Usage of open sets in the definition of manifold

general-topologymanifoldssoft-question

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.

Best Answer

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.

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