Equivalent formulation of $T_1$ condition.












0












$begingroup$


I was asked to prove the following theorem:




A topological space if $T_1$ if and only if the following holds:



For any subset $A$ of $X$, $x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$.




I know how to prove the $rightarrow$ direction, but given the equivalence of limit point vs. number of points it intersects with $A$, I cannot think of how we can link this to the $T_1$ condition.






general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is not true in general. Take $X$ a finite set with the discrete topology or more generally a space $X$ with an isolated point $x$.
    $endgroup$
    – yamete kudasai
    Dec 16 '18 at 1:09










  • $begingroup$
    This result is from a Wikipedia article: en.wikipedia.org/wiki/T1_space , at the point where it talks about the equivalent formulation of $T_1$ condition.
    $endgroup$
    – William Sun
    Dec 16 '18 at 1:12






  • 1




    $begingroup$
    @HeroKenzan A limit point can't be isolated ...
    $endgroup$
    – Noah Schweber
    Dec 16 '18 at 1:12










  • $begingroup$
    Suppose $xne y$. Are $x$ and $y$ limit points of the set $A={x,y}$?
    $endgroup$
    – bof
    Dec 16 '18 at 1:23






  • 1




    $begingroup$
    Note that, if $A={x,y}$, the point $x$ certainly does not satisfy the condition "every neighborhood of $x$ contains infinitely many points of $A$". Therefore, if the topology is such that "$x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$," then we can conclude that $x$ is not a limit point of the set ${x,y}$. What does that tell you about the topology?
    $endgroup$
    – bof
    Dec 16 '18 at 1:31


















0












$begingroup$


I was asked to prove the following theorem:




A topological space if $T_1$ if and only if the following holds:



For any subset $A$ of $X$, $x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$.




I know how to prove the $rightarrow$ direction, but given the equivalence of limit point vs. number of points it intersects with $A$, I cannot think of how we can link this to the $T_1$ condition.






general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is not true in general. Take $X$ a finite set with the discrete topology or more generally a space $X$ with an isolated point $x$.
    $endgroup$
    – yamete kudasai
    Dec 16 '18 at 1:09










  • $begingroup$
    This result is from a Wikipedia article: en.wikipedia.org/wiki/T1_space , at the point where it talks about the equivalent formulation of $T_1$ condition.
    $endgroup$
    – William Sun
    Dec 16 '18 at 1:12






  • 1




    $begingroup$
    @HeroKenzan A limit point can't be isolated ...
    $endgroup$
    – Noah Schweber
    Dec 16 '18 at 1:12










  • $begingroup$
    Suppose $xne y$. Are $x$ and $y$ limit points of the set $A={x,y}$?
    $endgroup$
    – bof
    Dec 16 '18 at 1:23






  • 1




    $begingroup$
    Note that, if $A={x,y}$, the point $x$ certainly does not satisfy the condition "every neighborhood of $x$ contains infinitely many points of $A$". Therefore, if the topology is such that "$x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$," then we can conclude that $x$ is not a limit point of the set ${x,y}$. What does that tell you about the topology?
    $endgroup$
    – bof
    Dec 16 '18 at 1:31
















0












0








0





$begingroup$


I was asked to prove the following theorem:




A topological space if $T_1$ if and only if the following holds:



For any subset $A$ of $X$, $x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$.




I know how to prove the $rightarrow$ direction, but given the equivalence of limit point vs. number of points it intersects with $A$, I cannot think of how we can link this to the $T_1$ condition.






general-topology







share|cite|improve this question











$endgroup$




I was asked to prove the following theorem:




A topological space if $T_1$ if and only if the following holds:



For any subset $A$ of $X$, $x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$.




