let $X$ be a topological and A be a nonempty subset of X.then choose the correct statement












0












$begingroup$


Let $X$ be a topological space and A be a nonempty subset of X. Then

choose the correct statement:



$1.$ $A$ is dense in $X$, if $(Xsetminus A)$ is nowhere dense in $X.$



$2.$ $(Xsetminus A)$ is nowhere dense , if $A$ is dense in $X.$



$3$.$A$ is dense in $X$, if the interior of $(Xsetminus A)$ is empty.



$4.$The interior of $(X setminus A)$ is empty , if $A$ is dense in $X$.



My attempt : my answer is option $1 ,2 $ and $ 4.$



Option $1$ is True. Take $A = mathbb{Q}.$



Option $2$ is true. Same logic in option $1.$



Option $3$ is false. Take $A =[0,1]$ , $A^0 =A text{interior} =(0,1) neq phi.$



Option $4$ is true. Take $A= mathbb{Q}.$



Is its correct or not ?



Any hints/solution will be appreciated.



thanks u






general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    what is $mathbb{R}backslash mathbb{Q}$?
    $endgroup$
    – Thomas
    Dec 15 '18 at 15:31










  • $begingroup$
    @Thomas complement of $mathbb{Q}$
    $endgroup$
    – jasmine
    Dec 15 '18 at 15:38








  • 1




    $begingroup$
    If you say true, an example is not an argument. Only for false statements you need to exhibit a counterexample.
    $endgroup$
    – Henno Brandsma
    Dec 15 '18 at 17:50






  • 1




    $begingroup$
    For 3 your example does not obey that the interior of the complement is empty. So it's irrelevant.
    $endgroup$
    – Henno Brandsma
    Dec 15 '18 at 17:51






  • 1




    $begingroup$
    The counter-examples to 1. and 2. in the answers are examples of a set $A$ that is called "dense & co-dense" . I.e., both $A$ and $X$ $A$ are dense in $X.$
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 18:56


















0












$begingroup$


Let $X$ be a topological space and A be a nonempty subset of X. Then

choose the correct statement:



$1.$ $A$ is dense in $X$, if $(Xsetminus A)$ is nowhere dense in $X.$



$2.$ $(Xsetminus A)$ is nowhere dense , if $A$ is dense in $X.$



$3$.$A$ is dense in $X$, if the interior of $(Xsetminus A)$ is empty.



$4.$The interior of $(X setminus A)$ is empty , if $A$ is dense in $X$.



My attempt : my answer is option $1 ,2 $ and $ 4.$



Option $1$ is True. Take $A = mathbb{Q}.$



Option $2$ is true. Same logic in option $1.$



Option $3$ is false. Take $A =[0,1]$ , $A^0 =A text{interior} =(0,1) neq phi.$



Option $4$ is true. Take $A= mathbb{Q}.$



Is its correct or not ?



Any hints/solution will be appreciated.



thanks u






general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    what is $mathbb{R}backslash mathbb{Q}$?
    $endgroup$
    – Thomas
    Dec 15 '18 at 15:31










  • $begingroup$
    @Thomas complement of $mathbb{Q}$
    $endgroup$
    – jasmine
    Dec 15 '18 at 15:38








  • 1




    $begingroup$
    If you say true, an example is not an argument. Only for false statements you need to exhibit a counterexample.
    $endgroup$
    – Henno Brandsma
    Dec 15 '18 at 17:50






  • 1




    $begingroup$
    For 3 your example does not obey that the interior of the complement is empty. So it's irrelevant.
    $endgroup$
    – Henno Brandsma
    Dec 15 '18 at 17:51






  • 1




    $begingroup$
    The counter-examples to 1. and 2. in the answers are examples of a set $A$ that is called "dense & co-dense" . I.e., both $A$ and $X$ $A$ are dense in $X.$
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 18:56
















0












0








0





$begingroup$


Let $X$ be a topological space and A be a nonempty subset of X. Then

choose the correct statement:



$1.$ $A$ is dense in $X$, if $(Xsetminus A)$ is nowhere dense in $X.$



$2.$ $(Xsetminus A)$ is nowhere dense , if $A$ is dense in $X.$



$3$.$A$ is dense in $X$, if the interior of $(Xsetminus A)$ is empty.



