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Prove the product of two Hausdorff spaces is Hausdorff

  1. Feb 19, 2017 #1
    1. The problem statement, all variables and given/known data

    Prove the product of two Hausdorff spaces is Hausdorff

    2. Relevant equations


    3. The attempt at a solution

    For ##X##, ##Y## Hausdorff spaces, need to find ##s, t \in X \times Y## where neighbourhoods ##S, T## of ##s## and ##t## are disjoint.

    Firstly, by the definition of the Hausdorff property, ##\exists \ x_1, x_2 \in X## whose neighbourhoods ##U_1## and ##U_2## are such that ##U_1 \cap U_2 = \emptyset##,

    and the same can be said for some ##y_1, y_2 \in Y##.

    The product ##X \times Y## must therefore contain points ##(x_1, y_1), (x_2, y_2)## made of these elements, whose neighbourhoods are the sets:

    ##T = \{ (x_1 - \delta , x_1 + \delta), (y_1 - \delta , y_1 + \delta) \}## with ##0 < \delta < min(x_1,y_1)##

    and

    ##S = \{ (x_1 - \epsilon , x_2 + \epsilon), (y_2 - \epsilon, y_2 + \epsilon) \}## with ##0 < \epsilon < min(x_2, y_2)##

    Now by original construction, ##x_1 \not= x_2## and ##y_1 \not= y_2##, so it can be said that ##x_1 < x_2## and ##y_1 < y_2## without loss of generality.

    and by the original condition for the Hausdorff spaces ##X## and ##Y##, the points ##x_1##, ##x_2##, ##y_1##, and ##y_2## all have open and separate neighbourhoods in their respective spaces, so there must exist ##\delta## and ##\epsilon## such that

    ##x_1 + \delta < x_2 - \epsilon##
    and
    ##y_1 + \delta < y_2 - \epsilon##

    which means there are two points ##t = (x_1, y_1)## and ##s = (x_2, y_2) ## in ##X \times Y## such that

    ##(x_1 + \delta , y_1 + \delta) < (x_2 - \epsilon , y_2 - \epsilon)##

    meaning ##t < s \ \forall \ s, t## under the above construction,

    whose neighbourhoods ##T## and ##S## cannot intersect.

    Hence the product of two Hausdorff spaces is Hausdorff.


    Well... that's my effort. Not sure how solid it is though. Feels like I invoked a circular argument along the way...

    Thanks.
     
  2. jcsd
  3. Feb 19, 2017 #2

    PeroK

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    A Hausdorff space means every pair of distinct points have disjoint neighbourhoods. And, they are general topological spaces, where you appear to have assumed they are well-ordered metric spaces.
     
  4. Feb 19, 2017 #3
    Bummer, yeah I did. I thought Hausdorff meant there was just one or 'some' pairs of distinct points that could be found, rather than every pair being distinct by definition.

    I'll have a rethink. Cheers.
     
  5. Feb 20, 2017 #4
    Answers have come out for the exercise sheet I was working on.

    The proof I was trying to get to is surprisingly simple. Just thought I'd add it here for closure to this thread, and for my own practice. Learning through repetition, heh.

    Suppose ##X## and ##Y## Hausdorff spaces.

    For ##x_1, \ x_2## distinct in ##X##, there exists ##U_1, \ U_2 \subset X## where ##U_1 \cap U_2 = \emptyset##

    Now make the sets ##V_1 = U_1 \times Y##
    and ## V_2 = U_2 \times Y##

    where ##V_1, \ V_2## are any two sets in ##X \times Y## which are distinct by construction via the usage of ##x_1## and ##x_2##.

    Then ##V_1 \cap V_2 = (U_1 \cap U_2) \times Y = \emptyset \times Y = \emptyset##

    Hence ##X \times Y## is Hausdorff.
     
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