Confused about open and closed set in $mathbb{R}^2$












0












$begingroup$


While this question is basically meant for all $mathbb{R}^n$ where $r > 1$, I suppose it would be the easiest to grasp the concept in $mathbb{R}^2$ and conclude the analog to $mathbb{R}^2$.



So I was wondering since we just started out with Topology, how to find out if a set is closed or open in $mathbb{R}^2$.



Let's say I have a set $A subset mathbb{R}$ defined as:



$$A=(0,1) times [0,1]$$



Since I now have a set "composed" by an open and a closed set I wonder if $A$ is now called open or closed.



Or is this a case where it neither is open nor closed?






general-topology







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It is neither open nor closed. This is true of most subsets in a sense.
    $endgroup$
    – Matt Samuel
    Dec 15 '18 at 12:53
















0












$begingroup$


While this question is basically meant for all $mathbb{R}^n$ where $r > 1$, I suppose it would be the easiest to grasp the concept in $mathbb{R}^2$ and conclude the analog to $mathbb{R}^2$.



So I was wondering since we just started out with Topology, how to find out if a set is closed or open in $mathbb{R}^2$.



Let's say I have a set $A subset mathbb{R}$ defined as:



$$A=(0,1) times [0,1]$$



Since I now have a set "composed" by an open and a closed set I wonder if $A$ is now called open or closed.



Or is this a case where it neither is open nor closed?






general-topology







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It is neither open nor closed. This is true of most subsets in a sense.
    $endgroup$
    – Matt Samuel
    Dec 15 '18 at 12:53














0












0








0





$begingroup$


While this question is basically meant for all $mathbb{R}^n$ where $r > 1$, I suppose it would be the easiest to grasp the concept in $mathbb{R}^2$ and conclude the analog to $mathbb{R}^2$.



So I was wondering since we just started out with Topology, how to find out if a set is closed or open in $mathbb{R}^2$.



Let's say I have a set $A subset mathbb{R}$ defined as:



$$A=(0,1) times [0,1]$$



Since I now have a set "composed" by an open and a closed set I wonder if $A$ is now called open or closed.



Or is this a case where it neither is open nor closed?






general-topology







share|cite|improve this question











$endgroup$




While this question is basically meant for all $mathbb{R}^n$ where $r > 1$, I suppose it would be the easiest to grasp the concept in $mathbb{R}^2$ and conclude the analog to $mathbb{R}^2$.



So I was wondering since we just started out with Topology, how to find out if a set is closed or open in $mathbb{R}^2$.



Let's say I have a set $A subset mathbb{R}$ defined as:



$$A=(0,1) times [0,1]$$



Since I now have a set "composed" by an open and a closed set I wonder if $A$ is now called open or closed.



Or is this a case where it neither is open nor closed?






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 15 '18 at 13:08









Andrés E. Caicedo

65.4k8158249




65.4k8158249










asked Dec 15 '18 at 12:51









MLKMLK

80112




80112








  • 1




    $begingroup$
    It is neither open nor closed. This is true of most subsets in a sense.
    $endgroup$
    – Matt Samuel
    Dec 15 '18 at 12:53














  • 1




    $begingroup$
    It is neither open nor closed. This is true of most subsets in a sense.
    $endgroup$
    – Matt Samuel
    Dec 15 '18 at 12:53








1




1




$begingroup$
It is neither open nor closed. This is true of most subsets in a sense.
$endgroup$
– Matt Samuel
Dec 15 '18 at 12:53




$begingroup$
It is neither open nor closed. This is true of most subsets in a sense.
$endgroup$
– Matt Samuel
Dec 15 '18 at 12:53










2 Answers
2






active

oldest

votes


















1












$begingroup$


  • Since $(0,0)notin A$ but every open ball centered at $(0,0)$ contains points of $A$, $A$ is not closed.

  • Since $left(frac12,0right)in A$ but every open ball centered at $left(frac12,0right)in A$ contains points out of $A$, $A$ is not open.


