Give an example of a closed set that not all its elements are limit points [closed]












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From the definition, a closed set contains all its limit points. But how can a closed set that not all its elements are limit points? My example is $S={1/n}$.






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closed as unclear what you're asking by Brahadeesh, Loreno Heer, Xander Henderson, Lord Shark the Unknown, Nosrati Dec 8 '18 at 10:49


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.















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    Closed in what? In your example, the set $S$ is closed in $(0,infty)$ but not in $[0,infty)$.
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    – MPW
    Dec 7 '18 at 17:21










  • $begingroup$
    You need to adjoint $0$ to your set for it to be closed. Then the example works. $0$ is the only limit point of $S$, so since closed if and only if contains all limit point you want $0in S.$
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    – Melody
    Dec 7 '18 at 17:21


















-1












$begingroup$


From the definition, a closed set contains all its limit points. But how can a closed set that not all its elements are limit points? My example is $S={1/n}$.






real-analysis







share|cite|improve this question











$endgroup$



closed as unclear what you're asking by Brahadeesh, Loreno Heer, Xander Henderson, Lord Shark the Unknown, Nosrati Dec 8 '18 at 10:49


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    Closed in what? In your example, the set $S$ is closed in $(0,infty)$ but not in $[0,infty)$.
    $endgroup$
    – MPW
    Dec 7 '18 at 17:21










  • $begingroup$
    You need to adjoint $0$ to your set for it to be closed. Then the example works. $0$ is the only limit point of $S$, so since closed if and only if contains all limit point you want $0in S.$
    $endgroup$
    – Melody
    Dec 7 '18 at 17:21
















-1












-1








-1





$begingroup$


From the definition, a closed set contains all its limit points. But how can a closed set that not all its elements are limit points? My example is $S={1/n}$.






real-analysis







share|cite|improve this question











$endgroup$




From the definition, a closed set contains all its limit points. But how can a closed set that not all its elements are limit points? My example is $S={1/n}$.






real-analysis




real-analysis






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share|cite|improve this question













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edited Dec 7 '18 at 17:35









klirk

2,619530




2,619530










asked Dec 7 '18 at 17:16









Zero to HeroZero to Hero

61




61




closed as unclear what you're asking by Brahadeesh, Loreno Heer, Xander Henderson, Lord Shark the Unknown, Nosrati Dec 8 '18 at 10:49


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






closed as unclear what you're asking by Brahadeesh, Loreno Heer, Xander Henderson, Lord Shark the Unknown, Nosrati Dec 8 '18 at 10:49


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • $begingroup$
    Closed in what? In your example, the set $S$ is closed in $(0,infty)$ but not in $[0,infty)$.
    $endgroup$
    – MPW
    Dec 7 '18 at 17:21










  • $begingroup$
    You need to adjoint $0$ to your set for it to be closed. Then the example works. $0$ is the only limit point of $S$, so since closed if and only if contains all limit point you want $0in S.$
    $endgroup$
    – Melody
    Dec 7 '18 at 17:21




















  • $begingroup$
    Closed in what? In your example, the set $S$ is closed in $(0,infty)$ but not in $[0,infty)$.
    $endgroup$
    – MPW
    Dec 7 '18 at 17:21










  • $begingroup$
    You need to adjoint $0$ to your set for it to be closed. Then the example works. $0$ is the only limit point of $S$, so since closed if and only if contains all limit point you want $0in S.$
    $endgroup$
    – Melody
    Dec 7 '18 at 17:21


















$begingroup$
Closed in what? In your example, the set $S$ is closed in $(0,infty)$ but not in $[0,infty)$.
$endgroup$
– MPW
Dec 7 '18 at 17:21




$begingroup$
Closed in what? In your example, the set $S$ is closed in $(0,infty)$ but not in $[0,infty)$.
$endgroup$
– MPW
Dec 7 '18 at 17:21












$begingroup$
You need to adjoint $0$ to your set for it to be closed. Then the example works. $0$ is the only limit point of $S$, so since closed if and only if contains all limit point you want $0in S.$
$endgroup$
– Melody
Dec 7 '18 at 17:21






