Topology of ordered sets

Discussion in 'Math Research' started by William Elliot, Oct 19, 2011.

  1. Let (S,<=) be a (partial) order.
    Assume S has a topological base of order convex sets.
    Let R be the relation { (x,y) in SxS | x <= y }.

    A set K, is order convex when for all a,b in K,
    if a <= x <= b, then x in K.

    If R is closed within SxS, then S is Hausdorff.

    If S is Hausdorff, is R closed?

    If S is a linear order then yes, R is closed.
    What's the situation when S isn't linear?
    William Elliot, Oct 19, 2011
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  2. William Elliot

    Nils Gösche Guest

    The answer is no then. Take S to be the set of real numbers, and define

    x <= y iff y - x is rational and non-negative

    Note that open intervals are still order convex with this order, so they form
    a basis of order convex sets for the usual topology of the real line,
    which is Hausdorff.

    Now choose x = 0, y = sqrt(2). Then (x, y) is not in R. Every
    neighborhood of (x, y) in SxS contains a product IxJ, where I is an open
    interval containing x, and J an open interval containing y. You can
    choose a rational number q from J such that (0, q) is in R, so IxJ
    intersects R. Hence (x, y) is not in R but in the closure of R, and R
    is not closed.

    Nils Gösche, Oct 20, 2011
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  3. Thank you for the counter example to the converse.
    I was suspecting there would be one. I have additional
    questions about ordered topological spaces as described here.

    Order and Topology,
    An ordered topological space is a topological space (X,T) equipped with
    a partial order <=. Usual compatibility conditions between the topology
    and order include convexity (T has a basis of order-convex sets) and the
    T_2-ordered property: ie { (x,y) | x <= y }, is closed in XxX.

    Now that it's clear that the T_2 ordered property is stronger than
    Hausdorff, why assume the T_2 ordered property instead of Hausdorff?
    What additional propositions does the T_2 ordered property give a
    space that merits the use of it instead of Hausdorff?

    Is there any more or less agreed upon topology for a lattice,
    based upon the order of the lattice?

    William Elliot, Oct 22, 2011
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