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Slight confusion in terminology:

What does $S=\{t \in I : \gamma(t)=\hat{\gamma}(t) \}$ closed in $I$ (open interval) by continuity mean? Why is it different from $S$ being a open set? $\gamma, \hat{\gamma}:J \rightarrow M$ is are integral curves on a smooth vector field.

This is the first part of the proof of "Fundamental Theorem on Flows" or the existence of maximal flow on Lee's Smooth Manifolds p. 213.

The reason $S$ is open is because the proof writes:

$\gamma(t_1)=p \implies \gamma(t_1)=\hat{\gamma}(t_1)=p\in M$

Each $p$ has a neighborhood, thus $\gamma, \hat{\gamma}$ agree on open sets $\implies$ $S$ is made of open sets.

mavavilj
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1 Answers1

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If $\gamma$ is continuously differentiable, then $S$ is the inverse image of the closed set $\{0\}$ under the map $t\mapsto \gamma(t)-\gamma'(t)$, hence is closed (as a subset of $I$).

Hagen von Eitzen
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  • But the proof I think shows that $S$ is an open set. If I read correctly, because the $\gamma=\gamma'$ agree on neighborhoods, which are open sets. So $S$ is made of open sets. So unless the proof has some error, then I think the closed thing and the open set thing mean different things, since they co-exist. – mavavilj Oct 22 '18 at 19:50
  • @mavavilj, I am fairly certain that $S$ is not open in general. For instance, consider $I=(-\pi/2,\pi/2)$ and $\gamma(t)=(-1+\cos t,1+\sin t)$ on $I$. Then $S={0}$. Alternatively, if $\gamma'(t)=\gamma(t)$ on an open interval, then this differential equation can be solved to yield $\gamma(t)=e^{t-t_0}\gamma(t_0)$ for some $t_0$ in that interval. – Sangchul Lee Oct 22 '18 at 20:04
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    @mavavilj: Isn't the point of the proof to show that $S$ is both open and closed in $I$, in order to conclude that $S=I$? It's closed in $I$ by continuity, and open in $I$ according to some other argument. – Hans Lundmark Oct 22 '18 at 20:23
  • @HansLundmark Okay so it's closed initially, then through reasoning it's also open, but since $I$ is connected then $S=I$, since $I$ was also open. But is the "closed part" necessary at all? – mavavilj Oct 22 '18 at 20:36
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    @mavavilj: Yes, since an open interval has plenty of open proper subintervals! The connectedness of $I$ enters into the open/closed argument like this: https://proofwiki.org/wiki/Connected_iff_no_Proper_Clopen_Sets – Hans Lundmark Oct 22 '18 at 20:40