0
$\begingroup$

In the book "An introduction to manifolds" by Tu, a topological manifold is defined to be a topological space $M$ that is Hausdorff, second countable and locally Euclidean.

Does this allow things like the disjoint union of a plane and a line? Then we have a component which is locally Euclidean of dimension $1$ and one of dimension $2$?

$\endgroup$
0
3
$\begingroup$

Tu allows manifolds having connected components of different dimensions. He explicitly says it in this post. Usually people talk about a space being "locally $\Bbb R^n$" or "locally Euclidean of dimension $n$" as opposed to just "locally Euclidean", as he does. But it is not hard to show that for each $n \geq 0$, the set $$\{ x \in M \mid x \mbox{ has an open neighborhood homeomorphic to }\Bbb R^n \}$$is both open and closed in $M$. So this means that the dimension is well defined on each connected component of $M$.

$\endgroup$
6
  • $\begingroup$ Thanks for the answer. Don't you mean "homeomorphic to an open subset of $\mathbb{R}^n$"? $\endgroup$ – user745578 Jul 4 '20 at 21:48
  • $\begingroup$ Also, could you briefly sketch why the set you wrote down is closed? This is not immediately clear to me. I tried taking a sequence $x_n \to x$ where $x_n$ is in the set and show that $x$ is in the set as well. $\endgroup$ – user745578 Jul 4 '20 at 21:55
  • $\begingroup$ @user745578 It is closed because the complement is open :) $\endgroup$ – Hagen von Eitzen Jul 4 '20 at 22:12
  • $\begingroup$ @HagenvonEitzen But why? If you take a point $x$ not in that set, then you know that $x$ has no open neighborhood homeomorphic to some open subset of $\Bbb{R}^n$. Then why is this the case for a nbh of $x$ as well? $\endgroup$ – user745578 Jul 4 '20 at 22:26
  • 1
    $\begingroup$ If $x_k\to x$ and each $x_k$ has a neighborhood homemorphic to $\Bbb R^n$, use that $x$ has a neighborhood homeomorphic to some $\Bbb R^m$. This contains one of the $x_k$'s by convergence. The transition between the charts is a homeomorphism between open subsets of $\Bbb R^n$ and $\Bbb R^m$. So $n=m$. $\endgroup$ – Ivo Terek Jul 4 '20 at 22:29
3
$\begingroup$

Yes, it does. In the remark right after Definition 5.2, Tu states "Of course, if a topological manifold has several connected components, it is possible for each component to have a different dimension." The disjoint union of a plane and a line is a valid example of such a space. Note that the dimension will be constant on each connected component, so nothing more egregious than this example can happen.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy