# # of intervals between irrational numbers

If we assume $\mathbb R$ is a totally ordered set, given two irrational numbers $a$ and $b$, by the density of irrational numbers in $\mathbb R$, we can find an irrationals $c$ between, and at the same time we find at least two rational $r_1$ and $r_2$ that has $a\leq r_1< c< r_2< b$ so the collection $A$ of intervals $i_\lambda$ between a and all irrational numbers in [a,b] for all irrational numbers in $[a,b]$, $\lambda \in \Lambda$, is total ordered. There is a map $\phi\colon A\rightarrow Q$ which maps i to a rational, my question: is this an injective map? or not, and why? any reference recommendation is welcomed.

Detail：we find this collection $A$ is totally ordered set under relation $<$ (not only has a minimal $\{b\}$ but a maximal $[a,b]$) and there is a map $\psi\colon A\rightarrow [a,b]\cap {\mathbb Q}^c$, by mapping an interval to its right end, and this is bijection. Also there is a bijection from $A$ and the induced $B$ contains $i'_\lambda$'s which are just $i_\lambda - \cup_{\beta<\lambda} i_\beta$. By axiom of choice we may have a map for $B$ ($B$ is disjoint now) to a representative rational $r$ such that there again exists a bijection between $B$ and a subset of $[a,b]\cap \mathbb Q$，so it is concluded that the number of irrationals in this closed interval $[a,b]$ is no larger than number of rationals in $[a,b]$. Can you help me find what is doing wrong here?

• Is $R^1$ just $\mathbb R$ or some other set? If it is some other set, what is it? – 5xum Jan 25 '18 at 10:38
• I think you want that $c$ is between $a$ and $b$, and not larger than both. – Asaf Karagila Jan 25 '18 at 10:52
• While not a word-for-word duplicate, ultimately, this is the same question as math.stackexchange.com/questions/2610098/… and the many suggested duplicates there. – Asaf Karagila Jan 25 '18 at 10:54
• Right, it is $\mathbb R$. – Cheerupbubble Jan 25 '18 at 12:03
• right, c is in between – Cheerupbubble Jan 25 '18 at 12:04

You assume that the map which assigns each irrational $\lambda$ a rational number from $[a,\lambda]$ is in fact injective. But there is no justification for that, and indeed you cannot prove this.
Or you can make the argument indirectly. Simply prove that $\Bbb R$ is uncountable, as an independent proof. And then use the fact that the union of two countable sets is countable, and therefore $\Bbb Q^c$ is not countable.