Curves-maps; surfaces-sets

I remarked in the classroom the differences between the definition of a curve and a surface: a curve is a differentiable map and a surface is a subset of Euclidean space where there do exist parametrizations. I return again with it.

If a curve $\alpha:I\rightarrow{\mathbb R}^3$ is regular $t_0$, then $\alpha'(t_0)\not=0$. If we write in terms of the differential map of $\alpha$, it means that $(d\alpha)_t:{\mathbb R}\rightarrow {\mathbb R}^3$ is a non-zero linear map. This is equivalent to say that $\mbox{rank}(d\alpha)_t=1$, because $$(d\alpha)_{t_0}(1)=\frac{d}{ds}{\Big |}_{s=0}\alpha(t_0+s)=(x'(t_0),y'(t_0),z'(t_0))\not=(0,0,0).$$ Thus the rank  of $(d\alpha)_t$ is the maximum possible (it would be $0$ or $1$). Furthermore, by using the inverse function theorem, ``the curve is a graph locally around $t_0$''. In fact, it was proved that there exists $\epsilon>0$ such that $$\alpha:J=(t_0-\epsilon,t_0+\epsilon)\rightarrow \alpha(t_0-\epsilon,t_0+\epsilon)$$ coincides with the graph of a function, that is, there exists a differentiable function $f:K\subset {\mathbb R}\rightarrow {\mathbb R}$ such that $\{(x,f(x)):x\in K\}=\alpha(J)$. As a consequence, $\alpha:J\rightarrow\alpha(J)$ is homeomorphic to an interval of ${\mathbb R}$.

Then the map $\alpha$ would play (almost) the same role of parametrizations in a surface. If we want to give the definition of the  analogous $1$-dimensional case of a surface, then a subset $C\subset{\mathbb R}^3$ is a $1$-surface (=curve) if for each point $p\in C$ there exists $I\subset {\mathbb R}$ and a map $X:I\rightarrow V\subset C$ a homeomorphism, where $V$ is an open of $C$ around $p$, $X:I\rightarrow {\mathbb R}^3$ is differentiable and $X'(t)\not=(0,0,0)$.

The question is the definition of curve given in chapter $1$ is now a $1$-surface, more precisely, if the trace $\alpha(I)$ is a such $1$-surface. Then one would think `yes' by taking around each point $p\in C$ the corresponding restriction of $\alpha$ to the suitable interval $J$. However, the only problem is the following: Is $\alpha(J)$ an open set of $C$? because the other properties have been showed. We find the answer in the curve $\alpha(t)=(\cos(t),\sin(2t))$, $t\in {\mathbb R}$.

This curve self-intersects at the origin. Thus it can not be a $1$-surface because this point has not a neighbourhood which is homeomorphic to ${\mathbb R}$. By the inverse function theorem, around $t_0=0$, $\alpha(J)\cong J$ is a graph, but $\alpha(J)$ is not an open set of $\alpha({\mathbb R})$, which it happens exactly in our example, as one can see in the next picture: the red color line is $\alpha(J)$ is not an open set in $C$.