How do I show that $2a$ is the diameter of $A=(-a,a)$?












1












$begingroup$


proof. Suppose that $A = (-a,a) = (-a,0]cup [0,a)$ and let $x in [0,a)$ and $yin (-a,0]$ . Then $x + epsilon = a = sup A$ for some $epsilon > 0$ and $y-beta = -a = inf A$ for some $beta >0$. So,



$2a=a-(-a)=x+epsilon -(y-beta)=|x-y| + (epsilon + beta)$. Then



$2a ge |x-y|$.



Would I have to show that $2a$ is the supremum of $|x-y|$ by also showing the cases when $x,y in (-a,0]$ , and $x,y in [0,a)$






real-analysis







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$endgroup$

















    1












    $begingroup$


    proof. Suppose that $A = (-a,a) = (-a,0]cup [0,a)$ and let $x in [0,a)$ and $yin (-a,0]$ . Then $x + epsilon = a = sup A$ for some $epsilon > 0$ and $y-beta = -a = inf A$ for some $beta >0$. So,



    $2a=a-(-a)=x+epsilon -(y-beta)=|x-y| + (epsilon + beta)$. Then



    $2a ge |x-y|$.



    Would I have to show that $2a$ is the supremum of $|x-y|$ by also showing the cases when $x,y in (-a,0]$ , and $x,y in [0,a)$






    real-analysis







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      proof. Suppose that $A = (-a,a) = (-a,0]cup [0,a)$ and let $x in [0,a)$ and $yin (-a,0]$ . Then $x + epsilon = a = sup A$ for some $epsilon > 0$ and $y-beta = -a = inf A$ for some $beta >0$. So,



      $2a=a-(-a)=x+epsilon -(y-beta)=|x-y| + (epsilon + beta)$. Then



      $2a ge |x-y|$.



      Would I have to show that $2a$ is the supremum of $|x-y|$ by also showing the cases when $x,y in (-a,0]$ , and $x,y in [0,a)$






      real-analysis







      share|cite|improve this question









      $endgroup$




      proof. Suppose that $A = (-a,a) = (-a,0]cup [0,a)$ and let $x in [0,a)$ and $yin (-a,0]$ . Then $x + epsilon = a = sup A$ for some $epsilon > 0$ and $y-beta = -a = inf A$ for some $beta >0$. So,



      $2a=a-(-a)=x+epsilon -(y-beta)=|x-y| + (epsilon + beta)$. Then



      $2a ge |x-y|$.



      Would I have to show that $2a$ is the supremum of $|x-y|$ by also showing the cases when $x,y in (-a,0]$ , and $x,y in [0,a)$






      real-analysis




      real-analysis






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Dec 27 '18 at 23:06









      K.MK.M

      734413




      734413






















          3 Answers
          3






          active

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          1












          $begingroup$

          First you can prove that the diameter is $le 2a$ like this. Take any $x,y in (-a,a)$. Now consider two cases.




          • If $x le y$ then $-a < x le y < a$ and so $|x-y|=y-x<a-(-a)=2a$.


          • If $y le x$ then $-a < y le x < a$ and so $|x-y|=x-y<a-(-a)=2a$.



          Next you can prove that the diameter is $ge 2a$ like this. For any $epsilon in (0,a)$ we have $-a < -a+frac{epsilon}{2} < a - frac{epsilon}{2} < a$ and so the diameter is $ge (a - frac{epsilon}{2}) - (-a + frac{epsilon}{2}) = 2a - epsilon$. Since the diameter is $ge 2a-epsilon$ for all $epsilon>0$ it follows that the diameter is $ge 2a$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            For any $x,y in A$ we have $|x-y| < 2a$. [If you to prove this claim more rigorously: Suppose without loss of generality that $x le y$. Then $-a < x le y < a$ so $y-x < a - (-a) = 2a$.] This shows $text{diameter}(A) le 2a$.



