Can two utility representations always be connected by a strictly monotonic function?

The Microeconomics Lecture notes by Rubinstein has the following question in Problem set two.

Let $$U, V: X \to \mathbb{R}$$ be two utility representations of the preference relation $$P$$ (preference relations are a total ordering and a utility representation is a map that preserves this order). Is there always a strictly monotonic function $$f: \mathbb{R} \to \mathbb{R}$$ s.t. $$V = f \circ U$$. The tone in which the question is asked seems to suggest that there is not, but for my reasoning there should be:

One can define $$\tilde{U}, \tilde{V}: \frac{X}{\sim} \to \mathbb{R}$$ which are injective and then define $$\hat{f} = \tilde{V} \circ \tilde{U}^{-1}$$ on some subset $$M$$ of $$\mathbb{R}$$. $$\hat{f}: M \to \mathbb{R}$$ is strictly montonic and can be linearly interpolated in areas where $$M$$ is not dense and continuously completed in areas where $$M$$ is dense.

Edit: On a second thought the completion procedure is not completly trivial since the left and the right limit could differ. But you can just choose one consistently and it would still work I guess, $$f$$ just would not be continuous anymore.

Do I have an error in my reasoning here?

No there need not be such a function $$f$$, it is possible that you run out of space in some sense. Suppose $$X=\mathbb N$$ and the preference relation is just the usual ordering on $$\mathbb N$$. One way to represent this is just $$V:\mathbb N\rightarrow \mathbb R$$ the inclusion map. Another way to do this is $$U:\mathbb N\rightarrow \mathbb R,\ n\mapsto 1- \frac{1}{n+1}$$ Any map $$f:\mathbb R\rightarrow\mathbb R$$ with $$V=f\circ U$$ must map $$1-\frac{1}{n+1}$$ to $$n$$. But if $$f$$ should be strictly monotonic, it would have to send $$1$$ to a number which is larger then any $$n\in\mathbb N$$, which is impossible.
If you apply your procedure here, you could define $$f$$ for any $$x<1$$, but then get stuck.