$4.$The interior of $(X setminus A)$ is empty , if $A$ is dense in $X$.



My attempt : my answer is option $1 ,2 $ and $ 4.$



Option $1$ is True. Take $A = mathbb{Q}.$



Option $2$ is true. Same logic in option $1.$



Option $3$ is false. Take $A =[0,1]$ , $A^0 =A text{interior} =(0,1) neq phi.$



Option $4$ is true. Take $A= mathbb{Q}.$



Is its correct or not ?



Any hints/solution will be appreciated.



thanks u






general-topology







share|cite|improve this question











$endgroup$




Let $X$ be a topological space and A be a nonempty subset of X. Then

choose the correct statement:



$1.$ $A$ is dense in $X$, if $(Xsetminus A)$ is nowhere dense in $X.$



$2.$ $(Xsetminus A)$ is nowhere dense , if $A$ is dense in $X.$



$3$.$A$ is dense in $X$, if the interior of $(Xsetminus A)$ is empty.



$4.$The interior of $(X setminus A)$ is empty , if $A$ is dense in $X$.



My attempt : my answer is option $1 ,2 $ and $ 4.$



Option $1$ is True. Take $A = mathbb{Q}.$



Option $2$ is true. Same logic in option $1.$



Option $3$ is false. Take $A =[0,1]$ , $A^0 =A text{interior} =(0,1) neq phi.$



Option $4$ is true. Take $A= mathbb{Q}.$



Is its correct or not ?



Any hints/solution will be appreciated.



thanks u






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 18:48









DanielWainfleet

34.6k31648




34.6k31648










asked Dec 15 '18 at 15:20









jasminejasmine

1,679416




1,679416












  • $begingroup$
    what is $mathbb{R}backslash mathbb{Q}$?
    $endgroup$
    – Thomas
    Dec 15 '18 at 15:31










  • $begingroup$
    @Thomas complement of $mathbb{Q}$
    $endgroup$
    – jasmine
    Dec 15 '18 at 15:38








  • 1




    $begingroup$
    If you say true, an example is not an argument. Only for false statements you need to exhibit a counterexample.
    $endgroup$
    – Henno Brandsma
    Dec 15 '18 at 17:50






  • 1




    $begingroup$
    For 3 your example does not obey that the interior of the complement is empty. So it's irrelevant.
    $endgroup$
    – Henno Brandsma
    Dec 15 '18 at 17:51






  • 1




    $begingroup$
    The counter-examples to 1. and 2. in the answers are examples of a set $A$ that is called "dense & co-dense" . I.e., both $A$ and $X$ $A$ are dense in $X.$
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 18:56




















  • $begingroup$
    what is $mathbb{R}backslash mathbb{Q}$?
    $endgroup$
    – Thomas
    Dec 15 '18 at 15:31










  • $begingroup$
    @Thomas complement of $mathbb{Q}$
    $endgroup$
    – jasmine
    Dec 15 '18 at 15:38








  • 1




    $begingroup$
    If you say true, an example is not an argument. Only for false statements you need to exhibit a counterexample.
    $endgroup$
    – Henno Brandsma
    Dec 15 '18 at 17:50






  • 1




    $begingroup$
    For 3 your example does not obey that the interior of the complement is empty. So it's irrelevant.
    $endgroup$
    – Henno Brandsma
    Dec 15 '18 at 17:51






  • 1




    $begingroup$
    The counter-examples to 1. and 2. in the answers are examples of a set $A$ that is called "dense & co-dense" . I.e., both $A$ and $X$ $A$ are dense in $X.$
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 18:56


















$begingroup$
what is $mathbb{R}backslash mathbb{Q}$?
$endgroup$
– Thomas
Dec 15 '18 at 15:31




$begingroup$
what is $mathbb{R}backslash mathbb{Q}$?
$endgroup$
– Thomas
Dec 15 '18 at 15:31












