I just started reading Spivak's "A Comprehensive Introduction to Differential Geometry" and there is something in the proof of his 2nd Theorem that really bugs me (probably because its obvious to most people). First, all the information for people who don't have access to the book:
Let $X$ be a connected, locally compact metric space. For every $x \in X$ there is a closed compact ball {$y \in X : d(x,y) \leq$ r} with $r>0$ by the defintion of local compactness. Now consider the set of all such $r > 0$ (exclude the trivial case that there is one $x \in X$ so that all $r > 0$ belong to this set). For each different $x$ these $r$ make up a specific interval.
Let $$r(x) := \frac{1}{2}\mathrm{sup}\{ r \in \rm I\!R^+ : \{ y \in X : d(x,y) \leq r \}\mathrm{}\ is \ compact\}$$ denote one-half the least upper bound of all such $r$ belonging to one point $x \in X$.
Now suppose $A_n \subset X$ is compact, let $A_{n+1}$ be the union of all closed balls of radius $r(y)$ and center $y$ for all $y \in A_n$. Set $A_1 = \{ x_0\}$ with $x_0 \in X$.
Spivak now says the countable union of all $A_n$ is "clearly open", but I just cant see why. Can anyone give me a clear and rigorous argument why this is the case? My first thought was that this union would make up the whole space $X$ (which is in general not the case), then it would be both open and closed, but he wants to prove that by showing first that this union is both open and closed. If any information is missing feel free to ask!
Take a point $x$ in the set $\bigcup_{n=1}^\infty A_n$. It follows that there exists $n = 1,\ldots,\infty$ such that $x \in A_n$.
By definition, $A_{n+1}$ contains the closed ball of radius $r(x)>0$ and center $x$, and so it also contains the open ball of radius $r(x)$ and center $x$. This open ball is contained in the set $\bigcup_{n=1}^\infty A_n$ and it contains $x$. Since $x$ is an arbitrary element of the set $\bigcup_{n=1}^\infty A_n$, it follows that this set is a union of open balls. So it is open.