Therefore, $A$ is neither closed nor open.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Interesting. Thank you. So basically for every $mathbb{R}^n$ one just has to look at every set that is making up the set $A$ and if one is open while one is closed it is neither the one nor the other
    $endgroup$
    – MLK
    Dec 15 '18 at 12:58








  • 1




    $begingroup$
    Yes, but that's an almost useless piece of information, since very few subsets of $mathbb{R}^2$ can be expressed as the product of two subsets of $mathbb R$.
    $endgroup$
    – José Carlos Santos
    Dec 15 '18 at 13:45



















1












$begingroup$

This set $(0,1) times [0,1]$ is not closed and not open.



Not closed e.g. because $(frac{1}{n}, frac{1}{n})$ is a sequence in it conervging to $(0,0) notin A$. Not open e.g. because $(frac{1}{2}, -frac{1}{n})$ is a sequence converging to $(frac12,0)$ in $A$ but all of whose points lie outside $A$.



Another way: projections are open maps so if $A$ were open, so would $pi_2[A]= [0,1]$ be in $mathbb{R}$, which it is not. If $A$ were closed, it would be compact (closed and bounded in $mathbb{R}^2$) but $pi_1[A] = (0,1)$ is not compact (not even 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









    1












    $begingroup$


    • Since $(0,0)notin A$ but every open ball centered at $(0,0)$ contains points of $A$, $A$ is not closed.

    • Since $left(frac12,0right)in A$ but every open ball centered at $left(frac12,0right)in A$ contains points out of $A$, $A$ is not open.


    Therefore, $A$ is neither closed nor open.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Interesting. Thank you. So basically for every $mathbb{R}^n$ one just has to look at every set that is making up the set $A$ and if one is open while one is closed it is neither the one nor the other
      $endgroup$
      – MLK
      Dec 15 '18 at 12:58








    • 1




      $begingroup$
      Yes, but that's an almost useless piece of information, since very few subsets of $mathbb{R}^2$ can be expressed as the product of two subsets of $mathbb R$.
      $endgroup$
      – José Carlos Santos
      Dec 15 '18 at 13:45
















    1












    $begingroup$


    • Since $(0,0)notin A$ but every open ball centered at $(0,0)$ contains points of $A$, $A$ is not closed.

    • Since $left(frac12,0right)in A$ but every open ball centered at $left(frac12,0right)in A$ contains points out of $A$, $A$ is not open.


    Therefore, $A$ is neither closed nor open.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Interesting. Thank you. So basically for every $mathbb{R}^n$ one just has to look at every set that is making up the set $A$ and if one is open while one is closed it is neither the one nor the other
      $endgroup$
      – MLK
      Dec 15 '18 at 12:58








    • 1




      $begingroup$
      Yes, but that's an almost useless piece of information, since very few subsets of $mathbb{R}^2$ can be expressed as the product of two subsets of $mathbb R$.
      $endgroup$
      – José Carlos Santos
      Dec 15 '18 at 13:45














    1












    1








    1





    $begingroup$


    • Since $(0,0)notin A$ but every open ball centered at $(0,0)$ contains points of $A$, $A$ is not closed.

    • Since $left(frac12,0right)in A$ but every open ball centered at $left(frac12,0right)in A$ contains points out of $A$, $A$ is not open.


    Therefore, $A$ is neither closed nor open.






    share|cite|improve this answer









    $endgroup$




    • Since $(0,0)notin A$ but every open ball centered at $(0,0)$ contains points of $A$, $A$ is not closed.

    • Since $left(frac12,0right)in A$ but every open ball centered at $left(frac12,0right)in A$ contains points out of $A$, $A$ is not open.


    Therefore, $A$ is neither closed nor open.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 15 '18 at 12:55









    José Carlos SantosJosé Carlos Santos

    160k22127232




    160k22127232












    • $begingroup$
      Interesting. Thank you. So basically for every $mathbb{R}^n$ one just has to look at every set that is making up the set $A$ and if one is open while one is closed it is neither the one nor the other
      $endgroup$
      – MLK
      Dec 15 '18 at 12:58








    • 1




      $begingroup$
      Yes, but that's an almost useless piece of information, since very few subsets of $mathbb{R}^2$ can be expressed as the product of two subsets of $mathbb R$.
      $endgroup$
      – José Carlos Santos
      Dec 15 '18 at 13:45


















    • $begingroup$
      Interesting. Thank you. So basically for every $mathbb{R}^n$ one just has to look at every set that is making up the set $A$ and if one is open while one is closed it is neither the one nor the other
      $endgroup$
      – MLK
      Dec 15 '18 at 12:58








