A plane curve without self-intersection has empty interior Let $\alpha:[0,1]\rightarrow \mathbb{R}^2$ be continuous and injective. I want to prove that $\alpha([0,1])$ has empty interior. One way to show this I have seen in a question here in MSE (but I can't find it now to link it to the post) is to suppose that there is $p\in \alpha([0,1])^\circ$ and since we have a disc $D\subset \alpha([0,1])$ one may define a continuous bijection $\beta:=\alpha_{|\alpha^{-1}(D)}:\alpha^{-1}(D)\rightarrow D$ and the claim was that $\beta$ was actually a homeomorphism and by an argument of connectivity we get a contradiction. However, it is not clear to me why $\beta^{-1}$ must be continuous... Moreover, is dimension $2$ essential here or the same result holds for $\mathbb{R}^n$?
 A: Since continuous bijection from a compact set onto a Hausdorff space is a homeomorphism, we have that $\alpha$ is a homeomorphism from $[0,1]$ to $A=\alpha([0,1])$.
Assume we have a disk $D\subset A$. Since $D$ is connected, $I=\alpha^{-1}(D)$ should be an interval in $[0,1]$.
Now, choose a point $p\in D\setminus\{\alpha(0),\alpha(1)\}$. One can check that $D'=D\setminus \{p\}$ is still connected, but $\alpha^{-1}(p)$ should lie somewhere in $I$, so $\alpha^{-1}(D')=I\setminus\{\alpha^{-1}(p)\}$ should be disconnected.
This is a contradiction, so we do not have such a disk $D\subset A$.
A: To say that $\beta^{-1}$ is continuous is equivalent to saying that $\beta$ is a closed mapping, i.e., it maps closed sets to closed sets. However a subset of the domain $[0, 1]$ of $\beta$ is closed iff it is compact. Continuous functions like $\beta$ map compact sets to compact sets. Compact subsets of $\Bbb{R}^2$ are closed by the Heine-Borel theorem, so $\beta$ is a closed mapping and $\beta^{-1}$ is, indeed, continuous. This part of the proof works in all dimensions, not just $2$, but in dimension $1$, the connectivity argument used in the rest of the proof won't go though (unsurprisingly, since any non-constant continuous function $\beta : [0, 1] \to \Bbb{R}$ has an image with a non-empty interior). The connectivity argument works for all dimensions greater than 1.