            To show $text{diameter}(A) = 2a$, we need to show $sup_{x,y in A} |x-y| = 2a$.
            For any fixed $epsilon > 0$, can you find $x,y in A$ such that $|x-y| ge 2a - epsilon$?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Do we get $text{diameter}(A) le 2a$ since $text{diameter}(A)-epsilon <|y-x|<2a$ and that implies that $text{diameter}(A) le |y-x|<2a$?
              $endgroup$
              – K.M
              Dec 28 '18 at 4:32








            • 1




              $begingroup$
              @K.M We showed that $2a$ is an upper bound for $|x-y|$ for all $x,y in A$. Since the diameter is the supremum (a.k.a. "least upper bound") of these $|x-y|$ values, we must have $text{diameter}(A) le 2a$.
              $endgroup$
              – angryavian
              Dec 28 '18 at 4:47



















            1












            $begingroup$

            You could do, but there are far easier ways. For example, simple note that for $x>yin A$, we have $|x-y| leq |a-x|+|x-y|+|y-(-a)| = |a-(-a)| = 2a$, so $2a$ is an upper bound for the diameter of $A$, and also that for any $varepsilon > 0$ with $delta := min(varepsilon,a)$, if we take $x = a - frac{delta}{2}$ and $y = frac{delta}{2}-a$, we have $x > y in A$, and $|x-y| = |2a-delta| geq 2a - deltageq 2a-varepsilon$, so $2a$ is also a lower bound for the diameter of $A$, hence is equal to the diameter of $A$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I was wondering how you got $|a-(-a)|$ from $|a-x| +|x-y|+|y-(-a)|$?
              $endgroup$
              – K.M
              Dec 28 '18 at 0:04






            • 1




              $begingroup$
              Because $-a < y < x < a$, all three of the terms inside the absolute value signs are positive, so $|a - x| + |x - y| + |y - (-a)| = a - x +x-y+y-(-a) = a-(-a) = |a-(-a)|$.
              $endgroup$
              – user3482749
              Dec 28 '18 at 14:04










            • $begingroup$
              Since $x>y$, I want to make sure that it follows that $|2a-delta|=2a-delta$?
              $endgroup$
              – K.M
              Dec 28 '18 at 18:27






            • 1




              $begingroup$
              Yes, that's true.
              $endgroup$
              – user3482749
              Dec 28 '18 at 19:03












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            3 Answers
            3






            active

            oldest

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            3 Answers
            3






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            First you can prove that the diameter is $le 2a$ like this. Take any $x,y in (-a,a)$. Now consider two cases.




            • If $x le y$ then $-a < x le y < a$ and so $|x-y|=y-x<a-(-a)=2a$.


            • If $y le x$ then $-a < y le x < a$ and so $|x-y|=x-y<a-(-a)=2a$.



            Next you can prove that the diameter is $ge 2a$ like this. For any $epsilon in (0,a)$ we have $-a < -a+frac{epsilon}{2} < a - frac{epsilon}{2} < a$ and so the diameter is $ge (a - frac{epsilon}{2}) - (-a + frac{epsilon}{2}) = 2a - epsilon$. Since the diameter is $ge 2a-epsilon$ for all $epsilon>0$ it follows that the diameter is $ge 2a$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              First you can prove that the diameter is $le 2a$ like this. Take any $x,y in (-a,a)$. Now consider two cases.




              • If $x le y$ then $-a < x le y < a$ and so $|x-y|=y-x<a-(-a)=2a$.


              • If $y le x$ then $-a < y le x < a$ and so $|x-y|=x-y<a-(-a)=2a$.



              Next you can prove that the diameter is $ge 2a$ like this. For any $epsilon in (0,a)$ we have $-a < -a+frac{epsilon}{2} < a - frac{epsilon}{2} < a$ and so the diameter is $ge (a - frac{epsilon}{2}) - (-a + frac{epsilon}{2}) = 2a - epsilon$. Since the diameter is $ge 2a-epsilon$ for all $epsilon>0$ it follows that the diameter is $ge 2a$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                First you can prove that the diameter is $le 2a$ like this. Take any $x,y in (-a,a)$. Now consider two cases.




                • If $x le y$ then $-a < x le y < a$ and so $|x-y|=y-x<a-(-a)=2a$.


