How to find supremum and infimum of $(2014,2016]$?












4















Let $A=(2014,2016]$. Which of the following statements is/are true?




(a) $sup A=2016$ and $inf A=2014$



(b) $sup A=2014$ and $inf A=2016$



(c) $sup A=2016$ and $inf A$ does not exist



(d) $sup A$ does not exist and $inf A=2014$



(e) Both $sup A$ and $inf A$ do not exist.



My try



Since there is no possible least number inf does not exist and $sup(A)=2016$.
Is my argument correct?






real-analysis







share|cite|improve this question




















  • 4




    You are not correct that there is no infimum. The infimum is not the smallest number of a set. It is the greatest lower bound for a set.
    – quanticbolt
    Dec 8 at 17:52


















4















Let $A=(2014,2016]$. Which of the following statements is/are true?




(a) $sup A=2016$ and $inf A=2014$



(b) $sup A=2014$ and $inf A=2016$



(c) $sup A=2016$ and $inf A$ does not exist



(d) $sup A$ does not exist and $inf A=2014$



(e) Both $sup A$ and $inf A$ do not exist.



My try



Since there is no possible least number inf does not exist and $sup(A)=2016$.
Is my argument correct?






real-analysis







share|cite|improve this question




















  • 4




    You are not correct that there is no infimum. The infimum is not the smallest number of a set. It is the greatest lower bound for a set.
    – quanticbolt
    Dec 8 at 17:52
















4












4








4








Let $A=(2014,2016]$. Which of the following statements is/are true?




(a) $sup A=2016$ and $inf A=2014$



(b) $sup A=2014$ and $inf A=2016$



(c) $sup A=2016$ and $inf A$ does not exist



(d) $sup A$ does not exist and $inf A=2014$



(e) Both $sup A$ and $inf A$ do not exist.



My try



Since there is no possible least number inf does not exist and $sup(A)=2016$.
Is my argument correct?






real-analysis







share|cite|improve this question
















Let $A=(2014,2016]$. Which of the following statements is/are true?




(a) $sup A=2016$ and $inf A=2014$



(b) $sup A=2014$ and $inf A=2016$



(c) $sup A=2016$ and $inf A$ does not exist



(d) $sup A$ does not exist and $inf A=2014$



(e) Both $sup A$ and $inf A$ do not exist.



My try



Since there is no possible least number inf does not exist and $sup(A)=2016$.
Is my argument correct?






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 Dec 9 at 4:29









user21820

38.7k543153




38.7k543153










asked Dec 8 at 17:50









emil

417410




417410








  • 4




    You are not correct that there is no infimum. The infimum is not the smallest number of a set. It is the greatest lower bound for a set.
    – quanticbolt
    Dec 8 at 17:52
















  • 4




    You are not correct that there is no infimum. The infimum is not the smallest number of a set. It is the greatest lower bound for a set.
    – quanticbolt
    Dec 8 at 17:52










4




4




You are not correct that there is no infimum. The infimum is not the smallest number of a set. It is the greatest lower bound for a set.
– quanticbolt
Dec 8 at 17:52






You are not correct that there is no infimum. The infimum is not the smallest number of a set. It is the greatest lower bound for a set.
– quanticbolt
Dec 8 at 17:52












3 Answers
3






active

oldest

votes


















7














You are confusing between infimum and minimum. Infimum is the largest lower bound of the set. So the definition doesn't say the infimum must belong to the set itself. If it does belong to the set then it is called minimum. So what is the greatest lower bound of your set? Obviously it is $2014$. Why? First of all it is a lower bound. And it is easy to see that any bigger number than $2014$ is already not a lower bound of $A$.






share|cite|improve this answer





















  • This clarifies my doubts. So here Maximum and Supremum are Both 2016. But only infimum is 2014 right?
    – emil
    Dec 8 at 18:02






  • 4




    Yes. In general, if a maximum exists then it is equal to the supremum. If minimum exists then it is equal to the infimum. So in your example the supremum is $2016$ and it is also a maximum. $2014$ is the infimum but it is not a minimum. The set has no minimum.
    – Mark
    Dec 8 at 18:04












  • I think this answer would be better if it addressed the OP's lack of argument for why $operatorname{sup}(A)=2016$.
    – Shaun
    Dec 8 at 18:20



















2















"Since there is no possible least number inf does not exist"




You're thinking of the minimum there, not the infimum; the infimum is the greatest lower bound of the set.



