Baby Rudin Definition 2.18











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This are few Definitions stated in Principles of Mathematical Analysis by Walter Rudin:




Now, Consider the interval (1,2).



This interval is closed because according to $(b)$ the every point of this set contains another point in its neighbourhood, thus a limit point, hence closed.



Open at the same time according to the $(e), (f)$ given in the definition.



Please clarify.






general-topology metric-spaces







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  • 4




    Please do not delete questions after having gotten an answer.
    – quid♦
    2 days ago










  • There is no problem in itself with a set being open and closed. That can actually happen too.
    – Henno Brandsma
    2 days ago










  • ... also, sets can be open and closed at the same time.
    – AccidentalFourierTransform
    2 days ago










  • @AccidentalFourierTransform can you please give one example except $mathbb{R}$?
    – HindShah
    2 days ago










  • In the metric space $[0,1] cup {2}$ with the inherited metric from the reals, ${2}$ is open and closed, e.g. $N_1(2) = {2}$ so it's open, and ${2}$ has no limit points, so it's closed.
    – Henno Brandsma
    yesterday















up vote
0
down vote

favorite
1












This are few Definitions stated in Principles of Mathematical Analysis by Walter Rudin:




Now, Consider the interval (1,2).



This interval is closed because according to $(b)$ the every point of this set contains another point in its neighbourhood, thus a limit point, hence closed.



Open at the same time according to the $(e), (f)$ given in the definition.



Please clarify.






general-topology metric-spaces







share|cite|improve this question









New contributor




HindShah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 4




    Please do not delete questions after having gotten an answer.
    – quid♦
    2 days ago










  • There is no problem in itself with a set being open and closed. That can actually happen too.
    – Henno Brandsma
    2 days ago










  • ... also, sets can be open and closed at the same time.
    – AccidentalFourierTransform
    2 days ago










  • @AccidentalFourierTransform can you please give one example except $mathbb{R}$?
    – HindShah
    2 days ago










  • In the metric space $[0,1] cup {2}$ with the inherited metric from the reals, ${2}$ is open and closed, e.g. $N_1(2) = {2}$ so it's open, and ${2}$ has no limit points, so it's closed.
    – Henno Brandsma
    yesterday













up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





This are few Definitions stated in Principles of Mathematical Analysis by Walter Rudin:




Now, Consider the interval (1,2).



This interval is closed because according to $(b)$ the every point of this set contains another point in its neighbourhood, thus a limit point, hence closed.



Open at the same time according to the $(e), (f)$ given in the definition.



Please clarify.






general-topology metric-spaces







share|cite|improve this question









New contributor




HindShah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











This are few Definitions stated in Principles of Mathematical Analysis by Walter Rudin:




Now, Consider the interval (1,2).



This interval is closed because according to $(b)$ the every point of this set contains another point in its neighbourhood, thus a limit point, hence closed.



Open at the same time according to the $(e), (f)$ given in the definition.



Please clarify.






general-topology metric-spaces




general-topology metric-spaces






share|cite|improve this question









New contributor




HindShah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




HindShah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Henno Brandsma

101k344107




101k344107






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asked 2 days ago









HindShah

74




74




New contributor




HindShah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





HindShah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






HindShah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 4




    Please do not delete questions after having gotten an answer.
    – quid♦
    2 days ago










  • There is no problem in itself with a set being open and closed. That can actually happen too.
    – Henno Brandsma
    2 days ago










  • ... also, sets can be open and closed at the same time.
    – AccidentalFourierTransform
    2 days ago










  • @AccidentalFourierTransform can you please give one example except $mathbb{R}$?
    – HindShah
    2 days ago










  • In the metric space $[0,1] cup {2}$ with the inherited metric from the reals, ${2}$ is open and closed, e.g. $N_1(2) = {2}$ so it's open, and ${2}$ has no limit points, so it's closed.
    – Henno Brandsma
    yesterday














  • 4




    Please do not delete questions after having gotten an answer.
    – quid♦
    2 days ago










  • There is no problem in itself with a set being open and closed. That can actually happen too.
    – Henno Brandsma
    2 days ago










  • ... also, sets can be open and closed at the same time.
    – AccidentalFourierTransform
    2 days ago










