Let $f:\mathbb{R} \to \mathbb{R}$ be any real function. Must there be a "non-negligible" set $S \subset \mathbb{R}$ such that $f(S)$ is bounded? "Non-negligible" is open to interpretation. We can take it to mean that the closure of $S$ has nonempty interior. Alternatively, we can take it to mean that $S$ has positive measure. A̶l̶t̶e̶r̶n̶a̶t̶i̶v̶e̶l̶y̶,̶ ̶w̶e̶ ̶c̶a̶n̶ ̶s̶i̶m̶p̶l̶y̶ ̶t̶a̶k̶e̶ ̶i̶t̶ ̶t̶o̶ ̶m̶e̶a̶n̶ ̶t̶h̶a̶t̶ ̶$̶S̶$̶ ̶i̶s̶ ̶u̶n̶c̶o̶u̶n̶t̶a̶b̶l̶e̶ (edit: solution to this below). The answerer may choose any of these interpretations (or an alternative one, if he feels it is relevant).
I'm aware of the existence of some pathological functions like the Conway Base $13$ function. This is a function $f:\mathbb{R} \to \mathbb{R}$ with the property that any nonempty open set is mapped to $\mathbb{R}$. However, I'm not sure whether or not this is a counterexample to any of the claims above.
We can indeed find an uncountable $S \subset \mathbb{R}$ with $f(S)$ bounded. Consider that $$\mathbb{R} = \bigcup_{n \in \mathbb{Z}} f^{-1}([n, n+1])$$ hence there must exist an $n$ such that $f^{-1}([n, n+1])$ is uncountable. Hoping for a stronger result.