                • If $y le x$ then $-a < y le x < a$ and so $|x-y|=x-y<a-(-a)=2a$.



                Next you can prove that the diameter is $ge 2a$ like this. For any $epsilon in (0,a)$ we have $-a < -a+frac{epsilon}{2} < a - frac{epsilon}{2} < a$ and so the diameter is $ge (a - frac{epsilon}{2}) - (-a + frac{epsilon}{2}) = 2a - epsilon$. Since the diameter is $ge 2a-epsilon$ for all $epsilon>0$ it follows that the diameter is $ge 2a$.






                share|cite|improve this answer









                $endgroup$



                First you can prove that the diameter is $le 2a$ like this. Take any $x,y in (-a,a)$. Now consider two cases.




                • If $x le y$ then $-a < x le y < a$ and so $|x-y|=y-x<a-(-a)=2a$.


                • If $y le x$ then $-a < y le x < a$ and so $|x-y|=x-y<a-(-a)=2a$.



                Next you can prove that the diameter is $ge 2a$ like this. For any $epsilon in (0,a)$ we have $-a < -a+frac{epsilon}{2} < a - frac{epsilon}{2} < a$ and so the diameter is $ge (a - frac{epsilon}{2}) - (-a + frac{epsilon}{2}) = 2a - epsilon$. Since the diameter is $ge 2a-epsilon$ for all $epsilon>0$ it follows that the diameter is $ge 2a$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 27 '18 at 23:14









                Lee MosherLee Mosher

                52.4k33891




                52.4k33891























                    1












                    $begingroup$

                    For any $x,y in A$ we have $|x-y| < 2a$. [If you to prove this claim more rigorously: Suppose without loss of generality that $x le y$. Then $-a < x le y < a$ so $y-x < a - (-a) = 2a$.] This shows $text{diameter}(A) le 2a$.



                    To show $text{diameter}(A) = 2a$, we need to show $sup_{x,y in A} |x-y| = 2a$.
                    For any fixed $epsilon > 0$, can you find $x,y in A$ such that $|x-y| ge 2a - epsilon$?






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Do we get $text{diameter}(A) le 2a$ since $text{diameter}(A)-epsilon <|y-x|<2a$ and that implies that $text{diameter}(A) le |y-x|<2a$?
                      $endgroup$
                      – K.M
                      Dec 28 '18 at 4:32








                    • 1




                      $begingroup$
                      @K.M We showed that $2a$ is an upper bound for $|x-y|$ for all $x,y in A$. Since the diameter is the supremum (a.k.a. "least upper bound") of these $|x-y|$ values, we must have $text{diameter}(A) le 2a$.
                      $endgroup$
                      – angryavian
                      Dec 28 '18 at 4:47
















                    1












                    $begingroup$

                    For any $x,y in A$ we have $|x-y| < 2a$. [If you to prove this claim more rigorously: Suppose without loss of generality that $x le y$. Then $-a < x le y < a$ so $y-x < a - (-a) = 2a$.] This shows $text{diameter}(A) le 2a$.



                    To show $text{diameter}(A) = 2a$, we need to show $sup_{x,y in A} |x-y| = 2a$.
                    For any fixed $epsilon > 0$, can you find $x,y in A$ such that $|x-y| ge 2a - epsilon$?






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Do we get $text{diameter}(A) le 2a$ since $text{diameter}(A)-epsilon <|y-x|<2a$ and that implies that $text{diameter}(A) le |y-x|<2a$?
                      $endgroup$
                      – K.M
                      Dec 28 '18 at 4:32








                    • 1




                      $begingroup$
                      @K.M We showed that $2a$ is an upper bound for $|x-y|$ for all $x,y in A$. Since the diameter is the supremum (a.k.a. "least upper bound") of these $|x-y|$ values, we must have $text{diameter}(A) le 2a$.
                      $endgroup$
                      – angryavian
                      Dec 28 '18 at 4:47














                    1












                    1








                    1





                    $begingroup$

                    For any $x,y in A$ we have $|x-y| < 2a$. [If you to prove this claim more rigorously: Suppose without loss of generality that $x le y$. Then $-a < x le y < a$ so $y-x < a - (-a) = 2a$.] This shows $text{diameter}(A) le 2a$.



