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







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  • 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













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edited Dec 9 at 4:29









user21820

38.6k542153




38.6k542153










asked Dec 8 at 17:50









emil

415410




415410








  • 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





















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      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

      5,895415




      5,895415












      • 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

          8,214113578




          8,214113578























              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

                  41.3k137199




                  41.3k137199






























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