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Singleton sets closed in T_1 and Hausdorff spaces

  1. Apr 17, 2010 #1
    If [tex](X,\tau)[/tex] is either a [tex]T_1[/tex] space or Hausdorff space then for any [tex]x \in X[/tex] the singleton set [tex]\{ x \}[/tex] is closed.

    Why is this the case? I can't see the reason from the definitions of the spaces.

    Definition:
    Let [tex](X,\tau)[/tex] be a topological space and let [tex]x,y \in X[/tex] be any two
    distinct points, if there exists any two open sets [tex]A,B \in \tau[/tex]
    such that [tex]x \in A[/tex] but [tex]x \notin B[/tex] and [tex]y \in B[/tex] but [tex]y \notin A[/tex],
    then [tex](X, \tau)[/tex] is a [tex]T_1[/tex] space.

    Definition:
    A topological space [tex](X, \tau)[/tex] is Hausdorff if for any
    [tex]x,y \in X[/tex], [tex]x \ne y[/tex], [tex]\exists \text{ neighborhoods } U \ni x[/tex] and
    [tex]V \ni y[/tex] such that [tex]U \cap V = \varnothing[/tex].
     
  2. jcsd
  3. Apr 17, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    {x} is closed if the complement of {x} is open. Try to show X/{x} is open using the definitions. That's not so hard, is it?
     
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