Let $\gamma: \mathbb{R} \to \mathbb{R}^2$, $\gamma$ continuously differentiable with $\gamma'(t) \neq 0$, $\forall t \in \mathbb{R}$. Then, for all $t_1 \in \mathbb{R}$, there is a $\epsilon>0$ such that does not exists open set $A \subset \mathbb{R}^2$, with $ A \subset$ $\gamma(t_1 – \epsilon, t_1+ \epsilon)$.
My attempt:
For assumptions I guess I need to use the Inverse Function Theorem, so fixing $t_1 \in \mathbb{R}$ There is neighborhood $U$ of $t_1$ and a neighborhood $V$ of $\gamma(t_1)$ such that $\gamma: U \to V $ is a bijection and it's inverse $\gamma^{-1}:V \to U$ is differentiable.
Ok, $\gamma(U)$ is a open set, so I need to build a set $B$ inside of $\gamma(U)$ such that there is not open set inside of $B$,maybe Implicit function theorem? I dunno.
Can you help me?
Best Answer
Pick $t_1^*$. Choose $u \neq 0$ such that $u \bot \gamma'(t_1^*)$. Define $f(t) = \gamma(t_1)+t_2 u$.
Let $t^* = (t_1^*,0) \in \mathbb{R}^2$ and note that $f'(t^*) $ is invertible.
The inverse function theorem gives an open neighbourhoods $V$ containing $f(t^*) = \gamma(t_1^*)$, an open neighbourhood $U$ containing $t^*$ and a homeomorphism $g:V \to U$ that is a local inverse of $f$.
Note that for $t \in U$, $f(t) \in \gamma(\mathbb{R}) \cap V $ iff $t_2 = 0$.
In particular, for $n$ large enough, $t_n = (t_1^*, {1 \over n}) \in U$, $t_n \to t^*$ and $f(t_n) \notin \gamma(\mathbb{R}) \cap V$. Hence $\gamma(\mathbb{R}) \cap V$ cannot contain an open set (containing $\gamma(t_1^*)$).