I understand that the very definition of a Jordan Curve includes that it be a closed curve. However, I am reading a paper written by E. Hopf in 1950 in which he proves a theorem by Bernstein.
In this paper, we are working on the $\mathbb{R^2}$ plane and he mentions the existence of a Jordan curve $\big( x(t), y(t) \big)$ for which $x(t) \rightarrow \pm \infty$ as $t\rightarrow \pm\infty$. What confuses me is the fact that this curve will not be closed under normal circumstances. (I can show more of the context of the paper but I have chosen not to to keep things short)
My thoughts on possible explanations:
- This is a closed curve if we include the point at infinity, however it is not explicitly mentioned that we are working on the plane $\mathbb{R^2}\cup \{ \infty \}$.
- Hopf simply means a continuous curve when he talks about a Jordan curve (ie, not necessarily smooth).
What do you think?
Such a curve actually is necessarily closed. First, since $x(t) \to \pm\infty$ as $t \to \pm\infty$, we know that the curve extends to a continuous function $\gamma : S^1 \to S^2$ (where $S^1 = \mathbb{R} \cup \{\infty\}$ is the one-point compactification of $\mathbb{R}$ and $S^2 = \mathbb{R}^2 \cup \{\infty\}$ is the one-point compactification of $\mathbb{R}^2$) such that $\gamma(\infty) = \infty$. Since $S^1$ is compact and $S^2$ is Hausdorff, $\gamma(S^1)$ is closed in $S^2$. The image of the original curve is just $\gamma(S^1) \cap (S^2 \setminus \{\infty\})$ as a subset of $S^2 \setminus \{\infty\} = \mathbb{R}^2$. By definition of the subspace topology, the image of the original curve is closed in $\mathbb{R}^2$.