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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.

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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 $D$.

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  • $\begingroup$ Sorry, but I can't see why $\phi'^{-1}$ is continuous $\endgroup$
    – Curious
    Jul 26, 2021 at 2:29
  • $\begingroup$ @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. $\endgroup$ Jan 28, 2022 at 9:28

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