It is a very basic question. Given a metric space defined by $(X,d)$ where $X$ is a set and $d$ is a metric. Let's assume $X = \{x_{1},x_{2}\}$ and mark $s=d(x_{1},x_{2})$. Can I define a ball with a radius bigger than the distance $s$? Like this one: $B(x_{1},2s)=\{y\in X|d(x_{1},y) < 2s\}$. Is there any meaning to defining a ball with a radius larger than the largest distance in the metric space? Can a ball "cover" areas that dosen't have members of the set $X$ in them?
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If $X$ is a subspace of another ambient space, then there could be some differences, but if you are only considering the space $X$ itself, then there isn't much difference since all the balls you will be describing is simply $X$ itself. – user340297 Oct 24 '17 at 07:42
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Can I define a ball with a radius bigger than the distance $s$?
Yes, of course. For every $r$, the set $B(x, r)$ is defined as $$B(x_0,r)=\{x\in X| d(x, x_0) < r\}$$
There is nothing in the definition that would require you to only select values of $r$ that can be obtained in $X$.
That said, of course it doesn't make much sense in spending a lot of time picking the $r>s$, since $B(x_0, r)$ is the same for all $r>s$.
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