In the advanced calculus course my lecturer introduced an interesting thing to us saying
The exterior of rational number set $\text{Ext}(\mathbb Q) = \varnothing.$ Meanwhile, the interior of rational number set $\text{Int}({\mathbb Q}) = \varnothing$ as well.
We can use the Archimedean property to get a proof of $\text{Ext}(\mathbb Q) =\varnothing.$
Suppose $\text{Ext}({\mathbb Q}) \ne\varnothing.$ Then, $\exists r \in \mathbb R \: s.t. \: r \in \text{Ext}({\mathbb Q}).$ So for some positive real number $\epsilon$ we have $B_\epsilon(r) \subset \mathbb R \text{\\} \mathbb Q.$ However, according to the Archimedean property, choosing $r$ and $r-\epsilon$, $\exists q \in \mathbb Q, q \in (r-\epsilon, r),$ which is a contradiction.
So, I am wondering whether there exists such so call Anti-Archimedean property saying
$r_1,r_2 \in \mathbb R,r_1 < r_2, \exists I \in \mathbb R \text{\\} \mathbb Q \: \ \ s.t. \: I \in ( r_1,r_2).$
I know this claim is absolutely right but the proof is hard to make out. Here is my imagination.
If we choose an arbitrary small irrational number $\epsilon < r_2 - r_1$ and let $\epsilon \neq -r_1$ to avoid the situation of $\epsilon + r_1 = 0$. Then, we have $r_1 + \epsilon$ is an irrational number.
So, my question is whether my imagination fits the situation and does there exist some ways to prove this claim and prove $\text{Int}({\mathbb Q}) =\varnothing$?