                    To show $text{diameter}(A) = 2a$, we need to show $sup_{x,y in A} |x-y| = 2a$.
                    For any fixed $epsilon > 0$, can you find $x,y in A$ such that $|x-y| ge 2a - epsilon$?






                    share|cite|improve this answer









                    $endgroup$



                    For any $x,y in A$ we have $|x-y| < 2a$. [If you to prove this claim more rigorously: Suppose without loss of generality that $x le y$. Then $-a < x le y < a$ so $y-x < a - (-a) = 2a$.] This shows $text{diameter}(A) le 2a$.



                    To show $text{diameter}(A) = 2a$, we need to show $sup_{x,y in A} |x-y| = 2a$.
                    For any fixed $epsilon > 0$, can you find $x,y in A$ such that $|x-y| ge 2a - epsilon$?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 27 '18 at 23:12









                    angryavianangryavian

                    42.6k23481




                    42.6k23481












                    • $begingroup$
                      Do we get $text{diameter}(A) le 2a$ since $text{diameter}(A)-epsilon <|y-x|<2a$ and that implies that $text{diameter}(A) le |y-x|<2a$?
                      $endgroup$
                      – K.M
                      Dec 28 '18 at 4:32








                    • 1




                      $begingroup$
                      @K.M We showed that $2a$ is an upper bound for $|x-y|$ for all $x,y in A$. Since the diameter is the supremum (a.k.a. "least upper bound") of these $|x-y|$ values, we must have $text{diameter}(A) le 2a$.
                      $endgroup$
                      – angryavian
                      Dec 28 '18 at 4:47


















                    • $begingroup$
                      Do we get $text{diameter}(A) le 2a$ since $text{diameter}(A)-epsilon <|y-x|<2a$ and that implies that $text{diameter}(A) le |y-x|<2a$?
                      $endgroup$
                      – K.M
                      Dec 28 '18 at 4:32








                    • 1




                      $begingroup$
                      @K.M We showed that $2a$ is an upper bound for $|x-y|$ for all $x,y in A$. Since the diameter is the supremum (a.k.a. "least upper bound") of these $|x-y|$ values, we must have $text{diameter}(A) le 2a$.
                      $endgroup$
                      – angryavian
                      Dec 28 '18 at 4:47
















                    $begingroup$
                    Do we get $text{diameter}(A) le 2a$ since $text{diameter}(A)-epsilon <|y-x|<2a$ and that implies that $text{diameter}(A) le |y-x|<2a$?
                    $endgroup$
                    – K.M
                    Dec 28 '18 at 4:32






                    $begingroup$
                    Do we get $text{diameter}(A) le 2a$ since $text{diameter}(A)-epsilon <|y-x|<2a$ and that implies that $text{diameter}(A) le |y-x|<2a$?
                    $endgroup$
                    – K.M
                    Dec 28 '18 at 4:32






                    1




                    1




                    $begingroup$
                    @K.M We showed that $2a$ is an upper bound for $|x-y|$ for all $x,y in A$. Since the diameter is the supremum (a.k.a. "least upper bound") of these $|x-y|$ values, we must have $text{diameter}(A) le 2a$.
                    $endgroup$
                    – angryavian
                    Dec 28 '18 at 4:47




                    $begingroup$
                    @K.M We showed that $2a$ is an upper bound for $|x-y|$ for all $x,y in A$. Since the diameter is the supremum (a.k.a. "least upper bound") of these $|x-y|$ values, we must have $text{diameter}(A) le 2a$.
                    $endgroup$
                    – angryavian
                    Dec 28 '18 at 4:47











