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?
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.