Is $(0,1]$ a closed or open set?












3












$begingroup$


Is $A=(0,1]$ a closed or open set?



I think it's an open set because it is not a subset of its interior points. Mainly, $1in A$ but $1notin A^circ$.



If A is closed, then the complement is open. However, the complement $A^c$ is not open because it is not a subset of its interior points. Mainly, $0 in A^c$ but $0notin (A^c)^circ$






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    $(0, 1]$ is a semi-open or semi-closed set.
    $endgroup$
    – Paras Khosla
    9 hours ago






  • 5




    $begingroup$
    Depends on the topology!
    $endgroup$
    – Jakobian
    9 hours ago






  • 9




    $begingroup$
    Unlike doors, subsets of topological spaces may be both open and closed, and they may be neither open nor closed. This is an example of how using every day words to name precise mathematical definitions can be misleading.
    $endgroup$
    – Simon
    9 hours ago








  • 1




    $begingroup$
    The answer to your question is no.
    $endgroup$
    – Robert Shore
    9 hours ago






  • 2




    $begingroup$
    @Simon doors can also be open and closed at the same time! You just need them to be adjoint to two different entrances at once.
    $endgroup$
    – John Dvorak
    7 hours ago
















3












$begingroup$


Is $A=(0,1]$ a closed or open set?



I think it's an open set because it is not a subset of its interior points. Mainly, $1in A$ but $1notin A^circ$.



If A is closed, then the complement is open. However, the complement $A^c$ is not open because it is not a subset of its interior points. Mainly, $0 in A^c$ but $0notin (A^c)^circ$






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    $(0, 1]$ is a semi-open or semi-closed set.
    $endgroup$
    – Paras Khosla
    9 hours ago






  • 5




    $begingroup$
    Depends on the topology!
    $endgroup$
    – Jakobian
    9 hours ago






  • 9




    $begingroup$
    Unlike doors, subsets of topological spaces may be both open and closed, and they may be neither open nor closed. This is an example of how using every day words to name precise mathematical definitions can be misleading.
    $endgroup$
    – Simon
    9 hours ago








  • 1




    $begingroup$
    The answer to your question is no.
    $endgroup$
    – Robert Shore
    9 hours ago






  • 2




    $begingroup$
    @Simon doors can also be open and closed at the same time! You just need them to be adjoint to two different entrances at once.
    $endgroup$
    – John Dvorak
    7 hours ago














3












3








3


1



$begingroup$


Is $A=(0,1]$ a closed or open set?



I think it's an open set because it is not a subset of its interior points. Mainly, $1in A$ but $1notin A^circ$.



If A is closed, then the complement is open. However, the complement $A^c$ is not open because it is not a subset of its interior points. Mainly, $0 in A^c$ but $0notin (A^c)^circ$






real-analysis







share|cite|improve this question











$endgroup$




Is $A=(0,1]$ a closed or open set?



I think it's an open set because it is not a subset of its interior points. Mainly, $1in A$ but $1notin A^circ$.



If A is closed, then the complement is open. However, the complement $A^c$ is not open because it is not a subset of its interior points. Mainly, $0 in A^c$ but $0notin (A^c)^circ$






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago









Asaf Karagila♦

307k33438769




307k33438769










asked 9 hours ago









QwertfordQwertford

316212




316212












  • $begingroup$
    $(0, 1]$ is a semi-open or semi-closed set.
    $endgroup$
    – Paras Khosla
    9 hours ago






  • 5




    $begingroup$
    Depends on the topology!
    $endgroup$
    – Jakobian
    9 hours ago






  • 9




    $begingroup$
    Unlike doors, subsets of topological spaces may be both open and closed, and they may be neither open nor closed. This is an example of how using every day words to name precise mathematical definitions can be misleading.
    $endgroup$
    – Simon
    9 hours ago








  • 1




    $begingroup$
    The answer to your question is no.
    $endgroup$
    – Robert Shore
    9 hours ago






  • 2




    $begingroup$
    @Simon doors can also be open and closed at the same time! You just need them to be adjoint to two different entrances at once.
    $endgroup$
    – John Dvorak
    7 hours ago


















  • $begingroup$
    $(0, 1]$ is a semi-open or semi-closed set.
    $endgroup$
    – Paras Khosla
    9 hours ago






  • 5




    $begingroup$
    Depends on the topology!
    $endgroup$
    – Jakobian
    9 hours ago






  • 9




    $begingroup$
    Unlike doors, subsets of topological spaces may be both open and closed, and they may be neither open nor closed. This is an example of how using every day words to name precise mathematical definitions can be misleading.
    $endgroup$
    – Simon
    9 hours ago








  • 1




    $begingroup$
    The answer to your question is no.
    $endgroup$
    – Robert Shore
    9 hours ago






