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In Wikipedia:

A subset $S$ of $\mathbb{R}^n$ is bounded with respect to the Euclidean distance if and only if it bounded as subset of $\mathbb{R}^n$ with the product order.

More generally, I was wondering for a set which is both a metric space and partially-ordered set, when the boundedness of a subset wrt the metric and wrt the order agree, or just one implies the other not the other way around?

Thanks and regards!

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    Good attempt at generalization. However, you need to assume some sort of interaction between the metric and the partial order. With no conditions whatsoever, one cannot expect to prove anything in either direction.2011-04-19
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    @user6312: The connection is what I am asking for. Are you aware of cases when they are related? For example, when the topology induced by the metric and the topology induced by the order are compatible in some way?2011-04-19
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    It should have been clear to me that this was the only thing you could be asking. The sought-for conditions may not be easy, and certainly cannot be purely *topological*, since any distance function $d(x,y)$ can be replaced by $d_1(x,y)=\max(d(x,y),1)$ without changing the topology.2011-04-19
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    Certainly if your metric is bounded you're going to have problems.2011-04-19

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