# Prove that a Curve has no interior points

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?

• Idea: if you remove a point from a ball interior, the set remains connected, but removing a point within a curve disconnects it. – Cheerful Parsnip Feb 10 '20 at 0:19
• 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) – Alexandar Solženjicin Feb 10 '20 at 0:22
• The key idea is that connectedness is a topological property, so it is invariant under homeomorphism. – Ken Hung Feb 10 '20 at 1:52

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.