If $f:mathbb{R}tomathbb{R}$ is continuous, and $f(mathbb{Q})subsetmathbb{N}$, then is $f$ unbounded,...












0












$begingroup$



Suppose $f:mathbb{R}tomathbb{R}$ is a continuous function such that $ f(mathbb{Q})subsetmathbb{N}$. What can be said about $f$? Is it (1) unbounded, (2) constant, or (3) non-constant bounded?




I think it should be constant, as $mathbb{Q}$ is dense in $mathbb{R}$.



Kindly help! Thanks & Regards










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    You are right .
    $endgroup$
    – Hagen von Eitzen
    Jan 22 at 18:54






  • 1




    $begingroup$
    Yes. By continuity, $f(mathbb{R})=f(overline{mathbb{Q}})subseteqoverline{f(mathbb{Q})}=mathbb{N}$. Since $f(mathbb{Q})$ is a non-empty connected subset of $mathbb{N}$, it is a singleton. Therefore $f$ is constant.
    $endgroup$
    – Sangchul Lee
    Jan 22 at 19:22


















0












$begingroup$



Suppose $f:mathbb{R}tomathbb{R}$ is a continuous function such that $ f(mathbb{Q})subsetmathbb{N}$. What can be said about $f$? Is it (1) unbounded, (2) constant, or (3) non-constant bounded?




I think it should be constant, as $mathbb{Q}$ is dense in $mathbb{R}$.



Kindly help! Thanks & Regards










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    You are right .
    $endgroup$
    – Hagen von Eitzen
    Jan 22 at 18:54






  • 1




    $begingroup$
    Yes. By continuity, $f(mathbb{R})=f(overline{mathbb{Q}})subseteqoverline{f(mathbb{Q})}=mathbb{N}$. Since $f(mathbb{Q})$ is a non-empty connected subset of $mathbb{N}$, it is a singleton. Therefore $f$ is constant.
    $endgroup$
    – Sangchul Lee
    Jan 22 at 19:22
















0












0








0





$begingroup$



Suppose $f:mathbb{R}tomathbb{R}$ is a continuous function such that $ f(mathbb{Q})subsetmathbb{N}$. What can be said about $f$? Is it (1) unbounded, (2) constant, or (3) non-constant bounded?




I think it should be constant, as $mathbb{Q}$ is dense in $mathbb{R}$.



Kindly help! Thanks & Regards










share|cite|improve this question











$endgroup$





Suppose $f:mathbb{R}tomathbb{R}$ is a continuous function such that $ f(mathbb{Q})subsetmathbb{N}$. What can be said about $f$? Is it (1) unbounded, (2) constant, or (3) non-constant bounded?




I think it should be constant, as $mathbb{Q}$ is dense in $mathbb{R}$.



Kindly help! Thanks & Regards







real-analysis functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 19:17









Blue

48.6k870156




48.6k870156










asked Jan 22 at 18:51









Devendra Singh RanaDevendra Singh Rana

7751416




7751416








  • 3




    $begingroup$
    You are right .
    $endgroup$
    – Hagen von Eitzen
    Jan 22 at 18:54






  • 1




    $begingroup$
    Yes. By continuity, $f(mathbb{R})=f(overline{mathbb{Q}})subseteqoverline{f(mathbb{Q})}=mathbb{N}$. Since $f(mathbb{Q})$ is a non-empty connected subset of $mathbb{N}$, it is a singleton. Therefore $f$ is constant.
    $endgroup$
    – Sangchul Lee
    Jan 22 at 19:22
















  • 3




    $begingroup$
    You are right .
    $endgroup$
    – Hagen von Eitzen
    Jan 22 at 18:54






  • 1




    $begingroup$
    Yes. By continuity, $f(mathbb{R})=f(overline{mathbb{Q}})subseteqoverline{f(mathbb{Q})}=mathbb{N}$. Since $f(mathbb{Q})$ is a non-empty connected subset of $mathbb{N}$, it is a singleton. Therefore $f$ is constant.
    $endgroup$
    – Sangchul Lee
    Jan 22 at 19:22










