True or False: No Parametrized Surface is Closed in $\mathbb{R}^3$

Question

Is no parametrized surface (topologically) closed in $\mathbb{R}^3$? I use the definition of Fitzpatrick's Advanced Calculus (2009). So far, all the examples I've seen have not been closed -- for example, $\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2=1,z>0\}$ or $\{(x,y,z)\in\mathbb{R}^3:x^2+y^2=z^2,z>0,(x-1)^2+y^2<1\}.$

Answer 1

Consider a $C^2$ curve $\alpha : [0,1] \to \mathbb{R^2}$ having the following shape:

$\hspace{10.5em}$

Then write $\alpha(t) = (\alpha_1(t), \alpha_2(t))$ and define $\mathbf{r} : \mathcal{R} = \{ (x,y) \in \mathbb{R}^2 : x^2+y^2 < 1\} \to \mathbb{R}^3$ by

$$ \mathbf{r}(x, y) = \left( \frac{\alpha_1(r)}{r}x, \frac{\alpha_1(r)}{r}y, \alpha_2(r) \right), \qquad r=\sqrt{x^2+y^2}. \tag{*} $$

The only point where $\mathbf{r}$ may fail to be regular is the origin, but the regularity of $\alpha_1$ guarantees that $\mathbf{r}$ is indeed $C^1$ on all of $\mathcal{R}$. Now its graph looks like

$\hspace{3.3em}$

Although the surface seems self-intersecting, the parametrization is in fact one-to-one since $\alpha$ is one-to-one on $[0, 1)$.

Moreover, it is clear from the graph that $\mathcal{S} = \mathbf{r}(\mathcal{R})$ is topologically closed. Of course one can give a proof: Extend $\mathbf{r}$ to all of $\bar{\mathcal{R}} = \{ (x,y)\in\mathbb{R}^2:x^2+y^2\leq 1\}$ by using the formula $\text{(*)}$. Then we have $\mathcal{S} = \mathbf{r}(\bar{\mathcal{R}}) $ since $\alpha(1) = \alpha(t_0)$ for some $t_0 \in [0, 1)$. Now $\mathcal{S}$ is an image of the compact set $\bar{\mathcal{R}}$ under a continuous map, it is compact and hence closed in $\mathbb{R}^3$.