# Are open sets and open balls the same thing?

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?

• 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 0:28
• 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 0:28
• @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 0:36
• @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 $\Bbb R$. – user98602 Dec 16 '14 at 0:40
• 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 1:41

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 $\Bbb R^n$.