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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3041596%2flet-x-be-a-topological-and-a-be-a-nonempty-subset-of-x-then-choose-the-correct%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3041596%2flet-x-be-a-topological-and-a-be-a-nonempty-subset-of-x-then-choose-the-correct%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Bressuire

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
            Read more

            Cabo Verde

            Image
              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
            Read more

            Gyllenstierna

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
            Read more