3




3




$begingroup$
You are right .
$endgroup$
– Hagen von Eitzen
Jan 22 at 18:54




$begingroup$
You are right .
$endgroup$
– Hagen von Eitzen
Jan 22 at 18:54




1




1




$begingroup$
Yes. By continuity, $f(mathbb{R})=f(overline{mathbb{Q}})subseteqoverline{f(mathbb{Q})}=mathbb{N}$. Since $f(mathbb{Q})$ is a non-empty connected subset of $mathbb{N}$, it is a singleton. Therefore $f$ is constant.
$endgroup$
– Sangchul Lee
Jan 22 at 19:22






$begingroup$
Yes. By continuity, $f(mathbb{R})=f(overline{mathbb{Q}})subseteqoverline{f(mathbb{Q})}=mathbb{N}$. Since $f(mathbb{Q})$ is a non-empty connected subset of $mathbb{N}$, it is a singleton. Therefore $f$ is constant.
$endgroup$
– Sangchul Lee
Jan 22 at 19:22












1 Answer
1






active

oldest

votes


















1












$begingroup$

Yes, $f$ must be a constant. Suppose not. Pick $q_1, q_2 in mathbb{Q}$ such that $f(q_1) > f(q_2)$. Without loss of generality, assume $q_1 < q_2$. Pick any noninteger $y$ between $f(q_1)$ and $f(q_2)$. Then by intermediate value theorem, there exists $x in (q_1, q_2)$ such that $f(x) = y$. Any rational number sufficiently close to $x$ should not take integer value, a contradiction.






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%2f3083545%2fif-f-mathbbr-to-mathbbr-is-continuous-and-f-mathbbq-subset-mathbbn%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Yes, $f$ must be a constant. Suppose not. Pick $q_1, q_2 in mathbb{Q}$ such that $f(q_1) > f(q_2)$. Without loss of generality, assume $q_1 < q_2$. Pick any noninteger $y$ between $f(q_1)$ and $f(q_2)$. Then by intermediate value theorem, there exists $x in (q_1, q_2)$ such that $f(x) = y$. Any rational number sufficiently close to $x$ should not take integer value, a contradiction.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Yes, $f$ must be a constant. Suppose not. Pick $q_1, q_2 in mathbb{Q}$ such that $f(q_1) > f(q_2)$. Without loss of generality, assume $q_1 < q_2$. Pick any noninteger $y$ between $f(q_1)$ and $f(q_2)$. Then by intermediate value theorem, there exists $x in (q_1, q_2)$ such that $f(x) = y$. Any rational number sufficiently close to $x$ should not take integer value, a contradiction.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Yes, $f$ must be a constant. Suppose not. Pick $q_1, q_2 in mathbb{Q}$ such that $f(q_1) > f(q_2)$. Without loss of generality, assume $q_1 < q_2$. Pick any noninteger $y$ between $f(q_1)$ and $f(q_2)$. Then by intermediate value theorem, there exists $x in (q_1, q_2)$ such that $f(x) = y$. Any rational number sufficiently close to $x$ should not take integer value, a contradiction.






        share|cite|improve this answer









        $endgroup$



        Yes, $f$ must be a constant. Suppose not. Pick $q_1, q_2 in mathbb{Q}$ such that $f(q_1) > f(q_2)$. Without loss of generality, assume $q_1 < q_2$. Pick any noninteger $y$ between $f(q_1)$ and $f(q_2)$. Then by intermediate value theorem, there exists $x in (q_1, q_2)$ such that $f(x) = y$. Any rational number sufficiently close to $x$ should not take integer value, a contradiction.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 22 at 19:08









        user295959user295959

        664310




        664310






























            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%2f3083545%2fif-f-mathbbr-to-mathbbr-is-continuous-and-f-mathbbq-subset-mathbbn%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

            Mario Kart Wii

            The Binding of Isaac: Rebirth/Afterbirth

            What does “Dominus providebit” mean?