$begingroup$
@Thomas complement of $mathbb{Q}$
$endgroup$
– jasmine
Dec 15 '18 at 15:38






$begingroup$
@Thomas complement of $mathbb{Q}$
$endgroup$
– jasmine
Dec 15 '18 at 15:38






1




1




$begingroup$
If you say true, an example is not an argument. Only for false statements you need to exhibit a counterexample.
$endgroup$
– Henno Brandsma
Dec 15 '18 at 17:50




$begingroup$
If you say true, an example is not an argument. Only for false statements you need to exhibit a counterexample.
$endgroup$
– Henno Brandsma
Dec 15 '18 at 17:50




1




1




$begingroup$
For 3 your example does not obey that the interior of the complement is empty. So it's irrelevant.
$endgroup$
– Henno Brandsma
Dec 15 '18 at 17:51




$begingroup$
For 3 your example does not obey that the interior of the complement is empty. So it's irrelevant.
$endgroup$
– Henno Brandsma
Dec 15 '18 at 17:51




1




1




$begingroup$
The counter-examples to 1. and 2. in the answers are examples of a set $A$ that is called "dense & co-dense" . I.e., both $A$ and $X$ $A$ are dense in $X.$
$endgroup$
– DanielWainfleet
Dec 15 '18 at 18:56






$begingroup$
The counter-examples to 1. and 2. in the answers are examples of a set $A$ that is called "dense & co-dense" . I.e., both $A$ and $X$ $A$ are dense in $X.$
$endgroup$
– DanielWainfleet
Dec 15 '18 at 18:56












2 Answers
2






active

oldest

votes


















3












$begingroup$


  1. If $Xsetminus A$ is nowhere dense, this means that $Xsetminus (X setminus A)) = A$ contains a dense and open subset, so indeed $A$ is dense. An example cannot answer this, you need an argument.


  2. If $A$ is dense, then $Xsetminus A$ need not be nowhere dense, as shown by the example $A = mathbb{Q}$.


  3. If the interior of $Xsetminus A$ is empty, then $A$ is dense: $overline{A} = Xsetminus operatorname{int}(Xsetminus A))$.


  4. If $A$ is dense in $X$, the interior of $X setminus A$ is empty; this follows by the same formula, or note that otherwise a non-empty interior of $Xsetminus A$ could not intersect the dense set $A$.







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    The Association Of Pedants deducts marks from both you and the proposer for not specifying in 2. that $X=Bbb R$ (with the usual topology)............... +1
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 18:41



















2












$begingroup$

1) is false, $mathbb{R}backslashmathbb{Q}=mathbb{Q}'$ which is dense in $mathbb{R}$. 2) is a reformulation of 1 so is false. 3) is true. think of 4 ;)






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$


    1. If $Xsetminus A$ is nowhere dense, this means that $Xsetminus (X setminus A)) = A$ contains a dense and open subset, so indeed $A$ is dense. An example cannot answer this, you need an argument.


    2. If $A$ is dense, then $Xsetminus A$ need not be nowhere dense, as shown by the example $A = mathbb{Q}$.


    3. If the interior of $Xsetminus A$ is empty, then $A$ is dense: $overline{A} = Xsetminus operatorname{int}(Xsetminus A))$.


    4. If $A$ is dense in $X$, the interior of $X setminus A$ is empty; this follows by the same formula, or note that otherwise a non-empty interior of $Xsetminus A$ could not intersect the dense set $A$.







    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      The Association Of Pedants deducts marks from both you and the proposer for not specifying in 2. that $X=Bbb R$ (with the usual topology)............... +1
      $endgroup$
      – DanielWainfleet
      Dec 15 '18 at 18:41
















    3












    $begingroup$


    1. If $Xsetminus A$ is nowhere dense, this means that $Xsetminus (X setminus A)) = A$ contains a dense and open subset, so indeed $A$ is dense. An example cannot answer this, you need an argument.


    2. If $A$ is dense, then $Xsetminus A$ need not be nowhere dense, as shown by the example $A = mathbb{Q}$.


    3. If the interior of $Xsetminus A$ is empty, then $A$ is dense: $overline{A} = Xsetminus operatorname{int}(Xsetminus A))$.