  • 2




    $begingroup$
    @Simon doors can also be open and closed at the same time! You just need them to be adjoint to two different entrances at once.
    $endgroup$
    – John Dvorak
    7 hours ago
















$begingroup$
$(0, 1]$ is a semi-open or semi-closed set.
$endgroup$
– Paras Khosla
9 hours ago




$begingroup$
$(0, 1]$ is a semi-open or semi-closed set.
$endgroup$
– Paras Khosla
9 hours ago




5




5




$begingroup$
Depends on the topology!
$endgroup$
– Jakobian
9 hours ago




$begingroup$
Depends on the topology!
$endgroup$
– Jakobian
9 hours ago




9




9




$begingroup$
Unlike doors, subsets of topological spaces may be both open and closed, and they may be neither open nor closed. This is an example of how using every day words to name precise mathematical definitions can be misleading.
$endgroup$
– Simon
9 hours ago






$begingroup$
Unlike doors, subsets of topological spaces may be both open and closed, and they may be neither open nor closed. This is an example of how using every day words to name precise mathematical definitions can be misleading.
$endgroup$
– Simon
9 hours ago






1




1




$begingroup$
The answer to your question is no.
$endgroup$
– Robert Shore
9 hours ago




$begingroup$
The answer to your question is no.
$endgroup$
– Robert Shore
9 hours ago




2




2




$begingroup$
@Simon doors can also be open and closed at the same time! You just need them to be adjoint to two different entrances at once.
$endgroup$
– John Dvorak
7 hours ago




$begingroup$
@Simon doors can also be open and closed at the same time! You just need them to be adjoint to two different entrances at once.
$endgroup$
– John Dvorak
7 hours ago










3 Answers
3






active

oldest

votes


















9












$begingroup$

Neither. It doesn't contain a neighbourhood of $1$, so it isn't open; nor is its complement, $(-infty,,0]cup (1,,infty)$, which doesn't contain a neighbourhood of $0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    note that "its complement" is only what it is if we assume reals as the topology. $(0, 1]$ is open in itself when using the $D = |x-y|$ metric because the complement is the empty set.
    $endgroup$
    – John Dvorak
    7 hours ago








  • 2




    $begingroup$
    @JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
    $endgroup$
    – J.G.
    7 hours ago



















1












$begingroup$

It's important that you specify where you are considering the subset $A$.



If $A subset X$ with $ X = mathbb{R}$, J.G. is absolutely right in the usual topology of $mathbb{R}$.



If $A subset X$ with $ X = [0,1]$, $A^c = {0} $, which is closed in the usual topology, then $A = (0,1]$ is open.



In other words, it's important to specify in what topologic space $X$ you are considering $A$ as a subset. There are some stranger metrics which may define some different open sets where things can be different.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    A set is not a door.



    It is not the case that a set is either open or closed. It can also be neither or both.



    Indeed, your arguments correctly establish that $(0,1]$ is neither open nor closed as a subset of $mathbb{R}$ with the usual topology. The empty set $emptyset$ is always both open and closed, no matter what the ambient space is.






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      9












      $begingroup$

      Neither. It doesn't contain a neighbourhood of $1$, so it isn't open; nor is its complement, $(-infty,,0]cup (1,,infty)$, which doesn't contain a neighbourhood of $0$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        note that "its complement" is only what it is if we assume reals as the topology. $(0, 1]$ is open in itself when using the $D = |x-y|$ metric because the complement is the empty set.
        $endgroup$
        – John Dvorak
        7 hours ago








      • 2




        $begingroup$
        @JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
        $endgroup$
        – J.G.
        7 hours ago
















      9












      $begingroup$

      Neither. It doesn't contain a neighbourhood of $1$, so it isn't open; nor is its complement, $(-infty,,0]cup (1,,infty)$, which doesn't contain a neighbourhood of $0$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        note that "its complement" is only what it is if we assume reals as the topology. $(0, 1]$ is open in itself when using the $D = |x-y|$ metric because the complement is the empty set.
        $endgroup$
        – John Dvorak
        7 hours ago








      • 2




        $begingroup$
        @JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
        $endgroup$
        – J.G.
        7 hours ago














      9












      9








      9





      $begingroup$

      Neither. It doesn't contain a neighbourhood of $1$, so it isn't open; nor is its complement, $(-infty,,0]cup (1,,infty)$, which doesn't contain a neighbourhood of $0$.






      share|cite|improve this answer









      $endgroup$



      Neither. It doesn't contain a neighbourhood of $1$, so it isn't open; nor is its complement, $(-infty,,0]cup (1,,infty)$, which doesn't contain a neighbourhood of $0$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 9 hours ago









      J.G.J.G.

