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Hausdorf space condition problem

  1. Jun 18, 2017 #1
    1. The problem statement, all variables and given/known data
    Show that X is a Hausdorff space IFF the 'diagnol of x' given by t = {(x,x) | X * X} is closed as a subspace of X*X

    2. Relevant equations


    3. The attempt at a solution
    So since X Is Hausdorff so is X*X and t, because the product of two Hausdorff spaces if Hausdoff and the subspace of any Hausdorff space is Hausdorf.

    So i've been doing a lot of thinking about this and have some ideas, so here we go.

    Since X is Hausdorff, for any x_1 and x_2 in X, there exists open sets (or neighborhoods) U_1 and U_2 that contain x_1 and x_2 respectively and their intersection is the empty set.

    To show that t is closed as a subset of X*X, we must show that t contains all it's limit points. It is perhaps worth noting that if A is a subset of a Hausdorff space Y, then b is a limit point of A iff every neighborhood of b intersects A at infinitely many points.

    First I will try to show that X being a Hausdorff space implies that t is closed:

    I will proceed by contradiction
    Suppse X is a Hausdorff space and that t is not closed. Then there exists an element in the closure of t (hereby known as ct) that is not in t, we will call this element y=(x_1,x_2) where x_1 does not equal x_2, for if it did then y would be in ct. Since y is in the closure of t, every neighborhood of y contains some element of t, or said otherwise, if U is a neighborhood of y, then the intersection of U and t is nonempty.

    Question: y must be of the form (x_1,x_2), correct? I am confused by that notion because that would mean that y is an element of X*X, which is not necesarily true I don't think, but rather it is only gaurenteed to be on the closure of X*X.

    If y must be of the form (x_1,x_2) where each x_i is an element of X then I smell a contradiction lingering closely....

    Anyone have any thoughts/concerns?
     
  2. jcsd
  3. Jun 18, 2017 #2

    fresh_42

    Staff: Mentor

    Life would be easier if you forgot about the limit points and prove ##X \text{ Hausdorff } \Longleftrightarrow (X \times X) - t \text{ open }##. Points ##(x_1,x_2)## outside ##t## automatically fulfill ##x_1 \neq x_2## and open neighborhoods occur in the definition of open sets as well as of Hausdorff spaces.
     
    Last edited: Jun 18, 2017
  4. Jun 19, 2017 #3
    X is Hausdorff so for any points in X, say x_1 and x_2, there exists neighborhoods U_1 and U_2 such that x_1 is in U_1 and x_2 is in U_2 and U_1 and U_2 are disjoint.

    (X * X) - t must be open because for any point (x_1,x_2) in this space there exists an open neighborhood U_1 (Union) U_2 the contains the point.

    Am I close?
     
  5. Jun 19, 2017 #4

    fresh_42

    Staff: Mentor

    The track is correct, but "union" false. How do open sets in ##X \times X## look like? And how can you use ##U_1\, , \,U_2## here?
     
  6. Jun 19, 2017 #5
    Open sets in X*X are tuples that look like (x_i,x_j) where each x is an element in X. perhaps (U_1,U_2) is an open neighborhood that contains the point (x_1,x_2), or I think the correct notation would actually be U_1 * U_2.
     
  7. Jun 19, 2017 #6

    fresh_42

    Staff: Mentor

    Yes, that's the key object for both directions. Correct would have been ##U_1 \times U_2 \subseteq X \times X## as an open set, not U_1 * U_2. One doesn't write the product space with a dot, at least I haven't ever seen it. Now why is ##U_1 \times U_2 \subseteq X \times X - t## if ##X## is Hausdorff? This is needed as we want to find an open neighborhood of ##(x_1,x_2)## that is completely in our presumably open set ##X \times X - t ##.

    The other direction is similar.
     
  8. Jun 19, 2017 #7

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Think of the projection maps ##\pi_1, \pi_2 ; \pi_1(x,y)=x , \pi_2(x,y)=y ##.
     
  9. Jun 20, 2017 #8
    U_1 x U_2 is Hausdorff because it is the subspace of a Hausdorff space, is that what you are asking?
     
  10. Jun 20, 2017 #9

    fresh_42

    Staff: Mentor

    No, I meant an open set in ##X \times X## looks like ##U_1 \times U_2## with open sets ##U_i \subseteq X## by the definition of the product topology. Now all you need is a connection that relates ##x_1 \in U_1## and ##x_2 \in U_2## as the open sets necessary to establish the Hausdorff property in one direction, and the open property of ##X \times X - t## necessary for the other direction. ##x_1 \neq x_2## resp. the fact that ##U_1 \cap U_2 = \emptyset## has to be somehow related to this diagonal, which by the way would have been better noted as ##\Delta## rather than by ##t##, which could be overlooked too easily, especially if you don't use LaTeX to write it.

    Maybe you should restart from scratch:

    Hausdorff ##\Longleftrightarrow \; \forall \;x_1\neq x_2 \;\exists\; U_1(x_1)\; , \;U_2(x_2) \text{ open }\, : \,U_1 \cap U_2 = \emptyset##

    1. Find an open set of ##U(x_1,x_2) \subseteq X \times X- \Delta## that doesn't contain any points of ##\Delta =t##.
    2. Given ##(x_1,x_2) \in X \times X - \Delta ## and ##X \times X - \Delta## is open, show there are open sets ##U_i(x_i) \subseteq X## with ##U_1(x_1) \cap U_2(x_2) = \emptyset##.
    The entire proof is to see, how the diagonal plays a role in here.
     
  11. Jun 20, 2017 #10

    WWGD

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    Think of this, if ##U_x ## and ##U_y ## are disjoint. Can there be an element ##(a,a)## in ## U_x \times U_y ## ? If this is the case then a is in....( Use projection maps here).
     
  12. Jun 20, 2017 #11
    Wow! you guys are such good explainers, thanks so much! I really need to learn LaTeX....

    So showing that every element in X*X - t, so a generic element of the form (x_i,x_k) where i does not equal k, is contained in an open neighborhood U_i*U_k that is completely contained in X*X - t will show that X*X - t is open in X*X ?
     
  13. Jun 20, 2017 #12

    WWGD

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    Yes; if you can show that a generic point is contained in an open set S where S is contained entirely in the complement, this is saying the complement is open and therefore the diagonal is closed. But note that S can be of any form as long as it is open.
     
    Last edited: Jun 20, 2017
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