Also, you have presented no argument as to why $operatorname{sup}(A)=2016$.






share|cite|improve this answer





























    1














    For a bounded set like the interval $A$ here, supremum and infimum always exist and are equal to the larger and smaller numbers defining the interval's endpoints respectively, so the answer must be A, not C.






    share|cite|improve this answer





















      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%2f3031407%2fhow-to-find-supremum-and-infimum-of-2014-2016%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7














      You are confusing between infimum and minimum. Infimum is the largest lower bound of the set. So the definition doesn't say the infimum must belong to the set itself. If it does belong to the set then it is called minimum. So what is the greatest lower bound of your set? Obviously it is $2014$. Why? First of all it is a lower bound. And it is easy to see that any bigger number than $2014$ is already not a lower bound of $A$.






      share|cite|improve this answer





















      • This clarifies my doubts. So here Maximum and Supremum are Both 2016. But only infimum is 2014 right?
        – emil
        Dec 8 at 18:02






      • 4




        Yes. In general, if a maximum exists then it is equal to the supremum. If minimum exists then it is equal to the infimum. So in your example the supremum is $2016$ and it is also a maximum. $2014$ is the infimum but it is not a minimum. The set has no minimum.
        – Mark
        Dec 8 at 18:04












      • I think this answer would be better if it addressed the OP's lack of argument for why $operatorname{sup}(A)=2016$.
        – Shaun
        Dec 8 at 18:20
















      7














      You are confusing between infimum and minimum. Infimum is the largest lower bound of the set. So the definition doesn't say the infimum must belong to the set itself. If it does belong to the set then it is called minimum. So what is the greatest lower bound of your set? Obviously it is $2014$. Why? First of all it is a lower bound. And it is easy to see that any bigger number than $2014$ is already not a lower bound of $A$.






      share|cite|improve this answer





















      • This clarifies my doubts. So here Maximum and Supremum are Both 2016. But only infimum is 2014 right?
        – emil
        Dec 8 at 18:02






      • 4




        Yes. In general, if a maximum exists then it is equal to the supremum. If minimum exists then it is equal to the infimum. So in your example the supremum is $2016$ and it is also a maximum. $2014$ is the infimum but it is not a minimum. The set has no minimum.
        – Mark
        Dec 8 at 18:04












      • I think this answer would be better if it addressed the OP's lack of argument for why $operatorname{sup}(A)=2016$.
        – Shaun
        Dec 8 at 18:20














      7












      7








      7






      You are confusing between infimum and minimum. Infimum is the largest lower bound of the set. So the definition doesn't say the infimum must belong to the set itself. If it does belong to the set then it is called minimum. So what is the greatest lower bound of your set? Obviously it is $2014$. Why? First of all it is a lower bound. And it is easy to see that any bigger number than $2014$ is already not a lower bound of $A$.






      share|cite|improve this answer












      You are confusing between infimum and minimum. Infimum is the largest lower bound of the set. So the definition doesn't say the infimum must belong to the set itself. If it does belong to the set then it is called minimum. So what is the greatest lower bound of your set? Obviously it is $2014$. Why? First of all it is a lower bound. And it is easy to see that any bigger number than $2014$ is already not a lower bound of $A$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 8 at 17:53









      Mark

      6,015415




      6,015415












      • This clarifies my doubts. So here Maximum and Supremum are Both 2016. But only infimum is 2014 right?
        – emil
        Dec 8 at 18:02






      • 4




        Yes. In general, if a maximum exists then it is equal to the supremum. If minimum exists then it is equal to the infimum. So in your example the supremum is $2016$ and it is also a maximum. $2014$ is the infimum but it is not a minimum. The set has no minimum.
        – Mark
        Dec 8 at 18:04












      • I think this answer would be better if it addressed the OP's lack of argument for why $operatorname{sup}(A)=2016$.
        – Shaun
        Dec 8 at 18:20


















