${ x in X : |f(x)| geq epsilon }$ is compact then $f$ is uniformly continuous on $X$.












4














Past year paper question.



Let $(X, d)$ be a metric space and let $f: X to mathbb{R}$ be a continuous function, where $mathbb{R}$ is given the standard metric. Assume that for any $epsilon > 0$, the set ${ x in X : |f(x)| geq epsilon }$ is a compact metric subspace of $X$. Show that $f$ is uniformly continuous on $X$.



Attempt:



Let $epsilon >0$ be given. Let $K := { x in X : |f(x)| geq frac{epsilon}{2} }$. $f$ is continuous on $K$ $implies f$ is uniform continuous on $K$. Then there exists $delta$ such that $d(x,y)<delta implies |f(x)-f(y)| < frac{epsilon}{2}$.



Take any $2$ points $x,y in X$, such that $d(x,y)<delta$. If $x,y in K$, then $|f(x)-f(y)| < frac{epsilon}{2} < epsilon$. If $x,y notin K$, then $|f(x)-f(y)| leq |f(x)|+|f(y)| leq frac{epsilon}{2} + frac{epsilon}{2} leq epsilon$.



But I am stuck here, because what if $x in K$ and $y notin K$.






analysis metric-spaces







share|cite|improve this question



























    4














    Past year paper question.



    Let $(X, d)$ be a metric space and let $f: X to mathbb{R}$ be a continuous function, where $mathbb{R}$ is given the standard metric. Assume that for any $epsilon > 0$, the set ${ x in X : |f(x)| geq epsilon }$ is a compact metric subspace of $X$. Show that $f$ is uniformly continuous on $X$.



    Attempt:



    Let $epsilon >0$ be given. Let $K := { x in X : |f(x)| geq frac{epsilon}{2} }$. $f$ is continuous on $K$ $implies f$ is uniform continuous on $K$. Then there exists $delta$ such that $d(x,y)<delta implies |f(x)-f(y)| < frac{epsilon}{2}$.



    Take any $2$ points $x,y in X$, such that $d(x,y)<delta$. If $x,y in K$, then $|f(x)-f(y)| < frac{epsilon}{2} < epsilon$. If $x,y notin K$, then $|f(x)-f(y)| leq |f(x)|+|f(y)| leq frac{epsilon}{2} + frac{epsilon}{2} leq epsilon$.



    But I am stuck here, because what if $x in K$ and $y notin K$.






    analysis metric-spaces







    share|cite|improve this question

























      4












      4








      4


      2





      Past year paper question.



      Let $(X, d)$ be a metric space and let $f: X to mathbb{R}$ be a continuous function, where $mathbb{R}$ is given the standard metric. Assume that for any $epsilon > 0$, the set ${ x in X : |f(x)| geq epsilon }$ is a compact metric subspace of $X$. Show that $f$ is uniformly continuous on $X$.



      Attempt:



      Let $epsilon >0$ be given. Let $K := { x in X : |f(x)| geq frac{epsilon}{2} }$. $f$ is continuous on $K$ $implies f$ is uniform continuous on $K$. Then there exists $delta$ such that $d(x,y)<delta implies |f(x)-f(y)| < frac{epsilon}{2}$.



      Take any $2$ points $x,y in X$, such that $d(x,y)<delta$. If $x,y in K$, then $|f(x)-f(y)| < frac{epsilon}{2} < epsilon$. If $x,y notin K$, then $|f(x)-f(y)| leq |f(x)|+|f(y)| leq frac{epsilon}{2} + frac{epsilon}{2} leq epsilon$.



      But I am stuck here, because what if $x in K$ and $y notin K$.






      analysis metric-spaces







      share|cite|improve this question













      Past year paper question.



      Let $(X, d)$ be a metric space and let $f: X to mathbb{R}$ be a continuous function, where $mathbb{R}$ is given the standard metric. Assume that for any $epsilon > 0$, the set ${ x in X : |f(x)| geq epsilon }$ is a compact metric subspace of $X$. Show that $f$ is uniformly continuous on $X$.



      Attempt:



      Let $epsilon >0$ be given. Let $K := { x in X : |f(x)| geq frac{epsilon}{2} }$. $f$ is continuous on $K$ $implies f$ is uniform continuous on $K$. Then there exists $delta$ such that $d(x,y)<delta implies |f(x)-f(y)| < frac{epsilon}{2}$.



