$mathbb{Q}$ with the metric $d(p,q) = |p-q|$












0












$begingroup$



Consider $mathbb{Q}$ ,the set of rational number , with metric $d(p,q) = |p-q|$.then which of the following statement is true ?



$a)$${q in mathbb{Q} : 2< q^2 < 3 }$ is closed



$b)$ ${ q in mathbb{Q} : 2 le q^2 le 4}$ is compact



$c)$ ${ qin mathbb{Q} : 2 le q^2 le 4 }$ is closed





My attempt : I think option $b)$ and option $c)$ is correct by Heine Boral theorem : closed + bounded = compact



Is this True ?



Any hints/solution will be appreciated



Thank you!






general-topology







share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    Consider $mathbb{Q}$ ,the set of rational number , with metric $d(p,q) = |p-q|$.then which of the following statement is true ?



    $a)$${q in mathbb{Q} : 2< q^2 < 3 }$ is closed



    $b)$ ${ q in mathbb{Q} : 2 le q^2 le 4}$ is compact



    $c)$ ${ qin mathbb{Q} : 2 le q^2 le 4 }$ is closed





    My attempt : I think option $b)$ and option $c)$ is correct by Heine Boral theorem : closed + bounded = compact



    Is this True ?



    Any hints/solution will be appreciated



    Thank you!






    general-topology







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Consider $mathbb{Q}$ ,the set of rational number , with metric $d(p,q) = |p-q|$.then which of the following statement is true ?



      $a)$${q in mathbb{Q} : 2< q^2 < 3 }$ is closed



      $b)$ ${ q in mathbb{Q} : 2 le q^2 le 4}$ is compact



      $c)$ ${ qin mathbb{Q} : 2 le q^2 le 4 }$ is closed





      My attempt : I think option $b)$ and option $c)$ is correct by Heine Boral theorem : closed + bounded = compact



      Is this True ?



      Any hints/solution will be appreciated



      Thank you!






      general-topology







      share|cite|improve this question











      $endgroup$





      Consider $mathbb{Q}$ ,the set of rational number , with metric $d(p,q) = |p-q|$.then which of the following statement is true ?



      $a)$${q in mathbb{Q} : 2< q^2 < 3 }$ is closed



      $b)$ ${ q in mathbb{Q} : 2 le q^2 le 4}$ is compact



      $c)$ ${ qin mathbb{Q} : 2 le q^2 le 4 }$ is closed





      My attempt : I think option $b)$ and option $c)$ is correct by Heine Boral theorem : closed + bounded = compact



      Is this True ?



      Any hints/solution will be appreciated



      Thank you!






      general-topology




      general-topology






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 '18 at 5:20









      Chinnapparaj R

      5,3131828




      5,3131828










      asked Dec 1 '18 at 3:02









      jasminejasmine

      1,674416




      1,674416






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$


          Hints.



          (a) True. First try to check whether the set ${q in Bbb{Q}|sqrt 2< q< sqrt 3}$ is closed or not. Well, it is closed in $Bbb{Q}$, w.r.to the subspace topology from $Bbb{R}$. Since see that, ${q in Bbb{Q}|sqrt 2< q< sqrt 3}=[sqrt 2, sqrt 3]cap Bbb{Q}$ where $[sqrt 2, sqrt 3]$ is closed in $Bbb{R}$.



          (b) False. See that there is a sequence in this set that converges to $sqrt 2$, which is not in the set (because of density property of $Bbb{Q}$ in that closed interval). So, the set is not even Sequentially Compact.



          (c) True. Same logic as in (a). Think about the set $[sqrt 2, 2]cap Bbb{Q}$







          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Hints:



            The Heine-Borel Theorem only applies to $mathbb R$. Since $mathbb Q$ is a different topological space, you cannot characterize the compact subspaces of $mathbb Q$ using the same criteria. Instead, use the fact that a compact metric space must be complete. Or you could use the fact that every sequence in a compact set has a convergent subsequence. Can you find a sequence in the set in (b) that does not have a convergent subsequence?



            For a set to be closed, its complement must be open. What do the complements of the sets in (a) and (c) look like? Are they open in $mathbb Q$?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              im not getting how can we find complement
              $endgroup$
              – jasmine
              Dec 1 '18 at 3:24






            • 1




              $begingroup$
              For (a), the set is all the rational numbers whose square is between 2 and 3. The complement is all rational numbers whose square is less than 2 or greater than 3. Note that the square roots of 2 and 3 are not rational, so the complement of this set does not contain its endpoints.
              $endgroup$
              – Alex S
              Dec 1 '18 at 3:29











            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$


            Hints.