      • This clarifies my doubts. So here Maximum and Supremum are Both 2016. But only infimum is 2014 right?
        – emil
        Dec 8 at 18:02






      • 4




        Yes. In general, if a maximum exists then it is equal to the supremum. If minimum exists then it is equal to the infimum. So in your example the supremum is $2016$ and it is also a maximum. $2014$ is the infimum but it is not a minimum. The set has no minimum.
        – Mark
        Dec 8 at 18:04












      • I think this answer would be better if it addressed the OP's lack of argument for why $operatorname{sup}(A)=2016$.
        – Shaun
        Dec 8 at 18:20
















      This clarifies my doubts. So here Maximum and Supremum are Both 2016. But only infimum is 2014 right?
      – emil
      Dec 8 at 18:02




      This clarifies my doubts. So here Maximum and Supremum are Both 2016. But only infimum is 2014 right?
      – emil
      Dec 8 at 18:02




      4




      4




      Yes. In general, if a maximum exists then it is equal to the supremum. If minimum exists then it is equal to the infimum. So in your example the supremum is $2016$ and it is also a maximum. $2014$ is the infimum but it is not a minimum. The set has no minimum.
      – Mark
      Dec 8 at 18:04






      Yes. In general, if a maximum exists then it is equal to the supremum. If minimum exists then it is equal to the infimum. So in your example the supremum is $2016$ and it is also a maximum. $2014$ is the infimum but it is not a minimum. The set has no minimum.
      – Mark
      Dec 8 at 18:04














      I think this answer would be better if it addressed the OP's lack of argument for why $operatorname{sup}(A)=2016$.
      – Shaun
      Dec 8 at 18:20




      I think this answer would be better if it addressed the OP's lack of argument for why $operatorname{sup}(A)=2016$.
      – Shaun
      Dec 8 at 18:20











      2















      "Since there is no possible least number inf does not exist"




      You're thinking of the minimum there, not the infimum; the infimum is the greatest lower bound of the set.



      Also, you have presented no argument as to why $operatorname{sup}(A)=2016$.






      share|cite|improve this answer


























        2















        "Since there is no possible least number inf does not exist"




        You're thinking of the minimum there, not the infimum; the infimum is the greatest lower bound of the set.



        Also, you have presented no argument as to why $operatorname{sup}(A)=2016$.






        share|cite|improve this answer
























          2












          2








          2







          "Since there is no possible least number inf does not exist"




          You're thinking of the minimum there, not the infimum; the infimum is the greatest lower bound of the set.



          Also, you have presented no argument as to why $operatorname{sup}(A)=2016$.






          share|cite|improve this answer













          "Since there is no possible least number inf does not exist"




          You're thinking of the minimum there, not the infimum; the infimum is the greatest lower bound of the set.



          Also, you have presented no argument as to why $operatorname{sup}(A)=2016$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 at 17:56









          Shaun

          8,686113680




          8,686113680























              1














              For a bounded set like the interval $A$ here, supremum and infimum always exist and are equal to the larger and smaller numbers defining the interval's endpoints respectively, so the answer must be A, not C.






              share|cite|improve this answer


























                1














                For a bounded set like the interval $A$ here, supremum and infimum always exist and are equal to the larger and smaller numbers defining the interval's endpoints respectively, so the answer must be A, not C.






                share|cite|improve this answer
























                  1












                  1








                  1






                  For a bounded set like the interval $A$ here, supremum and infimum always exist and are equal to the larger and smaller numbers defining the interval's endpoints respectively, so the answer must be A, not C.






                  share|cite|improve this answer












                  For a bounded set like the interval $A$ here, supremum and infimum always exist and are equal to the larger and smaller numbers defining the interval's endpoints respectively, so the answer must be A, not C.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 8 at 17:53









                  Parcly Taxel

                  41.2k137199




                  41.2k137199






























                      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%2f3031407%2fhow-to-find-supremum-and-infimum-of-2014-2016%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

                      Plaza Victoria

                      Image
                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
                      Read more

                      In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
                      Read more

                      How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

                      Image
                      1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
                      Read more