I'm doing an exercise problem where I'm supposed to say if a region in $\mathbb{R}^3$ spans curves or not.
There are two surfaces that I thought would span curves but according to the answer key, they don't. The surfaces are:
- "The interior of the torus in $\mathbb{R}^3$ obtained by rotating the circle $(x-2)^2 + y^2 = 1$ about the $z$-axis"
- "The set of all points in $\mathbb{R}^3$ except for those on the circle $x^2 + y^2 = 1$, $\,z = 0$ in the $xy$-plane."
Why don't these span curves?
(According to the textbook:
If $D$ is an open set in $\mathbb{R}^3$ and any simple closed curve $C$ in $D$ is the boundary of an orientable surface which lies entirely in $D$, then $D$ "spans curves.")
I recommend that you sketch pictures of these examples:
For the open torus, consider a closed curve that "goes once around the hole in the center", e.g. the circle of radius $2$ in the $xy$-plane, centered at the origin: $$ \{ (x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 = 4 \text{ and } z = 0 \}. $$
For the complement of the circle in $\mathbb{R}^3$, consider a closed curve that links with the missing circle, e.g. the circle in the $yz$-plane centered at the point $(0, 1, 0)$ on the missing circle: $$ \{ (x, y, z) \in \mathbb{R}^3 \mid (y - 1)^2 + z^2 = 1 \text{ and } x = 0 \}. $$
In either case, if the circle were the boundary of an oriented surface, then we could glue in a disk in the open set $D$ in question with the circle as its boundary. But this is not possible!