                    1












                    $begingroup$

                    You could do, but there are far easier ways. For example, simple note that for $x>yin A$, we have $|x-y| leq |a-x|+|x-y|+|y-(-a)| = |a-(-a)| = 2a$, so $2a$ is an upper bound for the diameter of $A$, and also that for any $varepsilon > 0$ with $delta := min(varepsilon,a)$, if we take $x = a - frac{delta}{2}$ and $y = frac{delta}{2}-a$, we have $x > y in A$, and $|x-y| = |2a-delta| geq 2a - deltageq 2a-varepsilon$, so $2a$ is also a lower bound for the diameter of $A$, hence is equal to the diameter of $A$.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      I was wondering how you got $|a-(-a)|$ from $|a-x| +|x-y|+|y-(-a)|$?
                      $endgroup$
                      – K.M
                      Dec 28 '18 at 0:04






                    • 1




                      $begingroup$
                      Because $-a < y < x < a$, all three of the terms inside the absolute value signs are positive, so $|a - x| + |x - y| + |y - (-a)| = a - x +x-y+y-(-a) = a-(-a) = |a-(-a)|$.
                      $endgroup$
                      – user3482749
                      Dec 28 '18 at 14:04










                    • $begingroup$
                      Since $x>y$, I want to make sure that it follows that $|2a-delta|=2a-delta$?
                      $endgroup$
                      – K.M
                      Dec 28 '18 at 18:27






                    • 1




                      $begingroup$
                      Yes, that's true.
                      $endgroup$
                      – user3482749
                      Dec 28 '18 at 19:03
















                    1












                    $begingroup$

                    You could do, but there are far easier ways. For example, simple note that for $x>yin A$, we have $|x-y| leq |a-x|+|x-y|+|y-(-a)| = |a-(-a)| = 2a$, so $2a$ is an upper bound for the diameter of $A$, and also that for any $varepsilon > 0$ with $delta := min(varepsilon,a)$, if we take $x = a - frac{delta}{2}$ and $y = frac{delta}{2}-a$, we have $x > y in A$, and $|x-y| = |2a-delta| geq 2a - deltageq 2a-varepsilon$, so $2a$ is also a lower bound for the diameter of $A$, hence is equal to the diameter of $A$.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      I was wondering how you got $|a-(-a)|$ from $|a-x| +|x-y|+|y-(-a)|$?
                      $endgroup$
                      – K.M
                      Dec 28 '18 at 0:04






                    • 1




                      $begingroup$
                      Because $-a < y < x < a$, all three of the terms inside the absolute value signs are positive, so $|a - x| + |x - y| + |y - (-a)| = a - x +x-y+y-(-a) = a-(-a) = |a-(-a)|$.
                      $endgroup$
                      – user3482749
                      Dec 28 '18 at 14:04










                    • $begingroup$
                      Since $x>y$, I want to make sure that it follows that $|2a-delta|=2a-delta$?
                      $endgroup$
                      – K.M
                      Dec 28 '18 at 18:27






                    • 1




                      $begingroup$
                      Yes, that's true.
                      $endgroup$
                      – user3482749
                      Dec 28 '18 at 19:03














                    1












                    1








                    1





                    $begingroup$

                    You could do, but there are far easier ways. For example, simple note that for $x>yin A$, we have $|x-y| leq |a-x|+|x-y|+|y-(-a)| = |a-(-a)| = 2a$, so $2a$ is an upper bound for the diameter of $A$, and also that for any $varepsilon > 0$ with $delta := min(varepsilon,a)$, if we take $x = a - frac{delta}{2}$ and $y = frac{delta}{2}-a$, we have $x > y in A$, and $|x-y| = |2a-delta| geq 2a - deltageq 2a-varepsilon$, so $2a$ is also a lower bound for the diameter of $A$, hence is equal to the diameter of $A$.






