If $T_t$ is a diffeomorphism with $T_0=\text{id}$ and $O$ is a neighborhood of $\Omega$, then $T_t(\Omega)\subseteq O$ for small $t$

Question

Let $\tau>0$, $d\in\mathbb N$ and $T_t$ be a $C^1$-diffeomorphism from $\mathbb R^d$ onto $\mathbb R^d$ with $T_0=\operatorname{id}_{\mathbb R^d}$.

If $\Omega\subseteq\mathbb R^d$ and $O$ is an open neighborhood of $\Omega$, are we able to show that there is a $\delta>0$ with $T_t(\Omega)\subseteq O$ for all $t\in[0,\delta)$?

Intuitively, the claim should be true, but how can we prove it? Feel free to assume that $[0,\tau]\times\mathbb R^d\ni(t,x)\mapsto T_t(x)$ is continuous.

Answer 1

It is not true. Consider $\Omega = \{(x,0) : x \in \mathbb R\} \subset \mathbb R^2$, $O = \{(x,y) : |y| < \frac1{1+x^2}\}$, and $T_t (x,y) = (x,y+t)$.

Or $O = \{(x,y) : |y| < 1\}$ and $T_t(x,y) = (x, y+tx)$.

Now if $\Omega$ is compact, it is true. First note that if $O \supset \Omega$ is open, then there exists $\epsilon>0$ such that the $\epsilon$-neighborhood of $\Omega$ is in $O$. Then $T:[0,1] \times \Omega \to \mathbb R^d$ is uniformly continuous. So there exists $\delta > 0$ such that if $x,y \in \Omega$ with $d(x,y) < \delta$, and if $s,t \in [0,1]$ with $|s-t| < \delta$, then $d(T_s(x),T_t(y)) < \epsilon$. In particular, $d(T_s(x), x) < \epsilon$, and so $T_s(x) \in O$.