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Which of the following are Hausdorff?

  1. Feb 3, 2017 #1
    1. The problem statement, all variables and given/known data

    Which of the following topologies are Hausdorff (if any)?

    {∅, {a,b} }

    {∅, {a}, {a,b} }

    {∅, {b}, {a,b} }

    {∅, {a}, {b}, {a,b} }


    2. Relevant equations

    Definitions:

    A neighbourhood U of x is an open set U⊂X such that xϵU

    A topological space is Hausdorff if for each pair x, y of distinct points in X there exist neighbourhoods U of x and V of y which are disjoint.


    3. The attempt at a solution

    Another description I've seen for the Hausdorff property is that for some two subsets U and V in X, the intersection of U and V is the empty set.

    Looking at the given topologies, it seems that only the last one is Hausdorff. One can take {a} and {b} and see that their intersection is ∅. .

    Hoping I'm correct with this? I just wanted to put it by the physics forum crowd because topology is a new and potentially worrying subject I've taken on at university, with a lot of new definitions and terminology being thrown around at quite a fast pace!

    Thanks :oldsmile:
     
  2. jcsd
  3. Feb 3, 2017 #2
    You are on the right track, but it might help to write out your reasoning. i.e. If you believe the first 3 are not Hausdorff, you should be able to say why, in symbols. You can use you "other description" to do so. i.e. Why (in symbols) do you believe the first isn't Hausdorff? Prove it!
     
  4. Feb 3, 2017 #3
    And yes topology definitions are going to come at you hard and fast. braceyourself.jpg
     
  5. Feb 4, 2017 #4
    Alright, I'd say for the topologies ##\tau_n ## in the above question statement,
    for ##n=[1,3] ## and taking any ##U, V \subset \tau## where ##U \not= V## and ##U, V \not= \emptyset##,
    ##U \cap V \not= \emptyset## for all ## U, V##, hence ##\tau_1, \tau_2, \tau_3## are not Hausorff

    whereas for ##\tau_4## there exists ## U, V \subset \tau## such that ##U \cap V = \emptyset##, hence ##\tau_4## is Hausdorff.
     
  6. Feb 5, 2017 #5

    micromass

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    Try to prove that a finite space is Hausdorff if and only if it is discrete. That's a fun exercise.
     
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