# I am not sure - a manifold is locally connected and has countable basis?

1. Jul 19, 2012

### mahmoud2011

I am not sure -- a manifold is locally connected and has countable basis?

There is an Exercise in a book as following :

Given a Manifold M , if N is a sub-manifold , an V is open set then V $\cap$ N is a countable collection of connected open sets .

I am asking why he put this exercise for only the case of sub-manifold , Is n't this an immediate consequence of the fact that a manifold is locally connected and has countable basis ?????

I am not sure from what I say ?? I think there can't be exercise as easy as I think , I think I am wrong .

Thanks

2. Jul 19, 2012

### mahmoud2011

Re: I am not sure -- a manifold is locally connected and has countable basis?

I am sorry , I was having a confusion in the definitions between what is the n-submanifold property and what is the sub-manifold .

Thanks

3. Jul 20, 2012

### Bacle2

Re: I am not sure -- a manifold is locally connected and has countable basis?

The definition I know has it that S is a

submanifold o fM if S is embedded in M in the same way as R^k is embedded /sits-in

R^{k+j} in the "standard way", i.e., where (x_1,..,x_k)-->(x_1,x_2,...,x_k, 0,0,..,0)

So S is a submanifold of M if there are subspace charts mapping points of S into

points of the form (x_1,...,x_k, 0,0,..,0).

Last edited: Jul 20, 2012