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Showing A Subset of the Product Space ##Y \times Y## is Open

  1. Dec 2, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ##Y## be some order spaced endowed with the order topology, and let ##P = \{(x,y) \in Y \times Y ~|~ x < y \}##. I would like to show that ##P## is open in ##Y \times Y##.

    2. Relevant equations


    3. The attempt at a solution

    Let us momentarily assume that ##Y## has neither a smallest nor largest element. In this case, the basis elements for the product topology are of the form ##(a,b) \times (c,d)##, where ##(a,b)## is an open interval in ##Y##.

    Let ##\langle x,y \rangle## be some point in ##P \subseteq Y \times Y##. ##x,y \in Y##, there exists basis elements ##(a,b)## and ##(c,d)## such that ##x \in (a,b)## and ##y \in (c,d)##. Now, if there is an element ##e## between ##x## and ##y##, we can "readjust" the basis elements to obtain ##(a,e)## and ##(e,d)##, and clearly every element ##(a,e)## is smaller than every element in ##(e,d)##. Hence, ##\langle x,y \rangle \in (a,e) \times (e,d) \subseteq P##.

    Now, suppose that there is no element between them. Then ##(a,y)## will contain ##x## and ##(x,d)## will contain ##y##. Again, it isn't difficult to see that every element the latter set is smaller than every element in the former set. In either case, we can conclude that ##P## is open.

    Does this sound right? I suspect that the when ##Y## has a smallest or largest element the proof will be very similar. If so, I will omit it.
     
  2. jcsd
  3. Dec 3, 2016 #2

    haruspex

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    I believe your argument for this case also works when there is an intervening element.
    From a quick reading of the subject (never met it before) you can suppose a basis consisting of open "rays" (-∞, x), (x, ∞). I assume the infinities represent min and max elements if they exist. So you might be able to write one argument that covers all cases.
     
    Last edited: Dec 3, 2016
  4. Dec 3, 2016 #3
    Are you sure? I need an open set about ##U_x## about ##x## and an open set ##U_y## about ##y## such that everything in ##U_x## is strictly smaller than everything in ##U_y##. If I choose the sets I did above and there is an intervening element won't we have ##e < e##, which is a problem. I am sorry; I am not thinking clearly at the moment, and its pretty noisy where I am.

    In my book, basis elements for the order topology are defined as ##(a,b)##, ##[a_0,b)##, where ##a_0## is the smallest element (if it exists), and ##(a,b_0]##, where ##b_0## is the largest element (if it exists). But I think you are right that we can knock out all cases by just looking at ##(-\infty, a)## and ##(a,\infty)## and letting the infinities represent the largest/smallest elements or finite elements. , since the mechanics of the proof will be the same in each case. I would like verification, of course.
     
  5. Dec 3, 2016 #4

    haruspex

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    You are right, and thinking more clearly than I am.
     
  6. Dec 4, 2016 #5
    Okay, I think I may have made progress towards slightly more understanding. In my textbook, it is shown that elements of the form ##(a, \infty)## and ##(- \infty, a)## form a subbasis for the order topology, and a subbasis generates the topology through arbitrary unions of finite intersections of subbasis elements; i..e., letting ##\mathcal{S}## denote the set of all open rays, then a open set will be of the form ##\bigcup_{i \in I} S_i##, where ##S_i## is a finite intersection of elements in ##\mathcal{S}##. So you are right that we can just look at these open rays.

    This actually helps with another problem I am working on.
     
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