Infimum of the set ${xin mathbb{Q};|;x^2<2}$?












3












$begingroup$


My first year Analysis textbook at university includes examples to grasp the concepts of infimum, supremum, maximum, minimum, lower bound and upper bound in set theory for subsets of the real numbers $mathbb{R}$. One of such examples is the set:
$$S={xin mathbb{Q};|;x^2<2}$$



for which the textbook says that "there does not exist a maximum, the upper bounds are all real numbers greater than or equal to $sqrt{2}$, and the supremum is $sqrt2$".



Now, it also says that for this same subset, "there is no minimum, there are no lower bounds and thus there is no infimum": to me, it would seem like there is no minimum (since one could approach $-sqrt2$ infinitely by smaller and smaller rational numbers part of the set), with the infimum being $-sqrt2$ and lower bounds all real numbers smaller than or equal to $-sqrt2$. Am I wrong, or is the textbook wrong?






supremum-and-infimum







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  • 2




    $begingroup$
    infimum of the set ${xin mathbb{Q};|;x^2<2}$ is $-sqrt {2} $
    $endgroup$
    – Darman
    Dec 23 '18 at 9:38
















3












$begingroup$


My first year Analysis textbook at university includes examples to grasp the concepts of infimum, supremum, maximum, minimum, lower bound and upper bound in set theory for subsets of the real numbers $mathbb{R}$. One of such examples is the set:
$$S={xin mathbb{Q};|;x^2<2}$$



for which the textbook says that "there does not exist a maximum, the upper bounds are all real numbers greater than or equal to $sqrt{2}$, and the supremum is $sqrt2$".



Now, it also says that for this same subset, "there is no minimum, there are no lower bounds and thus there is no infimum": to me, it would seem like there is no minimum (since one could approach $-sqrt2$ infinitely by smaller and smaller rational numbers part of the set), with the infimum being $-sqrt2$ and lower bounds all real numbers smaller than or equal to $-sqrt2$. Am I wrong, or is the textbook wrong?






supremum-and-infimum







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    infimum of the set ${xin mathbb{Q};|;x^2<2}$ is $-sqrt {2} $
    $endgroup$
    – Darman
    Dec 23 '18 at 9:38














3












3








3





$begingroup$


My first year Analysis textbook at university includes examples to grasp the concepts of infimum, supremum, maximum, minimum, lower bound and upper bound in set theory for subsets of the real numbers $mathbb{R}$. One of such examples is the set:
$$S={xin mathbb{Q};|;x^2<2}$$



for which the textbook says that "there does not exist a maximum, the upper bounds are all real numbers greater than or equal to $sqrt{2}$, and the supremum is $sqrt2$".



Now, it also says that for this same subset, "there is no minimum, there are no lower bounds and thus there is no infimum": to me, it would seem like there is no minimum (since one could approach $-sqrt2$ infinitely by smaller and smaller rational numbers part of the set), with the infimum being $-sqrt2$ and lower bounds all real numbers smaller than or equal to $-sqrt2$. Am I wrong, or is the textbook wrong?






supremum-and-infimum







share|cite|improve this question











$endgroup$




My first year Analysis textbook at university includes examples to grasp the concepts of infimum, supremum, maximum, minimum, lower bound and upper bound in set theory for subsets of the real numbers $mathbb{R}$. One of such examples is the set:
$$S={xin mathbb{Q};|;x^2<2}$$



for which the textbook says that "there does not exist a maximum, the upper bounds are all real numbers greater than or equal to $sqrt{2}$, and the supremum is $sqrt2$".



Now, it also says that for this same subset, "there is no minimum, there are no lower bounds and thus there is no infimum": to me, it would seem like there is no minimum (since one could approach $-sqrt2$ infinitely by smaller and smaller rational numbers part of the set), with the infimum being $-sqrt2$ and lower bounds all real numbers smaller than or equal to $-sqrt2$. Am I wrong, or is the textbook wrong?






supremum-and-infimum




supremum-and-infimum






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edited Dec 23 '18 at 10:25









Asaf Karagila♦

305k33436767




305k33436767










asked Dec 23 '18 at 9:31









MewMew

205




205








  • 2




    $begingroup$
    infimum of the set ${xin mathbb{Q};|;x^2<2}$ is $-sqrt {2} $
    $endgroup$
    – Darman
    Dec 23 '18 at 9:38














