# Image of continuous injective map has empty interior.

Let $\varphi :\left [ 0,1 \right ]\rightarrow \mathbb{R^2}$ be a continuous injective map. Let $I = \varphi \left ( \left [ 0,1 \right ] \right )$ be the image of this map. Prove that $I$ has empty interior.

This problem was on my Topology final. I can see that $\varphi \left ( \left [ 0,1 \right ] \right )$ is a path in $\mathbb{R^2}$ which does not intersect itself (since $\varphi$ is injective). Therefore it would have empty interior (as it will not contain any open ball in $\mathbb{R^2}$) but I could not think a way to rigorously prove it.

Assume that $\phi(I)$ had inhabited interior. Then we could find some closed ball $D=D(x,\epsilon)$ (homeomorphic to the disk $D^2$) within $\phi(I)$. Now the restriction $\phi':\phi^{-1}(D)\to D$ would be a homeomorphism from some subset of $I$ to the 2-dimensional disk. Can you show that no such homeomorphism exists? See what happens when we remove a point from $D$.