basic doubt about topological manifold

Question

In his book "Introduction to Smooth Manifolds", J.M. Lee defines a topological manifold to be a second countable, Hausdorff space with every point having a neighbourhood homeomorphic to an open subset of $\mathbb{R}^{n}$ for some $n$. I was wondering should it not say that it should be homeomorphic to a connected open subset of $\mathbb{R}^{n}$? He also mentions that equivalent definitions are obtained if one replaces open subset by whole of $\mathbb{R}^{n}$ or by the open unit ball.

I cannot see how this can happen as a connected open subset cannot be homeomorphic to open unit ball or whole of $\mathbb{R}^{n}$.

Answer 1

Well, as I was typing the question I got an insight! We could always choose the connected component of the open set of the manifold which is homeomorphic to an open subset of Euclidean space!

Should this question be closed?

Answer 2

In the version of Lee I own he answers your question with Exercise 1.1 which asks to show that the locally Euclidean condition is equivalent to asking that every point has a neighborhood homeomorphic to an open ball in $\mathbb R^n$ or just $\mathbb R^n$ itself. This condition is clearly stronger than the one originally given. To show the other direction, let $M$ be a topological manifold and let $p \in M$. Suppose $p$ has a neighborhood $U_p$ homeomorphic to $U \in \mathbb R^n$ open by the homeomorphism $\varphi: U_p \to U$. Then there is some open ball $V \subset U$ containing $\varphi(p)$. Then if we restrict $\varphi$ to $\varphi^{-1} (V)$, (which is open since $\varphi$ is a homeomorphism) we have that $\varphi$ maps $\varphi^{-1} (V)$ homeomorphically onto $V$, an open ball in $\mathbb R^n$.

The next part of the problem just requires showing that open balls in $\mathbb R^n$ are homeomorphic to $\mathbb R^n$, which I'll leave to you. You can explicitly write down the homeomorphism without too much trouble.