dimension of smooth connected manifold

Question

Let $M^n$ be a smooth, connected $n$-manifold. By definition, this means that every point $p\in M$ has a neighborhood $U$ that is homeomorphic to some open subset of $\mathbb{R}^n$. Now let us assume we only know that for some point $p\in M$ we know that it is in a neighborhood homeomorphic to $\mathbb R^n$ and for the other points we just know that they have a neighborhood that is homeomorphic to $\mathbb R^k$ for some $k\in \mathbb N$. I think that we can show using path-connectedness that $k$ is always equal to $n$ but I don't know how exactly.

Answer 1

Let $\sim$ be the equivalence relation which states that $x \sim y$ if there exists charts around them which go to the same dimension (they are not necessarily on a same chart. This only states that each one has a chart around them that both have the same dimension). This is clearly an equivalence relation, and each class is clearly open. By connectedness, we must have only one class. By invariance of domain (or by looking at the derivatives if you are assuming smoothness), there can be no two charts around the same point which go to different dimensions. Hence, the charts only go to a fixed dimension.