0
$\begingroup$

A topological $n$-manifold $M$ is locally Euclidean of dimension $n$ (each point of $M$ has a neighborhood that is homeomorphic to an open subset of $\mathbb{R}^n)$.

But what does locally Euclidean of dimension 0 mean?

$\endgroup$
7
  • 3
    $\begingroup$ Discrete${}{}$? $\endgroup$ – Angina Seng Jun 3 '20 at 13:58
  • $\begingroup$ @AnginaSeng What does that mean? $\endgroup$ – Filippo Jun 3 '20 at 14:04
  • $\begingroup$ It means that "locally Euclidean of dimension zero" is equivalent to "discrete". $\endgroup$ – Lee Mosher Jun 3 '20 at 14:45
  • $\begingroup$ @LeeMosher A topological space $M$ is discrete if (and only if) every subset of $M$ is open, right? $\endgroup$ – Filippo Jun 3 '20 at 15:18
  • $\begingroup$ @Filippo right, all subsets are open. However manifolds are requested to be second countable also, hence uncountable discrete spaces are not allowed. $\endgroup$ – InsideOut Jun 3 '20 at 15:28
3
$\begingroup$

$\Bbb R^0$ is just a point, isn't it? Hence locally Euclidean of dimension zero means that locally homeomorphic to $\Bbb R^0=\{pt\}$. A $0-$topological manifold is then a countable set endowed with the discrete topology.

$\endgroup$
3
  • 1
    $\begingroup$ Only countable family if you have having a countable base as a part of the definition of a manifold. The OP might well not assume that. $\endgroup$ – Henno Brandsma Jun 3 '20 at 15:31
  • $\begingroup$ Maybe we are using different definitions of manifolds. I have always assumed manifolds to be second-countable and then the existence of a countable basis. However, if we drop the second-countability condition also uncountable union of points endowed with the discrete topology are allowed. $\endgroup$ – InsideOut Jun 3 '20 at 15:35
  • $\begingroup$ I am currently reading John Lee's book about smooth manifolds and i just found the equation/definition $\mathbb{R}^0=\{0\}$ on page 25. Since $\{\{0\},\emptyset\}$ is the only existing topology on $\{0\}$, we can even make sense of the definition in my question for the case $n=0$ and we can prove what InsideOut said. $\endgroup$ – Filippo Jun 3 '20 at 19:52

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.