Is identity permutation of $\mathbb R$ the only permutation of $\mathbb R$ with which we can achieve what is described in the body of this question?

Question

If $P(\mathbb R, \mathbb R)$ is the set of all permutations of $\mathbb R$, that is, the set of all bijections of $\mathbb R$ onto $\mathbb R$, then, some of them tend to much preserve the order and some tend to not much preserve the order.

Here, the $<$ is the usual order over $\mathbb R$.

We could say that permutation $p$ preserves much of the order of $\mathbb R$, if:

For every $w \in p(\mathbb R)$ we have that there exist at least two sequences $a_{p,w}(n)$ and $b_{p,w}(n)$ such that $a_{p,w}(1)<...< a_{p,w}(k)<...$ (in $\mathbb R$ and also in $p(\mathbb R)$, that is, $a$ is strictly monotone as a sequence in $p(\mathbb R)$ and $a^{-1}$ is strictly monotone also as a sequence in $\mathbb R$) and $\lim_{n \to + \infty}a_{p,w}(n)=w$ and $b_{p,w}(1) >...>b_{p,w}(k)>...$(also, as $a$, also is $b$ strictly monotone in $p(\mathbb R)$ and $b^{-1}$ is strictly monotone in $\mathbb R$) and $\lim_{n \to + \infty}b_{p,w}(n)=w$.

To clarify what it means that $a$ and $b$ are strictly monotone in $\mathbb R$ and also in $p(\mathbb R)$ ,it means, for for example $a$, that $a_{p,w}(1)<...< a_{p,w}(k)<...$ (in $p(\mathbb R))$ but also that the inverse sequence $a^{-1}_{w,p} \subset \mathbb R$ of the sequence $a_{p,w} \subset p(\mathbb R)$ is strictly monotone in $\mathbb R$.

Is the identity permutation (up to scalings and reflections of the identity permutation) of $\mathbb R$ the only permutation of $\mathbb R$ which preserves much of the order of $\mathbb R$?

That is, is the identity permutation, the permutation which does not change the order of $\mathbb R$ at all, the only permutation which preserves much of the order of $\mathbb R$?

Answer 1

The answer is yes for uninteresting reasons: just pick any continuous order-preserving bijection $\mathbb{R}\rightarrow\mathbb{R}$. So, for eample, take $x\mapsto 2(x-3)^5$, or $x\mapsto\arctan(x)$, or etc.

But highly pathological bijections also exist ...

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As a tangential remark, remember that if $p$ is a permutation of $\mathbb{R}$ then $p(\mathbb{R})=\mathbb{R}$. (That said, maybe the notation indicates that you want to consider mere injections rather than bijections?)

This, and some other abbreviations/clarifications, let us rephrase your definition of "preserves much of the order" a bit more snappily as follows:

A bijection $p:\mathbb{R}\rightarrow\mathbb{R}$ preserves much of the order of $\mathbb{R}$ if for all $w\in\mathbb{R}$ there are (not necessarily unique) sequences of reals $a_1<a_2<...$ and $b_1>b_2>...$ such that:

• $p(a_1)<p(a_2)<...$ and $p(b_1)>p(b_2)>...$, and

• $\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n=w$.

And adding to your question, I'll say that $p$ strongly preserves much of the order of $\mathbb{R}$ if additionally we have

• $\lim_{n\rightarrow\infty}p(a_n)=\lim_{n\rightarrow\infty}p(b_n)=p(w)$.

There are lots of bijections which strongly preserve the order of $\mathbb{R}$. For example, let $\sim$ be the Vitali equivalence relation: $$a\sim b\quad\iff\quad a-b\in\mathbb{Q}.$$ Now any assignment $i$ of rationals to $\sim$-classes - that is, any map $i:\mathbb{R}/\sim\rightarrow\mathbb{Q}$ - yields a bijection $B_i:\mathbb{R}\rightarrow\mathbb{R}$ given by $$B_i(a)=a+i([a]_\sim).$$ (Basically, each $\sim$-class "shifts" by a rational amount - the fact that the shift is rational in each case guarantees that $B_i$ is a bijection.)

The point is that for each $w\in\mathbb{R}$ we can just pick $a_n,b_n$ appropriately from $[w]_\sim$; these get shifted forward by the same amount. Note that this doesn't require the axiom of choice (although choice does guarantee that there are "lots" of non-constant $i$s).