Let be integral curves in a manifold then is closed.

**Proof:**

Let be have a limit in . Let be a chart then converges to by continuity. But also converges to . Hence and so . Thus is closed.

Let be integral curves in a manifold then is closed.

**Proof:**

Let be have a limit in . Let be a chart then converges to by continuity. But also converges to . Hence and so . Thus is closed.

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does not mean .

For example, ; does have the right properties?

Yes because is a chart hence one-to-one.

and both cts is cts.

Let and

and

Claim is open.

where

is continuous given s.t.

s.t.

is open complement closed closed.

1. The minus function () doesn’t necessarily exist in the manifold so you can’t even define .

2. Allowing for this, why should and this always be ?

1 you are right need to apply to and then define as the difference of these two images now in . This is still cts and we now have a metric.

2 2 is my nonsense, now improved by pre-applying ; is now a non zero element of so has a non-zero length, if we call its length then halve this length to be our to be used in the argument.

intuitively any non zero has a cloud of non-zero around it and the pre-images belong to an open ball in

I see what you are saying. Wouldn’t it be better to argue that if then there is an open interval around such that this interval is in . Take the union of these open intervals and since the arbitrary unions of open sets is open then must be open. Hence must be closed.