The problem is this: let two balls closed be in the metric space $(\Bbb{R},d)$ with the trivial metric, show that if the two balls intersect, then one is included in the other. I conclude that both balls are equal, and therefore one is included in the other, but I'm not sure how to show it.
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What is a trivial metric? And is $IR$ a notation for irrationals? – Aniruddha Deshmukh Aug 12 '18 at 06:02
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I think $IR $ stands for $\Bbb R$ – xbh Aug 12 '18 at 06:05
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In the trivial metric, the ball $B_r (x) $ of radius $r $ centered at $x $ is $$B_r (x)=\begin{cases}\{x\},&\ r <1\\ \ \\ \mathbb R,&\ r\geq1\end {cases} $$ If you now consider the nonempty intersection $B_r (x) $ and $B_s (y) $: either at least one of them is $\mathbb R $, in which case the inclusion is obvious, or they are $\{x\}$ and $\{y\}$, so equal.
Martin Argerami
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Hint: Show there are two types of balls, the space itself and sets containing exactly one element.
Sil
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Jens Schwaiger
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