A question related to Jordan Curve Theorem

Question

Problem. Let $C$ be a smooth Jordan curve, and $ \gamma: [-1,1]\to\mathbb R^2,$ a segment crossing $C$ perpendicularly at $x_0\in C$, with $\gamma(0)=x_0$. Show that there exists an $\varepsilon>0,\,$ such that the segments $$ \ell_-=\big\{\gamma(t): t\in (-\varepsilon,0)\big\}\quad\text{and}\quad \ell_+=\big\{\gamma(t): t\in (0,\varepsilon)\big\} $$ lie each wholly in a different connected component of $\mathbb R^2\setminus C$.

This question, once we draw a curve and a perpendicular, looks obvious. Often in proofs, it is also considered obvious. Nevertheless, I have not managed to provide a rigorous proof/explanations. Any ideas?

Note. By "smooth curve" I mean that $\gamma$ is continuously differentiable with non-vanishing derivative.

Answer 1

Let $J: S^1 \hookrightarrow \mathbb{R}^2$ be the Jordan curve. Then, since $J$ is $C^1$ with non-vanishing derivative and a topological embedding, the image $C = J(S^1)$ is a $C^1$-submanifold of $\mathbb{R}^2$. So, there is a chart $\phi: U \to \mathbb{R}^2$ of $\mathbb{R}^2$ at $x_0 \in U \subseteq \mathbb{R^2}$ with $\phi(C \cap U) = \mathbb{R}\times\{0\}$ and $\phi(x_0) = (0,0)$. Now, since $\gamma(0) = x_0 \in U$ there is some $\delta > 0$ with $\gamma((-\delta, \delta)) \subseteq U$ and therefore $g:=\phi\circ\gamma|_{(-\delta,\delta)}: (-\delta,\delta) \to \mathbb{R}^2$ a $C^1$-curve with $g(0) = \phi(x_0) = (0,0)$.

Since $\gamma$ intersects $C$ orthogonally in $x_0$, the tangent vectors of $\gamma$ in $0$ and $J$ at $J^{-1}(x_0)$ are linearly independent. Thus $(a,b)=g'(0) \in T_{(0,0)} \mathbb{R}^2 = \mathbb{R}^2$ has $b \neq 0$, since $T_{(0,0)}\phi(C \cap U) = \mathbb{R}\times \{0\}$. W.l.o.g let $b > 0$. Therefore, there is some $\varepsilon > 0$ with $\varepsilon \leq \delta$ such that $g_2(t) > 0$ for $0<t<\varepsilon$ and $g_2(t) < 0$ for $-\varepsilon<t<0$ (where $g = (g_1,g_2)$).

Let $A_1,A_2$ be the connected components of $\mathbb{R}^2\setminus C$.

Let $U_0 = \phi^{-1}(\{x_2 = 0\})$, $U_- = \phi^{-1}(\{x_2 < 0\})$ and $U_+ = \phi^{-1}(\{x_2 > 0\})$. $U_\pm$ are connected and $U_\pm \subseteq A_1 \cup A_2$. Since $\partial A_1 = \partial A_2 = C$ (Jordan Curve Theorem) it is $U\cap A_1\neq \emptyset,U\cap A_2 \neq \emptyset$. Therefore w.l.o.g. $U_+ \subseteq A_1$ and $U_- \subseteq A_2$.

Now, $U_+\cup U_0 \cup U_-=U = (U\cap A_1)\cup(U\cap C) \cup (U\cap A_2) = (U\cap A_1)\cup U_0 \cup (U\cap A_2)$ disjoint, thus $U_+\cup U_-=(U\cap A_1) \cup (U\cap A_2)$. With $U_+ \subseteq A_1\cap U$ and $U_- \subseteq A_2\cap U$ it follows $U_+ = A_1\cap U$ and $U_- = A_2\cap U$.

This means that $\phi\circ \gamma(t) = g(t) \in \phi(U\cap A_1)$ for $0<t<\varepsilon$ and $\phi\circ \gamma(t) = g(t) \in \phi(U\cap A_2)$ for $-\varepsilon<t<0$. Thus $\gamma(t) \in A_1$ for $0<t<\varepsilon$ and $\gamma(t) \in A_2$ for $-\varepsilon<t<0$.