$begingroup$
You need to adjoint $0$ to your set for it to be closed. Then the example works. $0$ is the only limit point of $S$, so since closed if and only if contains all limit point you want $0in S.$
$endgroup$
– Melody
Dec 7 '18 at 17:21












3 Answers
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Your example does not work, if I understand it correctly (I presume you actually mean $S={{1over n}: ninmathbb{N}}$): $0$ is a limit point of $S$, but isn't in $S$, so $S$ is not closed. If we add $0$ to $S$, we get something that works *(since e.g. ${1over 42}$ is in the set but not a limit point of the set - indeed, every element of this set, other than $0$, is a non-limit-point element!)*, but in fact we can do much simpler: ${17}$, for example, is closed (in the usual topology on $mathbb{R}$) and has no limit points.






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    $begingroup$

    $S={frac 1n|n in mathbb N}$ is not closed. It has a limit point $0$ and $0 not in S$. But $Scup {0}$ is a fine example. So is $mathbb N$ itself. Or a set with singletons: ${1,3}$. Or a combination: $[0,1]cup {2}$.



    "But how can a closed set that not all its elements are limit points?"



    A closed set contains all of its limit points; but that doesn't mean that the limit points are the only points it has.



    Consider ${frac 1n}cup {0}$ It has one limit point: $0$ and $0$ is in the set. So all of its limit points are in the set. So it is closed. All the rest of the points that aren't limit points? They have nothing to do with anything.



    Note. If a set doesn't have any limit points at all, such as $mathbb N$ then is is closed by default. It doesn't have any limit points so all of its limits points are in the set (because there aren't any).






    share|cite|improve this answer









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      0












      $begingroup$

      S is not a closed set, but you can use $Scup {0}$. Or simply a finite set for instance ${1,2}$.






      share|cite|improve this answer









      $endgroup$




















        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        Your example does not work, if I understand it correctly (I presume you actually mean $S={{1over n}: ninmathbb{N}}$): $0$ is a limit point of $S$, but isn't in $S$, so $S$ is not closed. If we add $0$ to $S$, we get something that works *(since e.g. ${1over 42}$ is in the set but not a limit point of the set - indeed, every element of this set, other than $0$, is a non-limit-point element!)*, but in fact we can do much simpler: ${17}$, for example, is closed (in the usual topology on $mathbb{R}$) and has no limit points.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          Your example does not work, if I understand it correctly (I presume you actually mean $S={{1over n}: ninmathbb{N}}$): $0$ is a limit point of $S$, but isn't in $S$, so $S$ is not closed. If we add $0$ to $S$, we get something that works *(since e.g. ${1over 42}$ is in the set but not a limit point of the set - indeed, every element of this set, other than $0$, is a non-limit-point element!)*, but in fact we can do much simpler: ${17}$, for example, is closed (in the usual topology on $mathbb{R}$) and has no limit points.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            Your example does not work, if I understand it correctly (I presume you actually mean $S={{1over n}: ninmathbb{N}}$): $0$ is a limit point of $S$, but isn't in $S$, so $S$ is not closed. If we add $0$ to $S$, we get something that works *(since e.g. ${1over 42}$ is in the set but not a limit point of the set - indeed, every element of this set, other than $0$, is a non-limit-point element!)*, but in fact we can do much simpler: ${17}$, for example, is closed (in the usual topology on $mathbb{R}$) and has no limit points.






            share|cite|improve this answer









            $endgroup$



            Your example does not work, if I understand it correctly (I presume you actually mean $S={{1over n}: ninmathbb{N}}$): $0$ is a limit point of $S$, but isn't in $S$, so $S$ is not closed. If we add $0$ to $S$, we get something that works *(since e.g. ${1over 42}$ is in the set but not a limit point of the set - indeed, every element of this set, other than $0$, is a non-limit-point element!)*, but in fact we can do much simpler: ${17}$, for example, is closed (in the usual topology on $mathbb{R}$) and has no limit points.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 7 '18 at 17:21









            Noah SchweberNoah Schweber

            122k10149284




            122k10149284























                1












                $begingroup$

                $S={frac 1n|n in mathbb N}$ is not closed. It has a limit point $0$ and $0 not in S$. But $Scup {0}$ is a fine example. So is $mathbb N$ itself. Or a set with singletons: ${1,3}$. Or a combination: $[0,1]cup {2}$.



