I have heard of a function which is unbounded for all and any neighborhood of any real X. I can't seem to wrap my head around the possibility of such a function and my companion can't remember the exact function and proof. The only functions I can think of are non continuous functions in which the function is not defined in any neighborhood but he claims this is not the case.
Thank you for all and any assistance.
Here's one that is not continuous but is well defined. For irrational $x$, let $f(x)=0$. For rational $x=\frac{a}{b}$ where $\gcd(a,b)=1$, let $f(x)=b$.
The proof is simple: for irrational $x$, let $x_n$ be a sequence of rationals that approach $x$. Now, any sequence of rationals that approach an irrational must have unbounded denominators (and numerators but that isn't important here). Suppose this was not the case, and there exists irrational $\lambda$ and sequence of rationals $\lambda_n=\frac{a_n}{b_n}$ such that the denominators are bounded by $M\in\mathbb{N}$. Then
$$ \lim_{n\to\infty}M! \frac{a_n}{b_n}=M!\lambda$$
But $M! \frac{a_n}{b_n}$ is an integer as $b_n\leq M$ implies $b_n|M!$. Since a sequence of integers cannot approach an irrational number, we conclude $b_n$ is unbounded.
For rational $x=\frac{a}{b}$ where $\gcd(a,b)=1$, the proof is even easier.
Define
$$x_n=x+\frac{1}{p_n}=\frac{a}{b}+\frac{1}{p_n}=\frac{ap_n+b}{bp_n}$$
where $p_n$ is the $n$th prime. Clearly, $x_n$ goes to $x$ since $p_n$ goes to infinity. Just as clearly, $\gcd(p,b_n)=1$ for all but a finite number of $n$. Thus, as $n$ goes to infinity,
$$\gcd(ap_n+b,b)=\gcd(ap_n,b)=1$$
Additionally, as $n$ goes to infinity
$$\gcd(ap_n+b,p_n)=\gcd(b,p_n)=1$$
Thus
$$\gcd(ap_n+b,bp_n)=1$$
Thus,
$$f(x_n)=f\left(\frac{ap_n+b}{bp_n}\right)=bp_n$$
which is unbounded and we are done.
If a real number $\alpha$ can be expressed as $a_m a_{m-1} \ldots a_0. b_1 \ldots b_n$ ($b_n \neq 0$) in decimal notation, then let $f(\alpha) = n$. Otherwise, let $f(\alpha) = 0$.
For example, $f(2.3) = 1$, $f(3.25) =2$, $f(6.23462)=5$, $f(93)=0$, $f(\frac{1}{3}) =0$, $f(\pi)=0$, $\ldots$.
This mapping is one of functions that satisfy the condition of problem.
Let $f:\mathbb{R}\to\mathbb{R}$ be defined by $f(x)=0$ for irrational $x$ and $f\left(\dfrac pq\right)=q$ for all integers $p$, $q$ with $q>0$ and $\text{gcd}(p,q)=1$. Can you see why this is unbounded in every neighbourhood of every point?
