Increasing bijection $f\colon \mathbb{N} \to \mathbb{Q}$

Question

We know that there exists a bijection $f\colon \mathbb{N} \to \mathbb{Q}$ from the set of natural numbers to the set of rational numbers. Is there a increasing bijection? Given $m, n \in \mathbb{N}$ such as that $m < n$, then $f(m) < f(n)$.

I've tried to argue as following: for each bijection $f$, or $f$ is increasing or there is a smallest natural number $M_f$ such as that $f$ is increasing for all natural numbers less than $M_f$. Let's order the bijections as follows: $f' \leq f$ if $M_{f'} \leq M_f$ and if the restriction of $f$ to all integers less than $M_{f'}$ equals $f'$.

Now, given a non-increasing bijection $f$, I can build another bijection $f'$ with $M_f < M_{f'}$, by simply swapping $f(M_f)$ with another $f(y)$ when $y > M_f$. Also, given a chain of bijections $f_i$, or the $M_{f_i}$'s are limited or they are not. If they are limited, take their maximum and build a upper bound by the method above. If they are not, for every $n \in \mathbb{N}$, $f_i(n)$ must be constant for big enough $i$. Define $f'(n) = f_i(n)$, and we will have $f' \geq f_i$ for all $i$. By Zorn's Lemma, there must exist a maximal bijection, which must be increasing. Is this correct?

EDIT: I just thought, no number can be $f(1)$, because for each rational number there is one strictly smaller. But now, where is the error in the argument above?

Answer 1

You didn't prove that your function $f'$ is a bijection, and in general it is not. It will only by an increasing injection. For example, given a chain $f_1,f_2,\dots$ of bijections where each $f_k$ satisfies $f_k(i)=i$ for all $i\le k$, then $f'(i)=i$ for all $i$, so $f'$ is not surjective.

$f_1 = 1,\frac12,\frac{47}{55},\dots$

$f_2 = 1,2,-\frac35\dots$

$f_3 = 1,2,3,\frac{99}{77}\dots$

$\vdots$

$f'=1,2,3,4,5,\dots$ is not a bijection.

Answer 2

I think your mistake is believing that just because you can construct an infinite sequence of bijections that are increasing for longer and longer, there is a final, "limit" bijection, as if the sequence is somehow converging. The sequence is not converging, and there is no limit. Or, there is a limit, but it's not a bijection any more. Whichever interpretation you like.

Think about trying to make a bijection with the property that $f(n) = n$ for all natural numbers. It's easy to make a sequence $f_N$ of bijections for which the property holds for all $n

Most of the "work" of a bijection like this is done in the tail end of $\Bbb N$, and leaving the tail free in each term of the sequence means you have room to make them into bijections. In the limit, the tail end is not free, and thus has a harder time being a bijection.

In fact, it's easy to give other examples of sequences of bijections that converge to something that is not a bijection. For instance, the functions $f_n:\Bbb R\to \Bbb R$ (or $\Bbb Q\to \Bbb Q$ if you like) given by $f_n(x) = \frac xn$ converge pointwise to $f(x) = 0$.

Answer 3

In fact, there's an easy way to show that there is no such bijection: there is a rational $q$ in between $f(0)$ and $f(1)$! But we can't have $f(n)=q$ for any $n>1$, since $f$ has to be increasing.

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Alright, so now let's look at your argument.

First of all, I think you've misdefined $M_f$: I think you want the largest number such that $f$ is increasing below $M_f$ (otherwise we'll always have $M_f=0$).

But the real error is when you try to apply Zorn's lemma. In fact, your poset does not have the necessary property that every chain has an upper bound! Specifically, the function you try to define as the upper bound of an arbitrary chain, may not be a bijection (in fact you can set it up so that this function is whatever increasing injection from $\mathbb{N}$ to $\mathbb{Q}$ that you want it to be).

This is a common mistake: you have a poset $P_1$ which you want to apply Zorn to, and it sits inside another natural poset $P_2$. In this case, $P_1$ is the poset you've defined, and $P_2$ is the poset of all injections from $\mathbb{N}$ to $\mathbb{Q}$, ordered in the analogous way ($g\le h$ if $M_h\ge M_g$ and $h(n)=g(n)$ for $n\le M_g$). The problem is that, while any chain in $P_2$ has an upper bound in $P_2$, a chain in $P_1$ will only be guaranteed to have an upper bound in $P_2$.

(And just for clarity, here's an explicit maximal element of $P_2$: the identity map $f: n\mapsto n$. Why is this maximal? Well, for any $g$ to be $\ge f$, we'd have to have that $g$ agrees with $f$ for as long as $f$ is increasing. More symbolically, the problem is that $M_f=\infty$. But $f$ is always increasing, so we'd have to have $g=f$. In general, the maximal elements of $P_2$ are exactly the increasing injections from $\mathbb{N}$ to $\mathbb{Q}$.)