Let $U$ be an open set on $\mathcal{T}$. Let $x$ be any element of $U$. Is it true that we can find some set $V$ which is open in the topology of $\mathcal{T}'$ where $x \in V \subseteq U$?