An example about irrational numbers and their metric structure I was trying to understand the following example but there is a point a cannot understand.
Namely, when he says that if we restrict $d$ to the irrationals we can always find, for each $\epsilon$, an $\delta$ such that $B_\delta(p)\subseteq \Delta_\epsilon(p)$. 
Fix $\epsilon>0$. I have point $p$ and point $p+\delta$ and I must find $\delta$ in such a way that $d(p,\, p+\delta)<\epsilon$. 
How I can make an upper esteem of the summation term in the definition of $d(x,\, y)$? Every esteem I make depends on the index $i$ of the sum and so I cannot esteem all of them. 

 A: Given $\epsilon > 0$, choose a $k\in\mathbb{N}$ such that
$$\sum_{i=k+1}^\infty 2^{-i} = 2^{-k} < \frac{\epsilon}{3}.$$
Then consider the finite sets $F_i = \{ r_j : j \leqslant i\}$ for $1 \leqslant i \leqslant k$. The functions $\operatorname{dist}(\,\cdot\,,F_i)$ are continuous, and positive on $\mathbb{R}\setminus F_i \supset \mathbb{R}\setminus \mathbb{Q}$, so $$\alpha_i \colon x \mapsto \frac{1}{\operatorname{dist}(x,F_i)} = \max_{j\leqslant i}\frac{1}{\lvert x-r_j\rvert}$$ is continuous on $\mathbb{R}\setminus F_k$. For any fixed $p \in \mathbb{R}\setminus \mathbb{Q}$, there is hence an $\eta > 0$ such that $\lvert \alpha_i(p) - \alpha_i(y)\rvert < \frac{\epsilon}{3}$ for all $y\in B_\eta(p)$. Then we can for $\lvert p-y\rvert < \eta$ estimate
\begin{align}
d(p,y) &\leqslant \lvert p-y\rvert +\sum_{i=1}^k 2^{-i} \min \left(1,\left\lvert \alpha_i(p) - \alpha_i(y)\right\rvert\right) + \frac{\epsilon}{3}\\
&\leqslant \lvert p-y\rvert +\frac{\epsilon}{3} + \sum_{i=1}^k 2^{-i} \min \left(1,\frac{\epsilon}{3}\right)\\
&\leqslant \lvert p-y\rvert +\frac{\epsilon}{3} + \frac{\epsilon}{3}.
\end{align}
Then just take $\delta = \min \left\{ \eta , \frac{\epsilon}{3}\right\}$ to have
$$B_\delta(p) \subset \Delta_\epsilon(p).$$