  • @AccidentalFourierTransform can you please give one example except $mathbb{R}$?
    – HindShah
    2 days ago










  • In the metric space $[0,1] cup {2}$ with the inherited metric from the reals, ${2}$ is open and closed, e.g. $N_1(2) = {2}$ so it's open, and ${2}$ has no limit points, so it's closed.
    – Henno Brandsma
    yesterday








4




4




Please do not delete questions after having gotten an answer.
– quid♦
2 days ago




Please do not delete questions after having gotten an answer.
– quid♦
2 days ago












There is no problem in itself with a set being open and closed. That can actually happen too.
– Henno Brandsma
2 days ago




There is no problem in itself with a set being open and closed. That can actually happen too.
– Henno Brandsma
2 days ago












... also, sets can be open and closed at the same time.
– AccidentalFourierTransform
2 days ago




... also, sets can be open and closed at the same time.
– AccidentalFourierTransform
2 days ago












@AccidentalFourierTransform can you please give one example except $mathbb{R}$?
– HindShah
2 days ago




@AccidentalFourierTransform can you please give one example except $mathbb{R}$?
– HindShah
2 days ago












In the metric space $[0,1] cup {2}$ with the inherited metric from the reals, ${2}$ is open and closed, e.g. $N_1(2) = {2}$ so it's open, and ${2}$ has no limit points, so it's closed.
– Henno Brandsma
yesterday




In the metric space $[0,1] cup {2}$ with the inherited metric from the reals, ${2}$ is open and closed, e.g. $N_1(2) = {2}$ so it's open, and ${2}$ has no limit points, so it's closed.
– Henno Brandsma
yesterday










2 Answers
2






active

oldest

votes

















up vote
3
down vote



accepted










There is a difference between




Every limit point of $E$ is a point of $E$ . . . . .$(1)$




and




Every point of $E$ is a limit point of $E$ . . . . . .$(2)$




You have reasoned that $(2)$ is true and concluded that $E$ is closed. But this is incorrect, because you need to check whether $(1)$ is true, since $E$ is defined to be closed if $(1)$ holds.



And indeed, $1$ is a limit point of $E = (1,2)$ but $1 notin E$, so $E$ is not closed.






share|cite|improve this answer























  • Should $p$ be in the set $E$ for the definition $(b)$ ?
    – HindShah
    2 days ago










  • @HindShah no, it is not necessary that $p$ should be in the set $E$ in the definition (b). The definition does not assume that $p in E$.
    – Brahadeesh
    2 days ago










  • @HindShah without additional assumptions in the statement (b), we just assume that “all points and sets are understood to be elements and subsets of $X$”.
    – Brahadeesh
    2 days ago


















up vote
3
down vote













$1$ and $2$ are limit points of $E= (1,2)$ but are not points of $E$. So $E$ is not closed.






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    There is a difference between




    Every limit point of $E$ is a point of $E$ . . . . .$(1)$




    and




    Every point of $E$ is a limit point of $E$ . . . . . .$(2)$




    You have reasoned that $(2)$ is true and concluded that $E$ is closed. But this is incorrect, because you need to check whether $(1)$ is true, since $E$ is defined to be closed if $(1)$ holds.



    And indeed, $1$ is a limit point of $E = (1,2)$ but $1 notin E$, so $E$ is not closed.






    share|cite|improve this answer























    • Should $p$ be in the set $E$ for the definition $(b)$ ?
      – HindShah
      2 days ago










    • @HindShah no, it is not necessary that $p$ should be in the set $E$ in the definition (b). The definition does not assume that $p in E$.
      – Brahadeesh
      2 days ago










    • @HindShah without additional assumptions in the statement (b), we just assume that “all points and sets are understood to be elements and subsets of $X$”.
      – Brahadeesh
      2 days ago















    up vote
    3
    down vote



    accepted










    There is a difference between




    Every limit point of $E$ is a point of $E$ . . . . .$(1)$




    and




    Every point of $E$ is a limit point of $E$ . . . . . .$(2)$




    You have reasoned that $(2)$ is true and concluded that $E$ is closed. But this is incorrect, because you need to check whether $(1)$ is true, since $E$ is defined to be closed if $(1)$ holds.



