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Let $\varphi:X\times \mathbb{R}\to X$ be a continuous flow on compact metric space $X$ without singularity. For $\delta>0$ and $x\in X$, take \begin{equation} \Gamma_\delta(x)=\bigcup_{h\in\mathcal{C}}\bigcap_{t\in\mathbb{R}}\varphi_{-h(t)}(B(\varphi_t(x), \delta)) \end{equation} where $\mathcal{C}$ is the set of continuous functions $h:\mathbb{R}\to \mathbb{R}$ with $h(0)=0$ and $B(a, \delta)=\{b:d(a,b)<\delta\}$.

In the following, I give a proof of the following statement:

There is $\alpha>0$ such that $\varphi_{(-\alpha, \alpha)}(x)\subseteq \Gamma_\delta(x)$ for all $x\in X$.

If it is not true, then for every $n\in\mathbb{N}$, there is $x_n\in X$ such that

\begin{equation} \varphi_{(-\frac{1}{n}, \frac{1}{n})}(x_n)\nsubseteq \Gamma_\delta(x_n) \end{equation} Hence for every $n\in\mathbb{N}$, there is $y_n\in\varphi_{(-\frac{1}{n}, \frac{1}{n})}(x_n)$ such that $y_n\notin \Gamma_\delta(x_n)$. This implies that for every $h\in\mathcal{C}$, there is $t_n\in\mathbb{R}$ with $d(\varphi_{h(t_n)}(y_n), \varphi_{t_n}(x_n))>\delta$. By $y_n\in\varphi_{(-\frac{1}{n}, \frac{1}{n})}(x_n)$, we can say that $lim_{n\to\infty}y_n=lim_{n\to\infty}x_n=x$ that is a contradiction with $d(\varphi_{h(t_n)}(y_n), \varphi_{t_n}(x_n))>\delta$ if $h(t)=t$.

Please help me to know that my proof is true or not? Thanks

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