    4. If $A$ is dense in $X$, the interior of $X setminus A$ is empty; this follows by the same formula, or note that otherwise a non-empty interior of $Xsetminus A$ could not intersect the dense set $A$.







    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      The Association Of Pedants deducts marks from both you and the proposer for not specifying in 2. that $X=Bbb R$ (with the usual topology)............... +1
      $endgroup$
      – DanielWainfleet
      Dec 15 '18 at 18:41














    3












    3








    3





    $begingroup$


    1. If $Xsetminus A$ is nowhere dense, this means that $Xsetminus (X setminus A)) = A$ contains a dense and open subset, so indeed $A$ is dense. An example cannot answer this, you need an argument.


    2. If $A$ is dense, then $Xsetminus A$ need not be nowhere dense, as shown by the example $A = mathbb{Q}$.


    3. If the interior of $Xsetminus A$ is empty, then $A$ is dense: $overline{A} = Xsetminus operatorname{int}(Xsetminus A))$.


    4. If $A$ is dense in $X$, the interior of $X setminus A$ is empty; this follows by the same formula, or note that otherwise a non-empty interior of $Xsetminus A$ could not intersect the dense set $A$.







    share|cite|improve this answer









    $endgroup$




    1. If $Xsetminus A$ is nowhere dense, this means that $Xsetminus (X setminus A)) = A$ contains a dense and open subset, so indeed $A$ is dense. An example cannot answer this, you need an argument.


    2. If $A$ is dense, then $Xsetminus A$ need not be nowhere dense, as shown by the example $A = mathbb{Q}$.


    3. If the interior of $Xsetminus A$ is empty, then $A$ is dense: $overline{A} = Xsetminus operatorname{int}(Xsetminus A))$.


    4. If $A$ is dense in $X$, the interior of $X setminus A$ is empty; this follows by the same formula, or note that otherwise a non-empty interior of $Xsetminus A$ could not intersect the dense set $A$.








    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 15 '18 at 17:47









    Henno BrandsmaHenno Brandsma

    106k347114




    106k347114








    • 1




      $begingroup$
      The Association Of Pedants deducts marks from both you and the proposer for not specifying in 2. that $X=Bbb R$ (with the usual topology)............... +1
      $endgroup$
      – DanielWainfleet
      Dec 15 '18 at 18:41














    • 1




      $begingroup$
      The Association Of Pedants deducts marks from both you and the proposer for not specifying in 2. that $X=Bbb R$ (with the usual topology)............... +1
      $endgroup$
      – DanielWainfleet
      Dec 15 '18 at 18:41








    1




    1




    $begingroup$
    The Association Of Pedants deducts marks from both you and the proposer for not specifying in 2. that $X=Bbb R$ (with the usual topology)............... +1
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 18:41




    $begingroup$
    The Association Of Pedants deducts marks from both you and the proposer for not specifying in 2. that $X=Bbb R$ (with the usual topology)............... +1
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 18:41











    2












    $begingroup$

    1) is false, $mathbb{R}backslashmathbb{Q}=mathbb{Q}'$ which is dense in $mathbb{R}$. 2) is a reformulation of 1 so is false. 3) is true. think of 4 ;)






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      1) is false, $mathbb{R}backslashmathbb{Q}=mathbb{Q}'$ which is dense in $mathbb{R}$. 2) is a reformulation of 1 so is false. 3) is true. think of 4 ;)






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        1) is false, $mathbb{R}backslashmathbb{Q}=mathbb{Q}'$ which is dense in $mathbb{R}$. 2) is a reformulation of 1 so is false. 3) is true. think of 4 ;)






        share|cite|improve this answer









        $endgroup$



        1) is false, $mathbb{R}backslashmathbb{Q}=mathbb{Q}'$ which is dense in $mathbb{R}$. 2) is a reformulation of 1 so is false. 3) is true. think of 4 ;)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 15 '18 at 16:17









        MalikMalik

        1017




        1017






























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