Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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. – André Nicolas Apr 19 '11 at 15:20
@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? – Tim Apr 19 '11 at 15:30
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. – André Nicolas Apr 19 '11 at 15:59
Certainly if your metric is bounded you're going to have problems. – JSchlather Apr 19 '11 at 16:34

For a functional analyst, the natural setting for questions of this sort is the theory of Banach lattices, i.e., vector spaces with both a norm and a lattice structure. The typical examples are the $L^p$-spaces and these are very useful for looking at the kind of relationships between the two structures that seem to interest you. Particularly suggestive are the two extreme cases $p=1$ (where the norm is order continuous) and $p=\infty$ (where the property you mention---equivalence of boundedness in the topological and in the order sense---is valid).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.