- #1
complexnumber
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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].
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].