# Thread: Closed Submanifold, in Spivak's Vol 1, chap 2, exercise 24(b)

1. ## Closed Submanifold, in Spivak's Vol 1, chap 2, exercise 24(b)

The problem is indeed an exercise in spivak's 'a comprehensive introduction to differential geometry' vol 1, chap 2, exercise 24.
I can do 24(a) and a half part of (b), but got stuck on the other half....
I first gave the result of 24(a), you can just use it:
If M is a $\displaystyle C^\infty$ manifold, a set $\displaystyle M_1$ in M can be made into a k-dimensional submanifold of M if and only if around each point in $\displaystyle M_1$ there is a coordinate system $\displaystyle (x,U)$ on M such that $\displaystyle M_1 \cap U=\{p:\quad x^{k+1}(p)=\cdots=x^n(p)=0\}$.

Now here is the second part of the question which I got stuck:

If the subset $\displaystyle M_1$ can be made into a closed submanifold, then the coordinate systems mentioned above exist around every point of M (not just $\displaystyle M_1$).

My idea is that suppose M is connected, then let $\displaystyle W= \bigcup U$, where U is the coordinates mentioned above, then W is an also a submanifold in M containing $\displaystyle M_1$, W is open obviously, I want to show W is also closed so that W=M, then we are done. But I got stuck here, maybe my idea was wrong.....

Please note that in Spivak's book, a closed submanifold $\displaystyle M_1$ in M is defined as:
(a) an immersed submanifold whose inclusion map is also an embedding, and:
(b) the subset $\displaystyle M_1$ is a closed subset of M.

Namely, a submanifold is just an embedding submanifold, and a closed submanifold is an embedding submanifold with itself is a closed set in the whole space.

Thank you in advance! Hope some guys who are familar with Spivak's book can help me~

2. You can always assume your manifolds are connected, as you can always work on a connected component.

About the answer. Collect all neighbourhoods of points in $\displaystyle M$ with that property. Their union is exactly $\displaystyle M_1$ but also an open set in$\displaystyle M$. So $\displaystyle M_1$ is open and closed in $\displaystyle M$, giving $\displaystyle M=M_1$.