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

[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

 

[mahmoud2011]

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

 

[Bacle2]

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).