It seems to me correct, so if we have a diffeomorphism from this open neighbourhood $U$ containing the point, and a diffeomorphism $h$ and an open subset $V$ let $M$ be a submanifold. then $h(U \cap M ) = V \cap \mathbb{R}^k \times \{0\}$
i am interested in the shape of the subset $V \cap \mathbb{R}^k \times \{0\}$. In a book i am reading, it says, it is an open interval for $k=1$. However, if $V$ is disjoint, then we will have a union of open intervals, is this okay? or is the continity of the diffeomorphism vorbid that

I want to prove the statement with tools available till this chapter that I'm studying. So let me give a sketch of the proof in the book and say exactly which parts bother me:-
1) $E=\cup_n E_n$, a disjoint union such that $\mu^* E_j<\infty$ for all $n$. This can be done as $\mu$ is sigma finite (given).
2) Finding sets $F_j\in S(A), E_j\subset F_j$ and $\mu^* E_j=\mu^* F_j$ for every $j$.
3) Take any $G\in S(A)$ such that $G\subset F_j- E_j$ (this can be done as $G$ can be empty set also, which is the sigma algebra) and show that $\mu^*G=0$.

really okey... (so in this case I will try to do the sequence stuff first, as I wanted to). So I have the following in the moment:
If I take x^(n) a sequence in K then I know from
a) that ||x^(n)||_p<\infty
b) if I fix n then for all epsilon there exists n such that \sum_{k\geq N} |x^(n)_k|^p<\epsilon.

But now I need to find a convergent subsequence. Then using that K is closed I can conclude that the limit is in K and hence K is compact. but the problem is this subsequence.
So I see that Using b) I can somehow split up the sum in a) and get that (\sum_{k=1}^N |x^(n)_k|^p+\epsilon)^(1/p)<\…