# 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 $$\varphi([0,1])$$ had inhabited interior. Then we could find some closed ball $$D=D(x,\epsilon)$$ (homeomorphic to the disk $$D^2$$) within $$I$$. Now the restriction $$\varphi':\varphi^{-1}(D)\to D$$ would be a homeomorphism from some subset of $$[0,1]$$ to the 2-dimensional disk. Can you show that no such homeomorphism exists? See what happens when we remove a point from $$D$$.
• Sorry, but I can't see why $\phi'^{-1}$ is continuous Jul 26, 2021 at 2:29
• @Curious. $φ'^{-1}$ is continuous because $φ$ is closed (i.e. $φ(C)$ is closed whenever $C\subseteq [0,1]$ is). That's a consequence of the fact that every continuous map from a compact space to a Hausdorff space is closed. Jan 28, 2022 at 9:28