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I am trying to solve the first exercise in John Lee's Introduction to Smooth Manifolds and I am confused by the terminology in the question.

He says (paraphrased): Consider the usual definition of a topological manifold $M$ (ie Hausdorff, second countable, and locally homeomorphic to $\mathbb{R}^{n}$). Show that equivalent definitions of manifolds are obtained if instead of requiring an open subset in $M$ to be homeomorphic to any open subset of $R^{n}$, we require it to be homeomorphic to an open ball in $\mathbb{R}^{n}$, or to $\mathbb{R}^{n}$ itself.

What kind of ball does he mean? Does he mean $n$-ball? Is this the same as an open set?

BCLC
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Stan Shunpike
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    Open set can be represented as union of open balls, (this is not necessarily a ball). Each open ball is an open set. – Curious Dec 16 '14 at 00:28
  • Stan, an open set $A$ is a set such that for every $x\in A$ an open ball with radius $r>0$ can be build centered at $x$ and all the points of the ball are inside the set. An open ball is an open set, not necessarily an open set is an open ball. – Vladimir Vargas Dec 16 '14 at 00:28
  • @NotStrang can you give an example of an open set that is not an open ball? Would the empty set be an example? – Stan Shunpike Dec 16 '14 at 00:36
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    @StanShunpike That's up for debate. A good example is, say, the unit square in $\Bbb R^2$; or even $(-1,0) \cup (1,2)$ in $\Bbb R$. –  Dec 16 '14 at 00:40
  • Think of open balls as the building blocks for open sets. Open sets are precisely the sets that are the unions of some collections of open balls. Note that this means that open balls are among the open sets (use a trivial collection of open balls that consists of only that single open ball). It's like "every pinkie is a finger, but every finger isn't a pinkie". – MPW Dec 16 '14 at 01:08
  • @MikeMiller: Open balls are determined by a center and a positive radius. Unless you are thinking of allowing a nonpositive radius, every open ball $B(x,r)={ y\in X : d(x,y)<r}$ in $(X,d)$ is nonempty since it contains its center $x$, right? – MPW Dec 16 '14 at 01:14
  • @MPW I'm claiming it's up for debate whether one should include nonpositive radius. (I've seen both definitions before, though I can't give references off the top of my head. I don't really care either way.) –  Dec 16 '14 at 01:21
  • @MikeMiller: Ok, agreed. – MPW Dec 16 '14 at 01:22
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    Okay, so let me see if I have this correct: an open set is the union open balls. As such, any open ball is automatically an open set. However, the reverse is not true. – Stan Shunpike Dec 16 '14 at 01:41

1 Answers1

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Here, an open ball probably refers to something like an open $n$-ball under the Euclidean norm on $\Bbb R^n$. No, not all open sets in $\Bbb R^n$ are balls, though all balls are open sets. However, f every $p \in M$ has a neighborhood that is homeomorphic in $\Bbb R^n$, then it follows that every point has a neighborhood that is homeomorphic to an open ball (the converse holds trivially).

In particular, suppose $p \in U \subset M$ with $U$ open, and $h: U \to V \subset\Bbb R^n$ is a homeomorphism.

By the definition of openness in $\Bbb R^n$, there exists an open ball $N$ around $h(p)$ such that $h(p) \in N \subset V$. From there, it follows that $p \in h^{-1}(N) \subset M$. That is, $p$ has neighborhood $h^{-1}(N)$, which is homeomorphic to an open ball in $\Bbb R^n$.

Ben Grossmann
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