      Take any $2$ points $x,y in X$, such that $d(x,y)<delta$. If $x,y in K$, then $|f(x)-f(y)| < frac{epsilon}{2} < epsilon$. If $x,y notin K$, then $|f(x)-f(y)| leq |f(x)|+|f(y)| leq frac{epsilon}{2} + frac{epsilon}{2} leq epsilon$.



      But I am stuck here, because what if $x in K$ and $y notin K$.






      analysis metric-spaces




      analysis metric-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 20 at 3:05









      eatfood

      1827




      1827






















          1 Answer
          1






          active

          oldest

          votes


















          3














          Suppose $f$ is not uniformly continuous. Then there exists $delta >0$ and sequences ${x_n},{y_n}$ such that $d(x_n,y_n) to 0$ but $|f(x_n)-f(y_n)| geq delta$ for each $n$. For each $n$ either $|f(x_n)| geq delta /2$ or $|f(y_n)| geq delta /2$. One of these holds for infinitely many $n$. Suppose $|f(x_n)| geq delta/2 $ along a subsequence. The hypothesis tells you that the subsequence lies in a compact set, so it has a convergent subsequence. Along this subsequence both sequences ${x_n},{y_n}$ converge to the same limit $x$ because $d(x_n,y_n) to 0$. This contradicts continuity of $f$ at $x$.






          share|cite|improve this answer





















          • Writing out formally, So $(x_{n_k}) to x$ for some subsequence. Then $(y_{n_k}) to x$ too, since $|x_{n_k} - y_{n_k}| to 0$. But $lim_{kto infty}f(x_{n_k}) neq lim_{kto infty}f(y_{n_k})$, since $|f(x_{n_k})-f(y_{n_k})| geq delta$. Contradicting continuity of $f$ at $x$.
            – eatfood
            Nov 20 at 7:05












          • This has solved my problem, thank you!
            – eatfood
            Nov 20 at 7:06











          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%2f3005882%2fx-in-x-fx-geq-epsilon-is-compact-then-f-is-uniformly-continuo%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









          3














          Suppose $f$ is not uniformly continuous. Then there exists $delta >0$ and sequences ${x_n},{y_n}$ such that $d(x_n,y_n) to 0$ but $|f(x_n)-f(y_n)| geq delta$ for each $n$. For each $n$ either $|f(x_n)| geq delta /2$ or $|f(y_n)| geq delta /2$. One of these holds for infinitely many $n$. Suppose $|f(x_n)| geq delta/2 $ along a subsequence. The hypothesis tells you that the subsequence lies in a compact set, so it has a convergent subsequence. Along this subsequence both sequences ${x_n},{y_n}$ converge to the same limit $x$ because $d(x_n,y_n) to 0$. This contradicts continuity of $f$ at $x$.






          share|cite|improve this answer





















          • Writing out formally, So $(x_{n_k}) to x$ for some subsequence. Then $(y_{n_k}) to x$ too, since $|x_{n_k} - y_{n_k}| to 0$. But $lim_{kto infty}f(x_{n_k}) neq lim_{kto infty}f(y_{n_k})$, since $|f(x_{n_k})-f(y_{n_k})| geq delta$. Contradicting continuity of $f$ at $x$.
            – eatfood
            Nov 20 at 7:05












          • This has solved my problem, thank you!
            – eatfood
            Nov 20 at 7:06
















          3














          Suppose $f$ is not uniformly continuous. Then there exists $delta >0$ and sequences ${x_n},{y_n}$ such that $d(x_n,y_n) to 0$ but $|f(x_n)-f(y_n)| geq delta$ for each $n$. For each $n$ either $|f(x_n)| geq delta /2$ or $|f(y_n)| geq delta /2$. One of these holds for infinitely many $n$. Suppose $|f(x_n)| geq delta/2 $ along a subsequence. The hypothesis tells you that the subsequence lies in a compact set, so it has a convergent subsequence. Along this subsequence both sequences ${x_n},{y_n}$ converge to the same limit $x$ because $d(x_n,y_n) to 0$. This contradicts continuity of $f$ at $x$.






          share|cite|improve this answer





















          • Writing out formally, So $(x_{n_k}) to x$ for some subsequence. Then $(y_{n_k}) to x$ too, since $|x_{n_k} - y_{n_k}| to 0$. But $lim_{kto infty}f(x_{n_k}) neq lim_{kto infty}f(y_{n_k})$, since $|f(x_{n_k})-f(y_{n_k})| geq delta$. Contradicting continuity of $f$ at $x$.
            – eatfood
            Nov 20 at 7:05












