Open and Closed Relations: A Topological Approach to Evaluating Limits

[alexfloo]

"Open" and "closed" relations

We know that if we have convergent sequences (x_n) and (y_n) in simply ordered metric space, then x_n\leqy_n implies that the limits x and y have x\leqy. Also, x_n<y_n.

My instinct on noting this is to say that "<" is an "open relation" on that metric space, and that "\leq" is its "closure" in that this pair of relations shares a certain property of open sets and their closures in a topological sense.

More generally, I would define an "open" relation to be a relation such that, if it holds pairwise for two convergent sequences (or equivalently if it eventually holds pairwise), it does not necessarily hold in their limits. It is "closed" if this does imply that it holds in the limit. The "closure" of a relation R is the "strictest" (in some sense I'm not certain of) relation such that if R holds pairwise on two sequences, then the closure holds for the limits.

This seems like it would be a useful concept for evaluating limits, but a Google search didn't turn up anything with my particular terminology. Is this an actual developed topic? And if so, what it the common terminology for it?

 

[disregardthat]

A relation R can be seen as a subset of X x X, so what you are looking for is probably the closure of R as a subset of X x X in the induced product topology from X, where X is your ordered space. It is the smallest relation (as a set) which is closed (by your definition) and contains R. Closed relations correspond to closed subsets of X x X.

I would define open relations as relations corresponding to the open subsets of X x X, so that the space of open relations form a topology on the powerset of relations.

 

[alexfloo]

That was my intuition, but I wasn't sure about it. So it is the case that R being a closed subset of X x X is equivalent to R being a "closed relation" as above?

 

[micromass]

That was my intuition, but I wasn't sure about it. So it is the case that R being a closed subset of X x X is equivalent to R being a "closed relation" as above?

On metric spaces, yes.
In general topological spaces, no. The reason is that sequences are too weak to characterize the closure in topological spaces. So you need to work with nets. If you work with nets, then they become equivalent.

 

[alexfloo]

Thanks, I haven't really been exposed much to topologies in general, which is why I specified metric spaces in the original post. I'm not familiar with the idea of a net, but I definitely plan an looking it up. Thanks a lot, both of you!

 

[disregardthat]

Yes, I agree with micromass, I think we require the space to be first-countable in order for this to be equivalent. A closed relation (if seen as the smallest relation that contains the limit of its sequences), is not necessarily closed for a space that is not first-countable, but the closed subsets of X x X are certainly closed relations.

The open relations if defined as the complements of the closed relations is however not necessarily a topology on the powerset of relations.

But I think that my suggestion is the "best" way to define the closure if you want the open relations to be a topology (in that it is the finest topology that can be generated in such a way).