Topological manifolds (dimension) I am taking an introductory course to topology and the professor defined a topological manifold of dimension $n$ if it is hausdorff and if for every point $x$ there exists an open set $U$ around $x$ such that $U$ is homeomorphic to $\mathbb{R}^n$. My question is: By the way she defined it, one could have (apriori) a topological space $X$ being a manifold of dimensions $n$ and $m$? Meaning I could find open sets $U$ and $V$ around every $x$ and $U$ is homeomorphic to $\mathbb{R}^n$ and $V$ is homeomorphic to $\mathbb{R}^m$, and this would make the notion of dimension not well defined. My guess is that since it is called "topological manifold of dimension $n$" what I described probably cant happen, but I dont see why not. 
 A: You can use local homology groups to solve this problem. If $X$ is a topological space and if $x\in X$, then the local homology groups of $X$ at $x$ are defined by the rule $H_i^{\text{local}}(X,x)=H_i(X,X-x)$, where the right hand side denotes the relative homology groups of $X$ relative to $X-x$.
Exercise 1: Prove that if $X$ is an $n$-manifold, then the local homology groups of $X$ agree with the homology of an $n$-sphere. (Hint: use excision to reduce to computing the homology of an $n$-ball relative to its boundary, an $(n-1)$-sphere.)
Exercise 2: Prove that the dimension of a connected topological manifold is an invariant.
Exercise 3: What can you say about manifolds with boundary?
Hope this helps! Of course, tomasz's answer and the comments are excellent and address the question in other ways.
A: You are correct.
There are several ways to show that it is impossible, and it basically boils down to the fact that no open subset of ${\bf R}^n$ is homeomorphic to an open subset of ${\bf R}^m$ if $n<m$ (as $U\cap V$ would be in your case). You can assume without loss of generality that the sets in question are connected.
One way to show it is to use the fact that you can find an $(n-1)$-dimensional sphere which disconnects any open subset of ${\bf R}^n$, but it is a general fact that if a compact set disconnects ${\bf R}^m$, then it must admit a non-nullhomotopic map onto $S^{m-1}$ (in fact, it is an equivalent condition), while on the other hand, no proper compact subset of $S^{m-1}$ admits such a map.
So, summing it all up, a set homeomorphic to $S^{n-1}$ can't disconnect an open subset of ${\bf R}^m$ if $m>n$ and we're done.
(The results used can be shown by combinatorial or algebraic topology, but are not simple enough to show here.)