    • 1




      $begingroup$
      Yes, but that's an almost useless piece of information, since very few subsets of $mathbb{R}^2$ can be expressed as the product of two subsets of $mathbb R$.
      $endgroup$
      – José Carlos Santos
      Dec 15 '18 at 13:45
















    $begingroup$
    Interesting. Thank you. So basically for every $mathbb{R}^n$ one just has to look at every set that is making up the set $A$ and if one is open while one is closed it is neither the one nor the other
    $endgroup$
    – MLK
    Dec 15 '18 at 12:58






    $begingroup$
    Interesting. Thank you. So basically for every $mathbb{R}^n$ one just has to look at every set that is making up the set $A$ and if one is open while one is closed it is neither the one nor the other
    $endgroup$
    – MLK
    Dec 15 '18 at 12:58






    1




    1




    $begingroup$
    Yes, but that's an almost useless piece of information, since very few subsets of $mathbb{R}^2$ can be expressed as the product of two subsets of $mathbb R$.
    $endgroup$
    – José Carlos Santos
    Dec 15 '18 at 13:45




    $begingroup$
    Yes, but that's an almost useless piece of information, since very few subsets of $mathbb{R}^2$ can be expressed as the product of two subsets of $mathbb R$.
    $endgroup$
    – José Carlos Santos
    Dec 15 '18 at 13:45











    1












    $begingroup$

    This set $(0,1) times [0,1]$ is not closed and not open.



    Not closed e.g. because $(frac{1}{n}, frac{1}{n})$ is a sequence in it conervging to $(0,0) notin A$. Not open e.g. because $(frac{1}{2}, -frac{1}{n})$ is a sequence converging to $(frac12,0)$ in $A$ but all of whose points lie outside $A$.



    Another way: projections are open maps so if $A$ were open, so would $pi_2[A]= [0,1]$ be in $mathbb{R}$, which it is not. If $A$ were closed, it would be compact (closed and bounded in $mathbb{R}^2$) but $pi_1[A] = (0,1)$ is not compact (not even closed).






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      This set $(0,1) times [0,1]$ is not closed and not open.



      Not closed e.g. because $(frac{1}{n}, frac{1}{n})$ is a sequence in it conervging to $(0,0) notin A$. Not open e.g. because $(frac{1}{2}, -frac{1}{n})$ is a sequence converging to $(frac12,0)$ in $A$ but all of whose points lie outside $A$.



      Another way: projections are open maps so if $A$ were open, so would $pi_2[A]= [0,1]$ be in $mathbb{R}$, which it is not. If $A$ were closed, it would be compact (closed and bounded in $mathbb{R}^2$) but $pi_1[A] = (0,1)$ is not compact (not even closed).






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        This set $(0,1) times [0,1]$ is not closed and not open.



        Not closed e.g. because $(frac{1}{n}, frac{1}{n})$ is a sequence in it conervging to $(0,0) notin A$. Not open e.g. because $(frac{1}{2}, -frac{1}{n})$ is a sequence converging to $(frac12,0)$ in $A$ but all of whose points lie outside $A$.



        Another way: projections are open maps so if $A$ were open, so would $pi_2[A]= [0,1]$ be in $mathbb{R}$, which it is not. If $A$ were closed, it would be compact (closed and bounded in $mathbb{R}^2$) but $pi_1[A] = (0,1)$ is not compact (not even closed).






        share|cite|improve this answer









        $endgroup$



        This set $(0,1) times [0,1]$ is not closed and not open.



        Not closed e.g. because $(frac{1}{n}, frac{1}{n})$ is a sequence in it conervging to $(0,0) notin A$. Not open e.g. because $(frac{1}{2}, -frac{1}{n})$ is a sequence converging to $(frac12,0)$ in $A$ but all of whose points lie outside $A$.



        Another way: projections are open maps so if $A$ were open, so would $pi_2[A]= [0,1]$ be in $mathbb{R}$, which it is not. If $A$ were closed, it would be compact (closed and bounded in $mathbb{R}^2$) but $pi_1[A] = (0,1)$ is not compact (not even closed).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 15 '18 at 12:58









        Henno BrandsmaHenno Brandsma

        109k347114




        109k347114






























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