Exotic bijection from $\mathbb R$ to $\mathbb R$

Question

Clearly there is no continuous bijections $f,g~:~\mathbb R \to \mathbb R$ such that $fg$ is a bijection from $\mathbb R$ to $\mathbb R$.

If we omit the continuity assumption, is there such an example ?

Notes: to follow from Dustan's comments: Notes: By definition $fg~:~x \mapsto f(x)\times g(x)$ and not $f\circ g$. If there were continuous bijections just look at the limits of $f$ and $g$ at $+\infty$ and $-\infty$ to conclude that $fg$ can't be a bijection

Ah, sorry, does $fg$ mean multiplication? I read it as composition — Robert Chamberlain, Feb 23 '15 at 15:55
You are receiving close votes on this question. I'd recommend two things: 1. Make it clear, per Robert's comments, that you are referring to multiplication of functions, not composition. 2. To add some more context, explain what makes you say your first remark is clear. It took me a while to come up with a proof myself, though that might just be because I haven't thought about this kind of thing in a while. — Dustan Levenstein, Feb 23 '15 at 16:27
Perhaps try it with a smaller set, say, $\mathbb Z$, first? Perhaps it can give you an idea. (Note: I have not tried this yet.) — Akiva Weinberger, Feb 27 '15 at 01:41

echinodermata · Accepted Answer · 2015-02-25T03:29:16.923

Let $f(x)=x$, and define $g$ piecewise by $$ g(x) = \begin{cases} -x ,& x \in \cdots \cup (-16,-8] \cup (-4,-2] \cup \cdots &= \bigcup_k -[2 \cdot 4^k, 4 \cdot 4^k) \\ 4x ,& x \in \cdots \cup (-8,-4] \cup (-2,-1] \cup \cdots &= \bigcup_k -[4^k, 2 \cdot 4^k) \\ 0 ,& x = 0 \\ x ,& x \in \cdots \cup [1,2) \cup [4,8) \cup \cdots &= \bigcup_k [4^k, 2 \cdot 4^k) \\ -\frac14 x ,& x \in \cdots \cup [2,4) \cup [8,16) \cup \cdots &= \bigcup_k [2 \cdot 4^k, 4 \cdot 4^k) \\ \end{cases} $$ (where $-[b,a)$ just means $(-a,-b]$).

This $g$ is a bijection from $\mathbb{R}$ to $\mathbb{R}$, since

$g$ maps $\bigcup_k -[2 \cdot 4^k, 4 \cdot 4^k)$ one-to-one onto $\bigcup_k [2 \cdot 4^k, 4 \cdot 4^k)$.
$g$ maps $\bigcup_k [2 \cdot 4^k, 4 \cdot 4^k)$ one-to-one onto $\bigcup_k -[2 \cdot 4^k, 4 \cdot 4^k)$.
$g$ maps $\bigcup_k -[4^k, 2 \cdot 4^k)$ one-to-one onto itself.
$g$ maps $\bigcup_k [4^k, 2 \cdot 4^k)$ one-to-one onto itself.
$g$ maps $0$ to itself.

And $fg$ is a bijection from $\mathbb{R}$ to $\mathbb{R}$, since

$fg$ maps $\bigcup_k -[2 \cdot 4^k, 4 \cdot 4^k)$ one-to-one onto $\bigcup_k -[4 \cdot 16^k, 16 \cdot 16^k)$.
$fg$ maps $\bigcup_k [2 \cdot 4^k, 4 \cdot 4^k)$ one-to-one onto $\bigcup_k -[16^k, 4 \cdot 16^k)$.
$fg$ maps $\bigcup_k -[4^k, 2 \cdot 4^k)$ one-to-one onto $\bigcup_k [4 \cdot 16^k, 16 \cdot 16^k)$.
$fg$ maps $\bigcup_k [4^k, 2 \cdot 4^k)$ one-to-one onto $\bigcup_k [16^k, 4 \cdot 16^k)$.
$fg$ maps $0$ to itself.

Wouldn't it make more sense to set $-(b,a]=[-a,-b)$, not $-[b,a)=[-a,-b)$? I'm assuming that you mean the notation in the sense of applying the $x\mapsto -x$ function to each element of the set, as in $2\Bbb Z$. — Mario Carneiro, Feb 24 '15 at 23:16
@MarioCarneiro I meant to write $-[b,a) = (-a,-b]$ (where the notation is meant to suggest applying $x \mapsto -x$ to the set, as you said). It's now fixed. — echinodermata, Feb 25 '15 at 03:31
To cover the domain/range of $\pm(0,1)$, I assume $-\infty<k<+\infty$. — Mark Hurd, Mar 03 '15 at 04:52

bof · Answer 2 · 2015-02-27T01:24:52.927

With the axiom of choice we can prove a much more general result:

Theorem. If $X$ is an infinite group (or field), then there are bijections $f,g:X\to X$ such that $f(x)g(x)=x$ for all $x\in X$.

Proof. It will suffice to prove it for a group; if $X$ is an infinite field, we can apply the theorem to the multiplicative group $X\setminus\{0\}$ and then extend the bijections to $X$ by setting $f(0)=g(0)=0$. Moreover, it will suffice to construct three bijections $f,g,h:X\to X$ such that $f(x)g(x)=h(x)$ for all $x\in X$; then $f\circ h^{-1}$ and $g\circ h^{-1}$ will be bijections whose product is the identity map.

Let $X$ be an infinite group, $|X|=\kappa$, and choose a well-ordering $\prec$ of $X$ such that $|\{y\in X:y\prec x\}|\lt\kappa$ for each $x\in X$. Partition $X$ into disjoint sets $A,B,C$ with $|A|=|B|=|C|=\kappa$. We define $f,g,h:X\to X$ by transfinite induction. Consider $x\in X$, and suppose $f(y),g(y),h(y)$ have been defined for all $y\prec x$; we define $f(x),g(x),h(x)$ as follows.

If $x\in A$, let $f(x)$ be the least element of $X\setminus\{f(y):y\prec x\}$; then choose $g(x)\in X\setminus\bigcup_{y\prec x}\{g(y),\ f^{-1}(x)h(y)\}$ and let $h(x)=f(x)g(x)$.

If $x\in B$, let $g(x)$ be the least element of $X\setminus\{g(y):y\prec x\}$; then choose $f(x)\in X\setminus\bigcup_{y\prec x}\{f(y),\ h(y)g(x)^{-1}\}$ and let $h(x)=f(x)g(x)$.

If $x\in C$, let $h(x)$ be the least element of $X\setminus\{h(y):y\prec x\}$; then choose $f(x)\in X\setminus\bigcup_{y\prec x}\{f(y),\ h(x)g(y)^{-1}\}$ and let $g(x)=f(x)^{-1}h(x)$.

Now we have bijections $f,g,h:X\to X$ such that $f(x)g(x)=h(x)$ for all $x\in X$. (Injectivity is clear from the definition; surjectivity can be proved by transfinite induction.)

I would rephrase the theorem statement to hold for any well-ordered group $X$ since you don't seem to need choice except in the well-ordering at the beginning. (It is easy to modify a given well-ordering to get a well-ordering whose order type is a cardinal.) — Mario Carneiro, Feb 25 '15 at 09:33
@MarioCarneiro Good point. In particular, the proof for countably infinite groups is constructive. — bof, Feb 25 '15 at 11:02

Exotic bijection from $\mathbb R$ to $\mathbb R$

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