Here, the following definition is stated.

A topological space M is called an $n$-dimensional topological manifold (or $n$-manifold) if,

  1. Every point $x\in M$ has an open neighbourhood $U\subset M$ that is homeomorphic to an open subset of $\mathbb{R}^n$.
  2. $M$ is Hausdorff.
  3. $M$ is second-countable.

In condition 1, why should $U$ be open? Couldn't one just say that for each $x\in M$, there is a neighboorhood $U$ which is homeomorphic to a subset of $\mathbb{R}^n$? Is this equivalent to saying that for each $x\in M$ there is a closed neighborhood $U$ which is homeomorphic to a closed subset of $\mathbb R^n$?

In general, I am wondering why one so often speaks about open sets, when I have the feeling that one also could speak about closed sets, so it seems like an arbitrary decision to do so.

  • $\begingroup$ A neighbourhood of a point $p$ includes an open subset containing $p$ by definition, so there is no loss of generality in restricting to open neighbourhoods. $\endgroup$ – Gibbs Oct 19 '19 at 10:55
  • $\begingroup$ Okay. But I need to say neighbourhood (and not just say "set containing $x$"), because otherwise there trivially exists a set of points in euclidean space homeomorphic to it (consider singletons), right? $\endgroup$ – user716465 Oct 19 '19 at 11:01
  • $\begingroup$ A manifold (from the intuition from examples like surfaces) must be Euclidean locally and this is formalised with neighbourhoods. You're modelling things like surfaces/curves and their higher dimensional analogs. $\endgroup$ – Henno Brandsma Oct 19 '19 at 11:03
  • $\begingroup$ @user716465 The word "set" does not require the existence of any topology, so in general you cannot talk about homeomorphisms between sets. You can talk about homeomorphisms between two topological spaces, so sets with a topological structure. In the context of topological manifolds you have a topology both on the abstract manifold and on $\mathbb{R}^n$, and you want a transformation preserving the topological structure, that is a homeomorphism. $\endgroup$ – Gibbs Oct 19 '19 at 11:10

Suppose that $V$ is any neighbourhood of $x$ (in $M$) such that there is a homeomorphism $f: V \to O$ where $O \subseteq\Bbb R^n$ is open.

Being a neighbourhood of $x$ in $M$ means that there is an open set $V_x$ containing $x$ in $M$ such that $V_x \subseteq V$.

But then $h[V_x]$ is open in $O$ (and hence open in $\Bbb R^n$ too!) and the restriction $h: V_x \to h[V_x]$ is a homeomorphism between an open neighbourhood of $x$ and an open subset of $\Bbb R^n$, just as your definition requires.

So whether we ask for just a neighbourhood or an open neighbourhood, we get the same result, no extra generality is achieved. But it's nicer (in manifold theory) to work with open neighbourhoods, so we just assume that from the start.

  • $\begingroup$ Thanks. Why is it nicer to work with open neighbourhoods? $\endgroup$ – user716465 Oct 19 '19 at 11:00
  • 1
    $\begingroup$ @user716465 you can differentiate everywhere, e.g. (for smooth manifold), and when the neighbourhoods overlap, they're always neighbourhoods of their common points, so atlasses of manifolds are easier to work with etc. $\endgroup$ – Henno Brandsma Oct 19 '19 at 11:02
  • $\begingroup$ Ok, thank you very much! $\endgroup$ – user716465 Oct 19 '19 at 11:05

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