I know how to prove the $rightarrow$ direction, but given the equivalence of limit point vs. number of points it intersects with $A$, I cannot think of how we can link this to the $T_1$ condition.






general-topology




general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 1:37









bof

52k558121




52k558121










asked Dec 16 '18 at 1:00









William SunWilliam Sun

471211




471211












  • $begingroup$
    This is not true in general. Take $X$ a finite set with the discrete topology or more generally a space $X$ with an isolated point $x$.
    $endgroup$
    – yamete kudasai
    Dec 16 '18 at 1:09










  • $begingroup$
    This result is from a Wikipedia article: en.wikipedia.org/wiki/T1_space , at the point where it talks about the equivalent formulation of $T_1$ condition.
    $endgroup$
    – William Sun
    Dec 16 '18 at 1:12






  • 1




    $begingroup$
    @HeroKenzan A limit point can't be isolated ...
    $endgroup$
    – Noah Schweber
    Dec 16 '18 at 1:12










  • $begingroup$
    Suppose $xne y$. Are $x$ and $y$ limit points of the set $A={x,y}$?
    $endgroup$
    – bof
    Dec 16 '18 at 1:23






  • 1




    $begingroup$
    Note that, if $A={x,y}$, the point $x$ certainly does not satisfy the condition "every neighborhood of $x$ contains infinitely many points of $A$". Therefore, if the topology is such that "$x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$," then we can conclude that $x$ is not a limit point of the set ${x,y}$. What does that tell you about the topology?
    $endgroup$
    – bof
    Dec 16 '18 at 1:31




















  • $begingroup$
    This is not true in general. Take $X$ a finite set with the discrete topology or more generally a space $X$ with an isolated point $x$.
    $endgroup$
    – yamete kudasai
    Dec 16 '18 at 1:09










  • $begingroup$
    This result is from a Wikipedia article: en.wikipedia.org/wiki/T1_space , at the point where it talks about the equivalent formulation of $T_1$ condition.
    $endgroup$
    – William Sun
    Dec 16 '18 at 1:12






  • 1




    $begingroup$
    @HeroKenzan A limit point can't be isolated ...
    $endgroup$
    – Noah Schweber
    Dec 16 '18 at 1:12










  • $begingroup$
    Suppose $xne y$. Are $x$ and $y$ limit points of the set $A={x,y}$?
    $endgroup$
    – bof
    Dec 16 '18 at 1:23






  • 1




    $begingroup$
    Note that, if $A={x,y}$, the point $x$ certainly does not satisfy the condition "every neighborhood of $x$ contains infinitely many points of $A$". Therefore, if the topology is such that "$x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$," then we can conclude that $x$ is not a limit point of the set ${x,y}$. What does that tell you about the topology?
    $endgroup$
    – bof
    Dec 16 '18 at 1:31


















$begingroup$
This is not true in general. Take $X$ a finite set with the discrete topology or more generally a space $X$ with an isolated point $x$.
$endgroup$
– yamete kudasai
Dec 16 '18 at 1:09




$begingroup$
This is not true in general. Take $X$ a finite set with the discrete topology or more generally a space $X$ with an isolated point $x$.
$endgroup$
– yamete kudasai
Dec 16 '18 at 1:09












$begingroup$
This result is from a Wikipedia article: en.wikipedia.org/wiki/T1_space , at the point where it talks about the equivalent formulation of $T_1$ condition.
$endgroup$
– William Sun
Dec 16 '18 at 1:12




$begingroup$
This result is from a Wikipedia article: en.wikipedia.org/wiki/T1_space , at the point where it talks about the equivalent formulation of $T_1$ condition.
$endgroup$
– William Sun
Dec 16 '18 at 1:12




1




1




$begingroup$
@HeroKenzan A limit point can't be isolated ...
$endgroup$
– Noah Schweber
Dec 16 '18 at 1:12




$begingroup$
@HeroKenzan A limit point can't be isolated ...
$endgroup$
– Noah Schweber
Dec 16 '18 at 1:12












$begingroup$
Suppose $xne y$. Are $x$ and $y$ limit points of the set $A={x,y}$?
$endgroup$
– bof
Dec 16 '18 at 1:23




$begingroup$
Suppose $xne y$. Are $x$ and $y$ limit points of the set $A={x,y}$?
$endgroup$
– bof
Dec 16 '18 at 1:23




