# Inverse Function Theorem and open sets

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?

• Perhaps you could show that the range of $\gamma$ has measure zero? – copper.hat Dec 9 '19 at 18:42
• How can you apply the inverse function theorem on a function $\mathbb{R} \to \mathbb{R}^2$? $\gamma'(t)$ cannot be invertible. – 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 $$\gamma(t_1^*)$$).

• 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 $f(t^{*}) = (\gamma(t_{1}^{*}), 0 )$ – Joãonani Dec 10 '19 at 23:18
• $f$ is defined everywhere, and I have given the evaluation at a specific point for clarity (or not, as seems the case :-)). – copper.hat Dec 10 '19 at 23:21
• can you give more details of $f$? what did you mean with $(\gamma(t_{1}^*), 0)$?? – Joãonani Dec 10 '19 at 23:27
• Fixed my typos. – copper.hat Dec 10 '19 at 23:39
• thank ya, now I got it :) – Joãonani Dec 10 '19 at 23:43