Let $M$ be a topological space and to each point $x\in M$ let there be an open interval $I(x)=(I_-(x),I_+(x))\subset\mathbb{R}$. To make it shorter, we set $E:=\bigcup_{x\in M}I(x)\times\left\{x\right\}\subset\mathbb{R}\times M$. A function $\Phi\colon E\to M$ is called a flow on $M$ if it fullfills:
(i) $E\subset\mathbb{R}\times M$ is open
(ii) $\Phi\colon E\to M$ is continious
(iii) For all $x\in M$ it is $0\in I(x)$ and $\Phi(0,x)=x$
(iv) For all $x\in M$ and all $t\in I(x)$ and $s\in I(\Phi(t,x))$ it is $s+t\in I(x)$ and $\Phi(s,\Phi(t,x))=\Phi(s+t,x)$.
Let $\Phi$ be a flow on a topological space $M$. Let $x\in M$ and $t\in I(x)$. Show that $I(\Phi(t,x))=I(x)-t$.
Here is a proof I found in a book:
According to the fourth condition, for any $s\in I(\Phi(t,x))$ it is $s+t\in I(x)$, so it is $s\in I(x)-t$, i.e. $$ I(\Phi(t,x))\subset I(x)-t. (*) $$
So $I_+(\Phi(t,x))\leqslant I_+(x)-t$. Assume that $I_+(\Phi(t,x))<I_+(x)-t$.
Choose a sequence $(t_n)_n\subset I(\Phi(t,x))$ with $\lim_n t_n=I_+(\Phi(t,x))$ and define $$ x_n:=\Phi(t_n+t,x)~\forall n\in\mathbb{N}. $$ Because $\Phi$ is continious, the sequewnce $(x_n)_n$ converges. Moreover from (*) it follows that $$ I_+(x_n)\leqslant I_+(\Phi(t,x))-t_n~\forall n\in\mathbb{N}.~~(++) $$ Because of condition (iv) that $0\in I(x)$ for all $x\in M$, it follows that $$ 0<I_+(\lim_n x_n). $$ Moreover, $I_+\colon M\to (0,\infty]$ is lower semi continious, so it is with (++) $$ 0<I_+(\lim_n x_n)\leqslant\liminf_n I_+(x_n)=\liminf_n I_+(\Phi(t_n+t,x))\\ \leqslant\liminf_n (I_+(\Phi(t,x))-t_n)\\ =I_+(\Phi(t,x))-\lim_n t_n=0. $$
This is a contradiction, so it is $I_+(\Phi(t,x))=I_+(x)-t$.
In the same way one can show that $I_-(\Phi(t,x))=I_-(x)-t$.
I have mainly two questions to this proof: (1.) Where exactly is the assumption $I_+(\Phi(t,x))<I_+(x)-t$ used? (2.) Whats the idea behind using the sequences $(t_n)_n$ and $(x_n)_n$? Did not understand that. Do not understand this details.
Maybe you can explain that to me? Would really like to understand it...