1




1




$begingroup$
Note that, if $A={x,y}$, the point $x$ certainly does not satisfy the condition "every neighborhood of $x$ contains infinitely many points of $A$". Therefore, if the topology is such that "$x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$," then we can conclude that $x$ is not a limit point of the set ${x,y}$. What does that tell you about the topology?
$endgroup$
– bof
Dec 16 '18 at 1:31






$begingroup$
Note that, if $A={x,y}$, the point $x$ certainly does not satisfy the condition "every neighborhood of $x$ contains infinitely many points of $A$". Therefore, if the topology is such that "$x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$," then we can conclude that $x$ is not a limit point of the set ${x,y}$. What does that tell you about the topology?
$endgroup$
– bof
Dec 16 '18 at 1:31












1 Answer
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Assume that if $Asubseteq X$ and $x$ is a limit point of $A$ then every open neighbourhood $U$ of $x$ contains infinitely points of $A$.



Recall that one of the many equivalent formulations of $X$ being $T_{1}$ is that all singletons in $X$ are closed. Then let $xin X$ and consider the singleton set ${x}$. If ${x}$ is not closed then ${x}$ has some limit point $yneq x$ in $X$. Then every open neighbourhood of $y$ contains infinitely points of ${x}$, but thats clearly impossible. Therefore ${x}$ is closed and hence $X$ is $T_{1}$.






share|cite|improve this answer









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    $begingroup$

    Assume that if $Asubseteq X$ and $x$ is a limit point of $A$ then every open neighbourhood $U$ of $x$ contains infinitely points of $A$.



    Recall that one of the many equivalent formulations of $X$ being $T_{1}$ is that all singletons in $X$ are closed. Then let $xin X$ and consider the singleton set ${x}$. If ${x}$ is not closed then ${x}$ has some limit point $yneq x$ in $X$. Then every open neighbourhood of $y$ contains infinitely points of ${x}$, but thats clearly impossible. Therefore ${x}$ is closed and hence $X$ is $T_{1}$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Assume that if $Asubseteq X$ and $x$ is a limit point of $A$ then every open neighbourhood $U$ of $x$ contains infinitely points of $A$.



      Recall that one of the many equivalent formulations of $X$ being $T_{1}$ is that all singletons in $X$ are closed. Then let $xin X$ and consider the singleton set ${x}$. If ${x}$ is not closed then ${x}$ has some limit point $yneq x$ in $X$. Then every open neighbourhood of $y$ contains infinitely points of ${x}$, but thats clearly impossible. Therefore ${x}$ is closed and hence $X$ is $T_{1}$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Assume that if $Asubseteq X$ and $x$ is a limit point of $A$ then every open neighbourhood $U$ of $x$ contains infinitely points of $A$.



        Recall that one of the many equivalent formulations of $X$ being $T_{1}$ is that all singletons in $X$ are closed. Then let $xin X$ and consider the singleton set ${x}$. If ${x}$ is not closed then ${x}$ has some limit point $yneq x$ in $X$. Then every open neighbourhood of $y$ contains infinitely points of ${x}$, but thats clearly impossible. Therefore ${x}$ is closed and hence $X$ is $T_{1}$.






        share|cite|improve this answer









        $endgroup$



        Assume that if $Asubseteq X$ and $x$ is a limit point of $A$ then every open neighbourhood $U$ of $x$ contains infinitely points of $A$.



        Recall that one of the many equivalent formulations of $X$ being $T_{1}$ is that all singletons in $X$ are closed. Then let $xin X$ and consider the singleton set ${x}$. If ${x}$ is not closed then ${x}$ has some limit point $yneq x$ in $X$. Then every open neighbourhood of $y$ contains infinitely points of ${x}$, but thats clearly impossible. Therefore ${x}$ is closed and hence $X$ is $T_{1}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 2:56









        Robert ThingumRobert Thingum

        7981316




        7981316






























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