          • This has solved my problem, thank you!
            – eatfood
            Nov 20 at 7:06














          3












          3








          3






          Suppose $f$ is not uniformly continuous. Then there exists $delta >0$ and sequences ${x_n},{y_n}$ such that $d(x_n,y_n) to 0$ but $|f(x_n)-f(y_n)| geq delta$ for each $n$. For each $n$ either $|f(x_n)| geq delta /2$ or $|f(y_n)| geq delta /2$. One of these holds for infinitely many $n$. Suppose $|f(x_n)| geq delta/2 $ along a subsequence. The hypothesis tells you that the subsequence lies in a compact set, so it has a convergent subsequence. Along this subsequence both sequences ${x_n},{y_n}$ converge to the same limit $x$ because $d(x_n,y_n) to 0$. This contradicts continuity of $f$ at $x$.






          share|cite|improve this answer












          Suppose $f$ is not uniformly continuous. Then there exists $delta >0$ and sequences ${x_n},{y_n}$ such that $d(x_n,y_n) to 0$ but $|f(x_n)-f(y_n)| geq delta$ for each $n$. For each $n$ either $|f(x_n)| geq delta /2$ or $|f(y_n)| geq delta /2$. One of these holds for infinitely many $n$. Suppose $|f(x_n)| geq delta/2 $ along a subsequence. The hypothesis tells you that the subsequence lies in a compact set, so it has a convergent subsequence. Along this subsequence both sequences ${x_n},{y_n}$ converge to the same limit $x$ because $d(x_n,y_n) to 0$. This contradicts continuity of $f$ at $x$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 20 at 6:37









          Kavi Rama Murthy

          49.4k31854




          49.4k31854












          • Writing out formally, So $(x_{n_k}) to x$ for some subsequence. Then $(y_{n_k}) to x$ too, since $|x_{n_k} - y_{n_k}| to 0$. But $lim_{kto infty}f(x_{n_k}) neq lim_{kto infty}f(y_{n_k})$, since $|f(x_{n_k})-f(y_{n_k})| geq delta$. Contradicting continuity of $f$ at $x$.
            – eatfood
            Nov 20 at 7:05












          • This has solved my problem, thank you!
            – eatfood
            Nov 20 at 7:06


















          • Writing out formally, So $(x_{n_k}) to x$ for some subsequence. Then $(y_{n_k}) to x$ too, since $|x_{n_k} - y_{n_k}| to 0$. But $lim_{kto infty}f(x_{n_k}) neq lim_{kto infty}f(y_{n_k})$, since $|f(x_{n_k})-f(y_{n_k})| geq delta$. Contradicting continuity of $f$ at $x$.
            – eatfood
            Nov 20 at 7:05












          • This has solved my problem, thank you!
            – eatfood
            Nov 20 at 7:06
















          Writing out formally, So $(x_{n_k}) to x$ for some subsequence. Then $(y_{n_k}) to x$ too, since $|x_{n_k} - y_{n_k}| to 0$. But $lim_{kto infty}f(x_{n_k}) neq lim_{kto infty}f(y_{n_k})$, since $|f(x_{n_k})-f(y_{n_k})| geq delta$. Contradicting continuity of $f$ at $x$.
          – eatfood
          Nov 20 at 7:05






          Writing out formally, So $(x_{n_k}) to x$ for some subsequence. Then $(y_{n_k}) to x$ too, since $|x_{n_k} - y_{n_k}| to 0$. But $lim_{kto infty}f(x_{n_k}) neq lim_{kto infty}f(y_{n_k})$, since $|f(x_{n_k})-f(y_{n_k})| geq delta$. Contradicting continuity of $f$ at $x$.
          – eatfood
          Nov 20 at 7:05














          This has solved my problem, thank you!
          – eatfood
          Nov 20 at 7:06




          This has solved my problem, thank you!
          – eatfood
          Nov 20 at 7:06


















          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f3005882%2fx-in-x-fx-geq-epsilon-is-compact-then-f-is-uniformly-continuo%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

          mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

          Image
          0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
          Read more

          How to change which sound is reproduced for terminal bell?

          Image
          3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
          Read more

          Can I use Tabulator js library in my java Spring + Thymeleaf project?

          Image
          0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
          Read more