# Rigorous proof that a simple closed curve with a given parametrization in $\mathbb{R}^2$ either has positive or negative orientation

I am asking for a rigorous proof of the following:

Theorem: Let $R$ be a region and $C$ be a simple closed curve so that $C=\partial R$. If $\gamma:[a,b]\to \mathbb{R}^2$ is parametrization of $C$ then $\gamma$ is either positive or negative orientiated with respect to $R$.

Definition of positive orientation: Define $n(t)=(-y^{\prime}(t),x^{\prime}(t))$ where $\gamma(t)=(x(t),y(t))$. Then $\gamma$ is positive orientiated with respect to $R$ if $\forall t\in [a,b]\ \exists \epsilon>0$ so that $\gamma(t)+\epsilon n(t)\in R$. Negative orientation is defined similarly but with $n(t)=(y^{\prime}(t),-x^{\prime}(t))$.

Note that I request a proof that doesn't utilise the Jordan Curve Theorem and this is the reason the definitions and the theorem are stated this way.

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In general, the theorem is not true. For example, $\gamma$ need not be differentiable or the derivative might vanish sometimes.
But if $x'$ and $y'$ exist and are continuous and there is no $t$ with $x'(t)=y'(t)=0$, then we can do the following: Consider a point $t_0\in [a,b]$. Wlog. $a<t_0<b$, $\gamma(t_0)=(0,0)$, $h:=x'(t_0)>0$, $y'(t_0)=0$ (apply a suitable translation and rotation to the figure). Then there is $\delta>0$ such that $x'(t)>\frac h2$ for $t_0-\delta< t<t_0+\delta$. As a consequence $x(t)>\frac h4 (t-t_0)^2$ for $t_0<t<t_0+\delta$ and $x(t)<-\frac h4 (t-t_0)^2$ for $t_0-\delta< t<t_0$.
The compact set $\gamma([a,t_0-\delta])\cup \gamma([t_0+\delta,b])$ has a positive distance $d$ from $(0,0)$ because $C$ is simple.
Because $(0,0)\in\partial R$, there is a point $(\xi,\eta)\in R$ at distance $<\min\{\frac d2,\frac h4\delta^2\}$. There is exactly one $t_1\in(t_0-\delta,t_0+\delta)$ with $x(t)=\xi$ because $x$ is monotone there. If $\eta<y(t_1)$, then we can walk inside $R$: first vertically down to $\left(\xi, \sqrt{\frac {d^2}4-\xi^2}\right)$, then along the circle of radius $\frac d2$ to $(0,-\frac d2)$, then to $(0,u)$ with $0>u>-\frac d2$. Similarly, if $\eta>y(t_1)$, then we find a point $(0,u)\in R$ with $0<u<\frac d2$. This shows: For all $t\in[a,b]$ there is $\epsilon>0$ such that $\gamma(t)+\epsilon n(t)\in R$ or $\gamma(t)-\epsilon n(t)\in R$. Since the two maps $t\mapsto \gamma(t)\pm\epsilon n(t)$ are continuous, the set of points $t$ where $\gamma is locally positive/negative oriented, is open. The next step should be to show that these sets are also closed. - You can omit the last sentence: As$S^1\$ is connected one of the two open sets has to be empty. –  Christian Blatter Oct 25 '12 at 16:19