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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?

Asaf Karagila
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Cheerupbubble
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  • 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
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    I think you want that $c$ is between $a$ and $b$, and not larger than both. – Asaf Karagila Jan 25 '18 at 10:52
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    While not a word-for-word duplicate, ultimately, this is the same question as https://math.stackexchange.com/questions/2610098/on-finding-enough-rationals-countable-to-fill-the-uncountable-number-of-interv 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
  • I saw the posts you posted here an hour ago, and i think my question is not so accurate, let me do a prove and you try to find what is wrong in it: – Cheerupbubble Jan 25 '18 at 12:06
  • You never seem to define which ordering you apply to your $A$. You say the ordering is $<$, but what does that mean between intervals? Also, it is certainly not the case that mapping each interval to its right endpoint is a bijection -- for example $[\sqrt2,\pi]$ and $[e,\pi]$ would both map to $\pi$ so the mapping isn't injective. – hmakholm left over Monica Jan 25 '18 at 12:47
  • Are you trying to say something like: For each irrational $\lambda$ choose some closed interval $i_\lambda$ whose right endpoint is $\lambda$, and then let $i'\lambda$ be this interval minus all the intervals with endpoints to the left of $\lambda$? In that case $i'\lambda={\lambda}$ (at least for most of your $\lambda$s) which does not have any rational element. – hmakholm left over Monica Jan 25 '18 at 12:56
  • @Henning: It seems that all intervals have the form $[a,x]$ for some $x\in\Bbb R$. – Asaf Karagila Jan 25 '18 at 14:01
  • @AsafKaragila: Hmm, yes. That just makes it worse, of course: $i'\lambda$ is _always a singleton then. – hmakholm left over Monica Jan 25 '18 at 15:25

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You are begging the question.

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.

An argument can be made directly, using diagonalization, pick any countable sequence of irrationals, and produce another irrational which either is mapped to a different rational number than all; or the choice of representatives is not injective anymore. And that is what happens in reality, there is no "canonical" way to assign an interval a rational number such that this assignment is injective.

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.

Asaf Karagila
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