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?

  • $\begingroup$ Perhaps you could show that the range of $\gamma$ has measure zero? $\endgroup$ – copper.hat Dec 9 '19 at 18:42
  • 1
    $\begingroup$ How can you apply the inverse function theorem on a function $\mathbb{R} \to \mathbb{R}^2$? $\gamma'(t)$ cannot be invertible. $\endgroup$ – copper.hat Dec 9 '19 at 18:56

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 $\gamma(t_1^*)$).

| cite | improve this answer | |
  • $\begingroup$ I almost got your explanation, but what's the domain of $f$? $f:\mathbb{R}^2 \to \mathbb{R}^2$? you have defined $f(t) = \gamma(t_1) + t_2u$ and $f(t^{*}) = (\gamma(t_{1}^{*}), 0 )$ $\endgroup$ – Joãonani Dec 10 '19 at 23:18
  • $\begingroup$ $f$ is defined everywhere, and I have given the evaluation at a specific point for clarity (or not, as seems the case :-)). $\endgroup$ – copper.hat Dec 10 '19 at 23:21
  • $\begingroup$ can you give more details of $f$? what did you mean with $(\gamma(t_{1}^*), 0)$?? $\endgroup$ – Joãonani Dec 10 '19 at 23:27
  • $\begingroup$ Fixed my typos. $\endgroup$ – copper.hat Dec 10 '19 at 23:39
  • $\begingroup$ thank ya, now I got it :) $\endgroup$ – Joãonani Dec 10 '19 at 23:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.