Suppose $f : [0,1] \rightarrow \mathbb{R}^2$ is continuous and injective, then show that $f([0,1])$ has an empty interior in $(\mathbb{R}^2, d_e)$.

Question

Suppose that $f : [0,1] \rightarrow \mathbb{R}^2$ is continuous on $[0,1]$ and also one-to-one, then show that $f([0,1])$ has an empty interior in $(\mathbb{R}^2, d_e)$.

This is for the one-dimensional Euclidean metric on $[0,1]$ and the two-dimensional Euclidean metric on $\mathbb{R}^2$.

Other than the basic definition of interior, continuous, and one-to-one, I have no clue on how to show this at all, so any kind of hint would be appreciated.

Answer 1

Let K be a closed subset of [0,1].
Thus, since K is compact and f continuous,
f(K) is compact, hence closed.
Thus f is a homeomorphism between [0,1] and f([0,1]).

So cutpoints of [0,1] are mapped to cutpoints of f([0,1]).
If B is an open ball of the real plane contained in f([0,1]),
then, as B has numerous points, at least one of them has to be a cutpoint of f([0,1]). Show that is a contradiction.