Question on the definition of the parametrization of a curve

Question

I have in my lecture notes the following definition about the normal parametrization of a curve in $\Bbb{R}^3$.

Definition. (Normal) Parametrization of a curve in $\mathbb{R}^3$ is a vector fuction $$\vec{r}:(a,b)\longrightarrow \Bbb{R}^3,\ t\mapsto \vec{r}(t)=(x(t),y(t),z(t))$$ where $-\infty \leq a \leq b \leq +\infty$, which has the following properties:

1. $\vec{r}(t)$ is smooth and at least $C^1$

2. $\frac{d\vec{r}}{dt} \neq \vec{0},\ \forall t\in(a,b)$

3. $\vec{r}$ must be 1-1 with her image.

Is this definition completely right? I can not understand the property 3. What does this mean? Is there an intuitive way to think about it?

Thank you in advance and I apologize for my English.

Answer 1

Being "1-1" means being injective.

However, I don't really understand condition (1): being smooth is much stronger than being just $C^1$...

Answer 2

By 1-1 ness you are avoiding self-intersections of your map. This for example tells you that the a figure 8 intersecting itself in the center is not counted as a curve according to your definition. *

For (1) the author may have meant "smooth OR at least $C^1$.

Addition: This is quite restrictive as it does not include closed curves, such as a circle. But then again one can define closed curves as those that are injective on the interior of the intervals and are equal at the endpoints.