    And indeed, $1$ is a limit point of $E = (1,2)$ but $1 notin E$, so $E$ is not closed.






    share|cite|improve this answer























    • Should $p$ be in the set $E$ for the definition $(b)$ ?
      – HindShah
      2 days ago










    • @HindShah no, it is not necessary that $p$ should be in the set $E$ in the definition (b). The definition does not assume that $p in E$.
      – Brahadeesh
      2 days ago










    • @HindShah without additional assumptions in the statement (b), we just assume that “all points and sets are understood to be elements and subsets of $X$”.
      – Brahadeesh
      2 days ago













    up vote
    3
    down vote



    accepted







    up vote
    3
    down vote



    accepted






    There is a difference between




    Every limit point of $E$ is a point of $E$ . . . . .$(1)$




    and




    Every point of $E$ is a limit point of $E$ . . . . . .$(2)$




    You have reasoned that $(2)$ is true and concluded that $E$ is closed. But this is incorrect, because you need to check whether $(1)$ is true, since $E$ is defined to be closed if $(1)$ holds.



    And indeed, $1$ is a limit point of $E = (1,2)$ but $1 notin E$, so $E$ is not closed.






    share|cite|improve this answer














    There is a difference between




    Every limit point of $E$ is a point of $E$ . . . . .$(1)$




    and




    Every point of $E$ is a limit point of $E$ . . . . . .$(2)$




    You have reasoned that $(2)$ is true and concluded that $E$ is closed. But this is incorrect, because you need to check whether $(1)$ is true, since $E$ is defined to be closed if $(1)$ holds.



    And indeed, $1$ is a limit point of $E = (1,2)$ but $1 notin E$, so $E$ is not closed.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 days ago

























    answered 2 days ago









    Brahadeesh

    5,53941956




    5,53941956












    • Should $p$ be in the set $E$ for the definition $(b)$ ?
      – HindShah
      2 days ago










    • @HindShah no, it is not necessary that $p$ should be in the set $E$ in the definition (b). The definition does not assume that $p in E$.
      – Brahadeesh
      2 days ago










    • @HindShah without additional assumptions in the statement (b), we just assume that “all points and sets are understood to be elements and subsets of $X$”.
      – Brahadeesh
      2 days ago


















    • Should $p$ be in the set $E$ for the definition $(b)$ ?
      – HindShah
      2 days ago










    • @HindShah no, it is not necessary that $p$ should be in the set $E$ in the definition (b). The definition does not assume that $p in E$.
      – Brahadeesh
      2 days ago










    • @HindShah without additional assumptions in the statement (b), we just assume that “all points and sets are understood to be elements and subsets of $X$”.
      – Brahadeesh
      2 days ago
















    Should $p$ be in the set $E$ for the definition $(b)$ ?
    – HindShah
    2 days ago




    Should $p$ be in the set $E$ for the definition $(b)$ ?
    – HindShah
    2 days ago












    @HindShah no, it is not necessary that $p$ should be in the set $E$ in the definition (b). The definition does not assume that $p in E$.
    – Brahadeesh
    2 days ago




    @HindShah no, it is not necessary that $p$ should be in the set $E$ in the definition (b). The definition does not assume that $p in E$.
    – Brahadeesh
    2 days ago












    @HindShah without additional assumptions in the statement (b), we just assume that “all points and sets are understood to be elements and subsets of $X$”.
    – Brahadeesh
    2 days ago




    @HindShah without additional assumptions in the statement (b), we just assume that “all points and sets are understood to be elements and subsets of $X$”.
    – Brahadeesh
    2 days ago










    up vote
    3
    down vote













    $1$ and $2$ are limit points of $E= (1,2)$ but are not points of $E$. So $E$ is not closed.






    share|cite|improve this answer

























      up vote
      3
      down vote













      $1$ and $2$ are limit points of $E= (1,2)$ but are not points of $E$. So $E$ is not closed.






      share|cite|improve this answer























        up vote
        3
        down vote










        up vote
        3
        down vote









        $1$ and $2$ are limit points of $E= (1,2)$ but are not points of $E$. So $E$ is not closed.






        share|cite|improve this answer












        $1$ and $2$ are limit points of $E= (1,2)$ but are not points of $E$. So $E$ is not closed.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Henno Brandsma

        101k344107




        101k344107






















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