                "But how can a closed set that not all its elements are limit points?"



                A closed set contains all of its limit points; but that doesn't mean that the limit points are the only points it has.



                Consider ${frac 1n}cup {0}$ It has one limit point: $0$ and $0$ is in the set. So all of its limit points are in the set. So it is closed. All the rest of the points that aren't limit points? They have nothing to do with anything.



                Note. If a set doesn't have any limit points at all, such as $mathbb N$ then is is closed by default. It doesn't have any limit points so all of its limits points are in the set (because there aren't any).






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  $S={frac 1n|n in mathbb N}$ is not closed. It has a limit point $0$ and $0 not in S$. But $Scup {0}$ is a fine example. So is $mathbb N$ itself. Or a set with singletons: ${1,3}$. Or a combination: $[0,1]cup {2}$.



                  "But how can a closed set that not all its elements are limit points?"



                  A closed set contains all of its limit points; but that doesn't mean that the limit points are the only points it has.



                  Consider ${frac 1n}cup {0}$ It has one limit point: $0$ and $0$ is in the set. So all of its limit points are in the set. So it is closed. All the rest of the points that aren't limit points? They have nothing to do with anything.



                  Note. If a set doesn't have any limit points at all, such as $mathbb N$ then is is closed by default. It doesn't have any limit points so all of its limits points are in the set (because there aren't any).






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    $S={frac 1n|n in mathbb N}$ is not closed. It has a limit point $0$ and $0 not in S$. But $Scup {0}$ is a fine example. So is $mathbb N$ itself. Or a set with singletons: ${1,3}$. Or a combination: $[0,1]cup {2}$.



                    "But how can a closed set that not all its elements are limit points?"



                    A closed set contains all of its limit points; but that doesn't mean that the limit points are the only points it has.



                    Consider ${frac 1n}cup {0}$ It has one limit point: $0$ and $0$ is in the set. So all of its limit points are in the set. So it is closed. All the rest of the points that aren't limit points? They have nothing to do with anything.



                    Note. If a set doesn't have any limit points at all, such as $mathbb N$ then is is closed by default. It doesn't have any limit points so all of its limits points are in the set (because there aren't any).






                    share|cite|improve this answer









                    $endgroup$



                    $S={frac 1n|n in mathbb N}$ is not closed. It has a limit point $0$ and $0 not in S$. But $Scup {0}$ is a fine example. So is $mathbb N$ itself. Or a set with singletons: ${1,3}$. Or a combination: $[0,1]cup {2}$.



                    "But how can a closed set that not all its elements are limit points?"



                    A closed set contains all of its limit points; but that doesn't mean that the limit points are the only points it has.



                    Consider ${frac 1n}cup {0}$ It has one limit point: $0$ and $0$ is in the set. So all of its limit points are in the set. So it is closed. All the rest of the points that aren't limit points? They have nothing to do with anything.



                    Note. If a set doesn't have any limit points at all, such as $mathbb N$ then is is closed by default. It doesn't have any limit points so all of its limits points are in the set (because there aren't any).







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 7 '18 at 17:26









                    fleabloodfleablood

                    68.9k22685




                    68.9k22685























                        0












                        $begingroup$

                        S is not a closed set, but you can use $Scup {0}$. Or simply a finite set for instance ${1,2}$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          S is not a closed set, but you can use $Scup {0}$. Or simply a finite set for instance ${1,2}$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            S is not a closed set, but you can use $Scup {0}$. Or simply a finite set for instance ${1,2}$.






                            share|cite|improve this answer









                            $endgroup$



                            S is not a closed set, but you can use $Scup {0}$. Or simply a finite set for instance ${1,2}$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 7 '18 at 17:24









                            Gil AstudilloGil Astudillo

                            315




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