            (a) True. First try to check whether the set ${q in Bbb{Q}|sqrt 2< q< sqrt 3}$ is closed or not. Well, it is closed in $Bbb{Q}$, w.r.to the subspace topology from $Bbb{R}$. Since see that, ${q in Bbb{Q}|sqrt 2< q< sqrt 3}=[sqrt 2, sqrt 3]cap Bbb{Q}$ where $[sqrt 2, sqrt 3]$ is closed in $Bbb{R}$.



            (b) False. See that there is a sequence in this set that converges to $sqrt 2$, which is not in the set (because of density property of $Bbb{Q}$ in that closed interval). So, the set is not even Sequentially Compact.



            (c) True. Same logic as in (a). Think about the set $[sqrt 2, 2]cap Bbb{Q}$







            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$


              Hints.



              (a) True. First try to check whether the set ${q in Bbb{Q}|sqrt 2< q< sqrt 3}$ is closed or not. Well, it is closed in $Bbb{Q}$, w.r.to the subspace topology from $Bbb{R}$. Since see that, ${q in Bbb{Q}|sqrt 2< q< sqrt 3}=[sqrt 2, sqrt 3]cap Bbb{Q}$ where $[sqrt 2, sqrt 3]$ is closed in $Bbb{R}$.



              (b) False. See that there is a sequence in this set that converges to $sqrt 2$, which is not in the set (because of density property of $Bbb{Q}$ in that closed interval). So, the set is not even Sequentially Compact.



              (c) True. Same logic as in (a). Think about the set $[sqrt 2, 2]cap Bbb{Q}$







              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$


                Hints.



                (a) True. First try to check whether the set ${q in Bbb{Q}|sqrt 2< q< sqrt 3}$ is closed or not. Well, it is closed in $Bbb{Q}$, w.r.to the subspace topology from $Bbb{R}$. Since see that, ${q in Bbb{Q}|sqrt 2< q< sqrt 3}=[sqrt 2, sqrt 3]cap Bbb{Q}$ where $[sqrt 2, sqrt 3]$ is closed in $Bbb{R}$.



                (b) False. See that there is a sequence in this set that converges to $sqrt 2$, which is not in the set (because of density property of $Bbb{Q}$ in that closed interval). So, the set is not even Sequentially Compact.



                (c) True. Same logic as in (a). Think about the set $[sqrt 2, 2]cap Bbb{Q}$







                share|cite|improve this answer









                $endgroup$




                Hints.



                (a) True. First try to check whether the set ${q in Bbb{Q}|sqrt 2< q< sqrt 3}$ is closed or not. Well, it is closed in $Bbb{Q}$, w.r.to the subspace topology from $Bbb{R}$. Since see that, ${q in Bbb{Q}|sqrt 2< q< sqrt 3}=[sqrt 2, sqrt 3]cap Bbb{Q}$ where $[sqrt 2, sqrt 3]$ is closed in $Bbb{R}$.



                (b) False. See that there is a sequence in this set that converges to $sqrt 2$, which is not in the set (because of density property of $Bbb{Q}$ in that closed interval). So, the set is not even Sequentially Compact.



                (c) True. Same logic as in (a). Think about the set $[sqrt 2, 2]cap Bbb{Q}$








                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 1 '18 at 13:31









                Indrajit GhoshIndrajit Ghosh

                1,0591718




                1,0591718























                    2












                    $begingroup$

                    Hints:



                    The Heine-Borel Theorem only applies to $mathbb R$. Since $mathbb Q$ is a different topological space, you cannot characterize the compact subspaces of $mathbb Q$ using the same criteria. Instead, use the fact that a compact metric space must be complete. Or you could use the fact that every sequence in a compact set has a convergent subsequence. Can you find a sequence in the set in (b) that does not have a convergent subsequence?



                    For a set to be closed, its complement must be open. What do the complements of the sets in (a) and (c) look like? Are they open in $mathbb Q$?






