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we know that in discrete metric space every subset is bounded which means we can find a ball that contains that subset. we only need to take a radius greater that 1. is this the only space that has this property? is there a metric space where every subset is bounded but that space is not discrete??

Hodiii
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  • Consider $\Bbb R$ with the metric $\displaystyle d(x,y)=\frac{|x-y|}{1+|x-y|}$ for all $x,y\in \Bbb R$. – Sumanta Apr 09 '21 at 10:37

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Sure there is: Take $[0,1]$ with the usual metric.

By the way: For every metric space $(X,d)$ there exists a metric $\tilde d$ on $X$ which induces the same topology but makes the whole space bounded.

principal-ideal-domain
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