  • 2




    $begingroup$
    infimum of the set ${xin mathbb{Q};|;x^2<2}$ is $-sqrt {2} $
    $endgroup$
    – Darman
    Dec 23 '18 at 9:38








2




2




$begingroup$
infimum of the set ${xin mathbb{Q};|;x^2<2}$ is $-sqrt {2} $
$endgroup$
– Darman
Dec 23 '18 at 9:38




$begingroup$
infimum of the set ${xin mathbb{Q};|;x^2<2}$ is $-sqrt {2} $
$endgroup$
– Darman
Dec 23 '18 at 9:38










2 Answers
2






active

oldest

votes


















4












$begingroup$

Looking at $S$ as a subset of $mathbb Q$ it has lower bounds, but has no infimum (i.e. greatest lower bound).



Looking at $S$ as a subset of $mathbb R$ it does have an infimum which is $-sqrt2inmathbb R$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Would the first case not mean there is also no supremum, though?
    $endgroup$
    – Mew
    Dec 23 '18 at 9:39












  • $begingroup$
    Indeed if $S$ is looked at as a subset of $mathbb Q$ (ordered by $leq$) it has no supremum either. If in the textbook they say that there is a supremum but no infimum then they are not consistent.
    $endgroup$
    – drhab
    Dec 23 '18 at 9:41












  • $begingroup$
    Thus, as it is asserted that there exists a supremum, I may conclude that the textbook contradicts itself and is wrong.
    $endgroup$
    – Mew
    Dec 23 '18 at 9:44










  • $begingroup$
    I think so, yes.
    $endgroup$
    – drhab
    Dec 23 '18 at 9:44



















2












$begingroup$

To me it looks like the textbook is wrong, and you have it right. The set does have lower bounds, and therefore a greatest lower bound (among the reals), and that's $-sqrt2$.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Looking at $S$ as a subset of $mathbb Q$ it has lower bounds, but has no infimum (i.e. greatest lower bound).



    Looking at $S$ as a subset of $mathbb R$ it does have an infimum which is $-sqrt2inmathbb R$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Would the first case not mean there is also no supremum, though?
      $endgroup$
      – Mew
      Dec 23 '18 at 9:39












    • $begingroup$
      Indeed if $S$ is looked at as a subset of $mathbb Q$ (ordered by $leq$) it has no supremum either. If in the textbook they say that there is a supremum but no infimum then they are not consistent.
      $endgroup$
      – drhab
      Dec 23 '18 at 9:41












    • $begingroup$
      Thus, as it is asserted that there exists a supremum, I may conclude that the textbook contradicts itself and is wrong.
      $endgroup$
      – Mew
      Dec 23 '18 at 9:44










    • $begingroup$
      I think so, yes.
      $endgroup$
      – drhab
      Dec 23 '18 at 9:44
















    4












    $begingroup$

    Looking at $S$ as a subset of $mathbb Q$ it has lower bounds, but has no infimum (i.e. greatest lower bound).



    Looking at $S$ as a subset of $mathbb R$ it does have an infimum which is $-sqrt2inmathbb R$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Would the first case not mean there is also no supremum, though?
      $endgroup$
      – Mew
      Dec 23 '18 at 9:39












    • $begingroup$
      Indeed if $S$ is looked at as a subset of $mathbb Q$ (ordered by $leq$) it has no supremum either. If in the textbook they say that there is a supremum but no infimum then they are not consistent.
      $endgroup$
      – drhab
      Dec 23 '18 at 9:41












    • $begingroup$
      Thus, as it is asserted that there exists a supremum, I may conclude that the textbook contradicts itself and is wrong.
      $endgroup$
      – Mew
      Dec 23 '18 at 9:44










    • $begingroup$
      I think so, yes.
      $endgroup$
      – drhab
      Dec 23 '18 at 9:44














    4












    4








    4





    $begingroup$

    Looking at $S$ as a subset of $mathbb Q$ it has lower bounds, but has no infimum (i.e. greatest lower bound).