Answer 2

I work in $\mathbb C$ here not $\mathbb R^2$ and use different notation than OP.

Let $\gamma: [0,1]\longrightarrow \mathbb C$ be a Jordan curve which has a continuous non-vanishing derivative everywhere and $\sigma: [-1,1]\longrightarrow \mathbb C$ be a line segment crossing $\big[\gamma\big]$ (the image of $\gamma$) perpendicularly at a point $z_0 \in \big[\gamma\big]$.

We need only consider the problem in a particularly simple setting. First, to make things easy we may parameterize $\gamma$ such that $\gamma\big(\frac{1}{2}\big)=z_0$. Second the properties of interest will be preserved by any invertible affine map, so using translation we can take $z_0 =0$ and using rotation we can take $\big[\sigma\big]$ to be a segment on the real line which implies $\gamma'\big(\frac{1}{2}\big) \not \in \mathbb R$. Third we can also choose to parameterize $\sigma$ to be negative when $t\lt 0$ and positive when $t\gt 0$.

Suppose for contradiction that $\ell_{+}$ does not exist. This means there is a point $p_k \in \big[\gamma\big] \cap \big(0, \delta_k\big)$ for all $k\in \mathbb N$ where $\delta_k := 2^{-k}$. But this implies $\pm\frac{p_k}{\tau_k-\frac{1}{2}} = \pm\frac{\gamma(\tau_k) - \gamma(\frac{1}{2})}{\tau_k-\frac{1}{2}} \to \pm\gamma'\big(\frac{1}{2}\big)$ where $\tau_k\to \frac{1}{2}$ but the sequence is entirely real hence $\gamma'\big(\frac{1}{2}\big) \in \mathbb R$ which is a contradiction. For avoidance of doubt, $\gamma(0)=\gamma(1)\neq 0$ hence for $\delta \gt 0$ small enough, we see that $\gamma_{\big\vert [0+\delta, 1-\delta]}$ is a homeomorphism to its image. Thus a sequence $\gamma(\tau_k)= p_k\to 0 \implies \tau_k \to \frac{1}{2}$.

By an essentially identical [symmetric] argument $\ell_{-}$ must exist as well and we take $\epsilon$ to be the smaller value between these two cases.

It remains show that $\sigma\big(-\frac{\epsilon}{2}\big)$ and $\sigma\big(\frac{\epsilon}{2}\big)$ are in [the two] different components of $\mathbb C-\big[\gamma\big]$. Suppose for contradiction that $\sigma\big(-\frac{\epsilon}{2}\big)$ and $\sigma\big(\frac{\epsilon}{2}\big)$ are in the same component $U \in\mathbb C-\big[\gamma\big]$.

Now $U$ is an open connected set hence there is a polygonal path $\sigma_2$ from $\sigma\big(-\frac{\epsilon}{2}\big)$ to $\sigma\big(\frac{\epsilon}{2}\big)$ lying entirely in $U$. This implies a polygonal curve $\Sigma$ --or if preferred polygonal cycle $\Sigma =(\sigma, \sigma_2)$-- which has $\sigma\big(\frac{\epsilon}{2}\big), \sigma\big(\frac{-\epsilon}{2}\big) \in \big[\Sigma\big]$ and $\big[\Sigma\big] \cap \big[\gamma\big]=\big\{0\big\}$ . In terms of winding numbers this implies $\Big\vert n\big(\Sigma,\gamma(\frac{1}{2}-\delta_k)\big)- n\big(\Sigma,\gamma(\frac{1}{2}+\delta_k)\big)\Big\vert=1$ for $k\geq K$ large enough. [This 'crossing' property for polygonal paths is easily verifiable -- the only essential item is that $\Sigma^{-1}(p)$ has cardinality one for $p \in [-\delta', \delta']$ for $\delta'\gt 0$ small enough and we get that from part one of this problem]. But we also know $\gamma\Big(\big[0, \frac{1}{2}-\delta_K\big] \cup \big[\frac{1}{2}+\delta_K,1\big] \Big) \in \mathbb C-\big[\Sigma\big]$ hence $n\big(\Sigma,\gamma(\frac{1}{2}-\delta_K)\big)= n\big(\Sigma,\gamma(\frac{1}{2}+\delta_K)\big)$ by path connectivity and this is a contradiction.