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Let $C$ be a curve meaning $C=\sigma(U) \subseteq \mathbb{R}^2$ where $\sigma: U\subseteq \mathbb{R} \to \mathbb{R}^2$ such that $\sigma: U \to \sigma (U)$ is a homeomorphism. I want to prove it has no iterior points meaning for no $p \in C$ is there a ball (open in $\mathbb{R}^2$) $B(p,\epsilon)\subseteq C$ such that $p\in B(p,\epsilon)$

This makes perfect geometric sense to me, but I simply do not know how to prove it form the definition. (or anything else for that matter)

Could someone help?

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    $\begingroup$ Idea: if you remove a point from a ball interior, the set remains connected, but removing a point within a curve disconnects it. $\endgroup$ – Cheerful Parsnip Feb 10 '20 at 0:19
  • $\begingroup$ I see, but my followup question would be how can I relate connectedness with "interiorness" or the characteristics of a curve (I assume the homeomorphism part will play a role here, yet I don't see what exactly) $\endgroup$ – Alexandar Solženjicin Feb 10 '20 at 0:22
  • $\begingroup$ The key idea is that connectedness is a topological property, so it is invariant under homeomorphism. $\endgroup$ – Ken Hung Feb 10 '20 at 1:52
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Suppose that there exists a ball in the image of $\sigma$, take the pre image of such ball. It's connected and open in $\mathbb{R}$, so is is an open interval, and such interval is a homeomorphic to said ball (it is clearly induced by the curve). So take a point off the ball, it's connected, but the interval minus a point is not.

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