    Looking at $S$ as a subset of $mathbb R$ it does have an infimum which is $-sqrt2inmathbb R$.






    share|cite|improve this answer









    $endgroup$



    Looking at $S$ as a subset of $mathbb Q$ it has lower bounds, but has no infimum (i.e. greatest lower bound).



    Looking at $S$ as a subset of $mathbb R$ it does have an infimum which is $-sqrt2inmathbb R$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 23 '18 at 9:38









    drhabdrhab

    102k545136




    102k545136












    • $begingroup$
      Would the first case not mean there is also no supremum, though?
      $endgroup$
      – Mew
      Dec 23 '18 at 9:39












    • $begingroup$
      Indeed if $S$ is looked at as a subset of $mathbb Q$ (ordered by $leq$) it has no supremum either. If in the textbook they say that there is a supremum but no infimum then they are not consistent.
      $endgroup$
      – drhab
      Dec 23 '18 at 9:41












    • $begingroup$
      Thus, as it is asserted that there exists a supremum, I may conclude that the textbook contradicts itself and is wrong.
      $endgroup$
      – Mew
      Dec 23 '18 at 9:44










    • $begingroup$
      I think so, yes.
      $endgroup$
      – drhab
      Dec 23 '18 at 9:44


















    • $begingroup$
      Would the first case not mean there is also no supremum, though?
      $endgroup$
      – Mew
      Dec 23 '18 at 9:39












    • $begingroup$
      Indeed if $S$ is looked at as a subset of $mathbb Q$ (ordered by $leq$) it has no supremum either. If in the textbook they say that there is a supremum but no infimum then they are not consistent.
      $endgroup$
      – drhab
      Dec 23 '18 at 9:41












    • $begingroup$
      Thus, as it is asserted that there exists a supremum, I may conclude that the textbook contradicts itself and is wrong.
      $endgroup$
      – Mew
      Dec 23 '18 at 9:44










    • $begingroup$
      I think so, yes.
      $endgroup$
      – drhab
      Dec 23 '18 at 9:44
















    $begingroup$
    Would the first case not mean there is also no supremum, though?
    $endgroup$
    – Mew
    Dec 23 '18 at 9:39






    $begingroup$
    Would the first case not mean there is also no supremum, though?
    $endgroup$
    – Mew
    Dec 23 '18 at 9:39














    $begingroup$
    Indeed if $S$ is looked at as a subset of $mathbb Q$ (ordered by $leq$) it has no supremum either. If in the textbook they say that there is a supremum but no infimum then they are not consistent.
    $endgroup$
    – drhab
    Dec 23 '18 at 9:41






    $begingroup$
    Indeed if $S$ is looked at as a subset of $mathbb Q$ (ordered by $leq$) it has no supremum either. If in the textbook they say that there is a supremum but no infimum then they are not consistent.
    $endgroup$
    – drhab
    Dec 23 '18 at 9:41














    $begingroup$
    Thus, as it is asserted that there exists a supremum, I may conclude that the textbook contradicts itself and is wrong.
    $endgroup$
    – Mew
    Dec 23 '18 at 9:44




    $begingroup$
    Thus, as it is asserted that there exists a supremum, I may conclude that the textbook contradicts itself and is wrong.
    $endgroup$
    – Mew
    Dec 23 '18 at 9:44












    $begingroup$
    I think so, yes.
    $endgroup$
    – drhab
    Dec 23 '18 at 9:44




    $begingroup$
    I think so, yes.
    $endgroup$
    – drhab
    Dec 23 '18 at 9:44











    2












    $begingroup$

    To me it looks like the textbook is wrong, and you have it right. The set does have lower bounds, and therefore a greatest lower bound (among the reals), and that's $-sqrt2$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      To me it looks like the textbook is wrong, and you have it right. The set does have lower bounds, and therefore a greatest lower bound (among the reals), and that's $-sqrt2$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        To me it looks like the textbook is wrong, and you have it right. The set does have lower bounds, and therefore a greatest lower bound (among the reals), and that's $-sqrt2$.






        share|cite|improve this answer









        $endgroup$



        To me it looks like the textbook is wrong, and you have it right. The set does have lower bounds, and therefore a greatest lower bound (among the reals), and that's $-sqrt2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 23 '18 at 9:36









        ArthurArthur

        117k7116200




        117k7116200






























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