Let $(X,d)$ a metric space and $B_X$ the closed unit ball in $X$. It is possible to proof that $X\setminus B_X$ is not bounded (i.e. the diameter of the set is finite, $$\operatorname{diam}(K)=\sup\{d(x,y)\mid x\in K, y\in K\},$$ where $K\subseteq X$) using the fapt that is the complement of a bounded set?
In $\Bbb{R}^2$, with euclidean distance, it seems to be valid this argument. Or not?
Is there somewhere, maybe a General Topology book, results concerning the bounded set and their properties in metric space?
This result holds:
Proof:
$B_X$ is certainly bounded, with $\operatorname{diam}(B_X) \le 2$.
If $X \setminus B_X$ is bounded then $$X = B_X \cup X \setminus B_X $$
and is hence bounded as a union of two bounded sets. Namely: $$\operatorname{diam}(X) \le \underbrace{\operatorname{diam}(B_X)}_{\le 2} + \underbrace{d(B_X, X \setminus B_X)}_{<+\infty} + \underbrace{\operatorname{diam} (X \setminus B_X)}_{<+\infty} < +\infty$$
Conversely, if $X \setminus B_X$ is unbounded, then so is $X$ since $$+\infty = \operatorname{diam}(X \setminus B_X) \le \operatorname{diam}(X)$$
which implies $\operatorname{diam}(X) = +\infty$.