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      im not getting how can we find complement
                      $endgroup$
                      – jasmine
                      Dec 1 '18 at 3:24






                    • 1




                      $begingroup$
                      For (a), the set is all the rational numbers whose square is between 2 and 3. The complement is all rational numbers whose square is less than 2 or greater than 3. Note that the square roots of 2 and 3 are not rational, so the complement of this set does not contain its endpoints.
                      $endgroup$
                      – Alex S
                      Dec 1 '18 at 3:29
















                    2












                    $begingroup$

                    Hints:



                    The Heine-Borel Theorem only applies to $mathbb R$. Since $mathbb Q$ is a different topological space, you cannot characterize the compact subspaces of $mathbb Q$ using the same criteria. Instead, use the fact that a compact metric space must be complete. Or you could use the fact that every sequence in a compact set has a convergent subsequence. Can you find a sequence in the set in (b) that does not have a convergent subsequence?



                    For a set to be closed, its complement must be open. What do the complements of the sets in (a) and (c) look like? Are they open in $mathbb Q$?






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      im not getting how can we find complement
                      $endgroup$
                      – jasmine
                      Dec 1 '18 at 3:24






                    • 1




                      $begingroup$
                      For (a), the set is all the rational numbers whose square is between 2 and 3. The complement is all rational numbers whose square is less than 2 or greater than 3. Note that the square roots of 2 and 3 are not rational, so the complement of this set does not contain its endpoints.
                      $endgroup$
                      – Alex S
                      Dec 1 '18 at 3:29














                    2












                    2








                    2





                    $begingroup$

                    Hints:



                    The Heine-Borel Theorem only applies to $mathbb R$. Since $mathbb Q$ is a different topological space, you cannot characterize the compact subspaces of $mathbb Q$ using the same criteria. Instead, use the fact that a compact metric space must be complete. Or you could use the fact that every sequence in a compact set has a convergent subsequence. Can you find a sequence in the set in (b) that does not have a convergent subsequence?



                    For a set to be closed, its complement must be open. What do the complements of the sets in (a) and (c) look like? Are they open in $mathbb Q$?






                    share|cite|improve this answer









                    $endgroup$



                    Hints:



                    The Heine-Borel Theorem only applies to $mathbb R$. Since $mathbb Q$ is a different topological space, you cannot characterize the compact subspaces of $mathbb Q$ using the same criteria. Instead, use the fact that a compact metric space must be complete. Or you could use the fact that every sequence in a compact set has a convergent subsequence. Can you find a sequence in the set in (b) that does not have a convergent subsequence?



                    For a set to be closed, its complement must be open. What do the complements of the sets in (a) and (c) look like? Are they open in $mathbb Q$?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 1 '18 at 3:13









                    Alex SAlex S

                    17.9k12160




                    17.9k12160












                    • $begingroup$
                      im not getting how can we find complement
                      $endgroup$
                      – jasmine
                      Dec 1 '18 at 3:24






                    • 1




                      $begingroup$
                      For (a), the set is all the rational numbers whose square is between 2 and 3. The complement is all rational numbers whose square is less than 2 or greater than 3. Note that the square roots of 2 and 3 are not rational, so the complement of this set does not contain its endpoints.
                      $endgroup$
                      – Alex S
                      Dec 1 '18 at 3:29


















                    • $begingroup$
                      im not getting how can we find complement
                      $endgroup$
                      – jasmine
                      Dec 1 '18 at 3:24






                    • 1




                      $begingroup$
                      For (a), the set is all the rational numbers whose square is between 2 and 3. The complement is all rational numbers whose square is less than 2 or greater than 3. Note that the square roots of 2 and 3 are not rational, so the complement of this set does not contain its endpoints.
                      $endgroup$
                      – Alex S
                      Dec 1 '18 at 3:29
















                    $begingroup$
                    im not getting how can we find complement
                    $endgroup$
                    – jasmine
                    Dec 1 '18 at 3:24




                    $begingroup$
                    im not getting how can we find complement
                    $endgroup$
                    – jasmine
                    Dec 1 '18 at 3:24




                    1




                    1




                    $begingroup$
                    For (a), the set is all the rational numbers whose square is between 2 and 3. The complement is all rational numbers whose square is less than 2 or greater than 3. Note that the square roots of 2 and 3 are not rational, so the complement of this set does not contain its endpoints.
                    $endgroup$
                    – Alex S
                    Dec 1 '18 at 3:29




                    $begingroup$
                    For (a), the set is all the rational numbers whose square is between 2 and 3. The complement is all rational numbers whose square is less than 2 or greater than 3. Note that the square roots of 2 and 3 are not rational, so the complement of this set does not contain its endpoints.
                    $endgroup$
                    – Alex S
                    Dec 1 '18 at 3:29


















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