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, 2019 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, 2019 at 18:56

1 Answer 1


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 $\gamma(t_1^*)$).

  • $\begingroup$ 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, 2019 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, 2019 at 23:21
  • $\begingroup$ can you give more details of $f$? what did you mean with $(\gamma(t_{1}^*), 0)$?? $\endgroup$
    – Joãonani
    Dec 10, 2019 at 23:27
  • $\begingroup$ Fixed my typos. $\endgroup$
    – copper.hat
    Dec 10, 2019 at 23:39
  • $\begingroup$ thank ya, now I got it :) $\endgroup$
    – Joãonani
    Dec 10, 2019 at 23:43

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