Is there an injection $f:\mathbb{R}\to\mathbb{R}$ that is unbounded on every interval?

Question

I'm going to say a function $f:\mathbb R \to \mathbb R$ is unbounded on every interval if for all reals $a,b,M$ with $a<b$, there exists $x\in[a,b]$ such that $|f(x)| > M$.

It is easy to that such functions exist, by for instance exploiting the density of the rationals. It's not too much harder to construct surjective examples: using e.g. decimal expansions, one can construct a function $f:\mathbb{R}\to\mathbb{R}$ which is 'surjective on every interval' in the sense that for any $a<b$ and $c$ there exists $x\in[a,b]$ with $f(x)= c$.

However, any such construction doesn't give you much 'control' about how $f$ actually behaves. Is there such a function which is unbounded on every interval, and is injective over $\mathbb{R}$?

Answer 1

Here is a dull way to construct a bijection $f:\mathbb R\to\mathbb R$ which is unbounded on every interval.

Enumerate the intervals with rational endpoints as $I_1,I_2,I_3,\dots$.

Construct a sequence of pairwise disjoint Cantor sets (nonempty nowhere dense perfect sets) $X_1,X_2,X_3,\dots$ with $X_n\subset I_n$.

Partition $\mathbb Z$ into unbounded sets $A_0,A_1,A_2,A_3,\dots$ and define $Y_n=\bigcup\{[a,a+1):a\in A_n\}$;
so $\mathbb R$ is partitioned into unbounded sets $Y_0,Y_1,Y_2,Y_3,\dots$ each of cardinality $\mathfrak c$.

Define a bijection $f:\mathbb R\to\mathbb R$ which maps $X_n$ onto $Y_n$ for $n=1,2,3,\dots$ and maps $\mathbb R\setminus\bigcup_{n=1}^\infty X_n$ onto $Y_0$.

Answer 2

The classical example of a function $g:\mathbb{R}\to\mathbb{R}$ which is unbounded on every interval is as follows:

$$g(x)=\begin{cases} |n| &\text{if }x=\frac{m}{n}\text{ and }gcd(m,n)=1 \\ 0 &\text{otherwise} \end{cases}$$

This function is of course not injective, but we can tweak it. Define $f:\mathbb{R}\to\mathbb{R}$ as follows: $f(0)=0$ and $f(x)=x$ if $x$ is irrational. Now arrange all rationals into a sequence $(q_n)$ and then define recursively

$$f(q_1)=g(q_1)$$ $$f(q_n)=\max(f(q_1),\ldots,f(q_{n-1}), g(q_n))+1$$

Then $f$ is injective, and by the fact that $f(x)\geq g(x)$ for all rationals it is also unbounded on every interval.