      31.6k23149




      31.6k23149












      • $begingroup$
        note that "its complement" is only what it is if we assume reals as the topology. $(0, 1]$ is open in itself when using the $D = |x-y|$ metric because the complement is the empty set.
        $endgroup$
        – John Dvorak
        7 hours ago








      • 2




        $begingroup$
        @JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
        $endgroup$
        – J.G.
        7 hours ago


















      • $begingroup$
        note that "its complement" is only what it is if we assume reals as the topology. $(0, 1]$ is open in itself when using the $D = |x-y|$ metric because the complement is the empty set.
        $endgroup$
        – John Dvorak
        7 hours ago








      • 2




        $begingroup$
        @JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
        $endgroup$
        – J.G.
        7 hours ago
















      $begingroup$
      note that "its complement" is only what it is if we assume reals as the topology. $(0, 1]$ is open in itself when using the $D = |x-y|$ metric because the complement is the empty set.
      $endgroup$
      – John Dvorak
      7 hours ago






      $begingroup$
      note that "its complement" is only what it is if we assume reals as the topology. $(0, 1]$ is open in itself when using the $D = |x-y|$ metric because the complement is the empty set.
      $endgroup$
      – John Dvorak
      7 hours ago






      2




      2




      $begingroup$
      @JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
      $endgroup$
      – J.G.
      7 hours ago




      $begingroup$
      @JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
      $endgroup$
      – J.G.
      7 hours ago











      1












      $begingroup$

      It's important that you specify where you are considering the subset $A$.



      If $A subset X$ with $ X = mathbb{R}$, J.G. is absolutely right in the usual topology of $mathbb{R}$.



      If $A subset X$ with $ X = [0,1]$, $A^c = {0} $, which is closed in the usual topology, then $A = (0,1]$ is open.



      In other words, it's important to specify in what topologic space $X$ you are considering $A$ as a subset. There are some stranger metrics which may define some different open sets where things can be different.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        It's important that you specify where you are considering the subset $A$.



        If $A subset X$ with $ X = mathbb{R}$, J.G. is absolutely right in the usual topology of $mathbb{R}$.



        If $A subset X$ with $ X = [0,1]$, $A^c = {0} $, which is closed in the usual topology, then $A = (0,1]$ is open.



        In other words, it's important to specify in what topologic space $X$ you are considering $A$ as a subset. There are some stranger metrics which may define some different open sets where things can be different.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          It's important that you specify where you are considering the subset $A$.



          If $A subset X$ with $ X = mathbb{R}$, J.G. is absolutely right in the usual topology of $mathbb{R}$.



          If $A subset X$ with $ X = [0,1]$, $A^c = {0} $, which is closed in the usual topology, then $A = (0,1]$ is open.



          In other words, it's important to specify in what topologic space $X$ you are considering $A$ as a subset. There are some stranger metrics which may define some different open sets where things can be different.






          share|cite|improve this answer











          $endgroup$



          It's important that you specify where you are considering the subset $A$.



          If $A subset X$ with $ X = mathbb{R}$, J.G. is absolutely right in the usual topology of $mathbb{R}$.



          If $A subset X$ with $ X = [0,1]$, $A^c = {0} $, which is closed in the usual topology, then $A = (0,1]$ is open.



          In other words, it's important to specify in what topologic space $X$ you are considering $A$ as a subset. There are some stranger metrics which may define some different open sets where things can be different.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 7 hours ago









          Norrius

          1055




          1055










          answered 8 hours ago









          521124521124

          78110




          78110























              0












              $begingroup$

              A set is not a door.



              It is not the case that a set is either open or closed. It can also be neither or both.



              Indeed, your arguments correctly establish that $(0,1]$ is neither open nor closed as a subset of $mathbb{R}$ with the usual topology. The empty set $emptyset$ is always both open and closed, no matter what the ambient space is.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                A set is not a door.



                It is not the case that a set is either open or closed. It can also be neither or both.



                Indeed, your arguments correctly establish that $(0,1]$ is neither open nor closed as a subset of $mathbb{R}$ with the usual topology. The empty set $emptyset$ is always both open and closed, no matter what the ambient space is.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  A set is not a door.



                  It is not the case that a set is either open or closed. It can also be neither or both.



                  Indeed, your arguments correctly establish that $(0,1]$ is neither open nor closed as a subset of $mathbb{R}$ with the usual topology. The empty set $emptyset$ is always both open and closed, no matter what the ambient space is.






                  share|cite|improve this answer









                  $endgroup$



                  A set is not a door.



                  It is not the case that a set is either open or closed. It can also be neither or both.



                  Indeed, your arguments correctly establish that $(0,1]$ is neither open nor closed as a subset of $mathbb{R}$ with the usual topology. The empty set $emptyset$ is always both open and closed, no matter what the ambient space is.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 7 hours ago









                  ArnoArno

                  1,2381615




                  1,2381615






























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