                    share|cite|improve this answer









                    $endgroup$



                    You could do, but there are far easier ways. For example, simple note that for $x>yin A$, we have $|x-y| leq |a-x|+|x-y|+|y-(-a)| = |a-(-a)| = 2a$, so $2a$ is an upper bound for the diameter of $A$, and also that for any $varepsilon > 0$ with $delta := min(varepsilon,a)$, if we take $x = a - frac{delta}{2}$ and $y = frac{delta}{2}-a$, we have $x > y in A$, and $|x-y| = |2a-delta| geq 2a - deltageq 2a-varepsilon$, so $2a$ is also a lower bound for the diameter of $A$, hence is equal to the diameter of $A$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 27 '18 at 23:14









                    user3482749user3482749

                    4,3291119




                    4,3291119












                    • $begingroup$
                      I was wondering how you got $|a-(-a)|$ from $|a-x| +|x-y|+|y-(-a)|$?
                      $endgroup$
                      – K.M
                      Dec 28 '18 at 0:04






                    • 1




                      $begingroup$
                      Because $-a < y < x < a$, all three of the terms inside the absolute value signs are positive, so $|a - x| + |x - y| + |y - (-a)| = a - x +x-y+y-(-a) = a-(-a) = |a-(-a)|$.
                      $endgroup$
                      – user3482749
                      Dec 28 '18 at 14:04










                    • $begingroup$
                      Since $x>y$, I want to make sure that it follows that $|2a-delta|=2a-delta$?
                      $endgroup$
                      – K.M
                      Dec 28 '18 at 18:27






                    • 1




                      $begingroup$
                      Yes, that's true.
                      $endgroup$
                      – user3482749
                      Dec 28 '18 at 19:03


















                    • $begingroup$
                      I was wondering how you got $|a-(-a)|$ from $|a-x| +|x-y|+|y-(-a)|$?
                      $endgroup$
                      – K.M
                      Dec 28 '18 at 0:04






                    • 1




                      $begingroup$
                      Because $-a < y < x < a$, all three of the terms inside the absolute value signs are positive, so $|a - x| + |x - y| + |y - (-a)| = a - x +x-y+y-(-a) = a-(-a) = |a-(-a)|$.
                      $endgroup$
                      – user3482749
                      Dec 28 '18 at 14:04










                    • $begingroup$
                      Since $x>y$, I want to make sure that it follows that $|2a-delta|=2a-delta$?
                      $endgroup$
                      – K.M
                      Dec 28 '18 at 18:27






                    • 1




                      $begingroup$
                      Yes, that's true.
                      $endgroup$
                      – user3482749
                      Dec 28 '18 at 19:03
















                    $begingroup$
                    I was wondering how you got $|a-(-a)|$ from $|a-x| +|x-y|+|y-(-a)|$?
                    $endgroup$
                    – K.M
                    Dec 28 '18 at 0:04




                    $begingroup$
                    I was wondering how you got $|a-(-a)|$ from $|a-x| +|x-y|+|y-(-a)|$?
                    $endgroup$
                    – K.M
                    Dec 28 '18 at 0:04




                    1




                    1




                    $begingroup$
                    Because $-a < y < x < a$, all three of the terms inside the absolute value signs are positive, so $|a - x| + |x - y| + |y - (-a)| = a - x +x-y+y-(-a) = a-(-a) = |a-(-a)|$.
                    $endgroup$
                    – user3482749
                    Dec 28 '18 at 14:04




                    $begingroup$
                    Because $-a < y < x < a$, all three of the terms inside the absolute value signs are positive, so $|a - x| + |x - y| + |y - (-a)| = a - x +x-y+y-(-a) = a-(-a) = |a-(-a)|$.
                    $endgroup$
                    – user3482749
                    Dec 28 '18 at 14:04












                    $begingroup$
                    Since $x>y$, I want to make sure that it follows that $|2a-delta|=2a-delta$?
                    $endgroup$
                    – K.M
                    Dec 28 '18 at 18:27




                    $begingroup$
                    Since $x>y$, I want to make sure that it follows that $|2a-delta|=2a-delta$?
                    $endgroup$
                    – K.M
                    Dec 28 '18 at 18:27




                    1




                    1




                    $begingroup$
                    Yes, that's true.
                    $endgroup$
                    – user3482749
                    Dec 28 '18 at 19:03




                    $begingroup$
                    Yes, that's true.
                    $endgroup$
                    – user3482749
                    Dec 28 '18 at 19:03


















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