Supremum Infimum argument: what did I do wrong?











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Suppose $f:[a,b]tomathbb R$ be a function such that $|f(x)-f(y)|<epsilon_0$ for all $x,yin[a,b]$. Then $M-mleepsilon_0$, where $M=sup{f(x):xin[a,b]}$ and $m=inf{f(x):xin[a,b]}$.




I went this way: suppose $M-m>epsilon_0$, hence $M>epsilon_0+m$, hence there is an $xin[a,b]$ such that$$M>f(x)>epsilon_0+m,tag1$$and similarly $M-epsilon_0>m$ implies$$M-epsilon_0>f(y)>m.tag2$$Subtracting $(1)$ from $(2)$ says $-epsilon_0>f(y)-f(x)>-epsilon_0$, which is absurd.



But on a second thought, I realized that my argument says that $M-m>varepsilon$ is false for any $varepsilon>0$, not just for $epsilon_0$, implying $M-m=0$. What did I do wrong?










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  • I hope that you don't mind about the way I've edited your question.
    – José Carlos Santos
    Dec 4 at 8:58










  • Also, you only use the fact that $M$ is an upper-bound, not a supremum. Same remark for $m$. To solve you exercise, remark that for all $x,yin [a,b]$ $f(y)-varepsilon_0 <f(x)<f(y)+varepsilon _0$. The conclusion is straightforward.
    – Surb
    Dec 4 at 8:58












  • @JoséCarlosSantos, no, in fact, this seems awesome. Thank you
    – Silent
    Dec 4 at 8:59










  • An additional minor error in your proof (unrelated to the problem you noticed): You are guaranteed an $x$ such that $M ge f(x) > epsilon_0 + m$, but you are not guaranteed that $M > f(x)$. It could be that $f(x_0) = M$ and for every other $x ne x_0, f(x) < m + epsilon_0$.
    – Paul Sinclair
    Dec 4 at 17:24















up vote
2
down vote

favorite
2













Suppose $f:[a,b]tomathbb R$ be a function such that $|f(x)-f(y)|<epsilon_0$ for all $x,yin[a,b]$. Then $M-mleepsilon_0$, where $M=sup{f(x):xin[a,b]}$ and $m=inf{f(x):xin[a,b]}$.




I went this way: suppose $M-m>epsilon_0$, hence $M>epsilon_0+m$, hence there is an $xin[a,b]$ such that$$M>f(x)>epsilon_0+m,tag1$$and similarly $M-epsilon_0>m$ implies$$M-epsilon_0>f(y)>m.tag2$$Subtracting $(1)$ from $(2)$ says $-epsilon_0>f(y)-f(x)>-epsilon_0$, which is absurd.



But on a second thought, I realized that my argument says that $M-m>varepsilon$ is false for any $varepsilon>0$, not just for $epsilon_0$, implying $M-m=0$. What did I do wrong?










share|cite|improve this question
























  • I hope that you don't mind about the way I've edited your question.
    – José Carlos Santos
    Dec 4 at 8:58










  • Also, you only use the fact that $M$ is an upper-bound, not a supremum. Same remark for $m$. To solve you exercise, remark that for all $x,yin [a,b]$ $f(y)-varepsilon_0 <f(x)<f(y)+varepsilon _0$. The conclusion is straightforward.
    – Surb
    Dec 4 at 8:58












  • @JoséCarlosSantos, no, in fact, this seems awesome. Thank you
    – Silent
    Dec 4 at 8:59










  • An additional minor error in your proof (unrelated to the problem you noticed): You are guaranteed an $x$ such that $M ge f(x) > epsilon_0 + m$, but you are not guaranteed that $M > f(x)$. It could be that $f(x_0) = M$ and for every other $x ne x_0, f(x) < m + epsilon_0$.
    – Paul Sinclair
    Dec 4 at 17:24













up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2






Suppose $f:[a,b]tomathbb R$ be a function such that $|f(x)-f(y)|<epsilon_0$ for all $x,yin[a,b]$. Then $M-mleepsilon_0$, where $M=sup{f(x):xin[a,b]}$ and $m=inf{f(x):xin[a,b]}$.




I went this way: suppose $M-m>epsilon_0$, hence $M>epsilon_0+m$, hence there is an $xin[a,b]$ such that$$M>f(x)>epsilon_0+m,tag1$$and similarly $M-epsilon_0>m$ implies$$M-epsilon_0>f(y)>m.tag2$$Subtracting $(1)$ from $(2)$ says $-epsilon_0>f(y)-f(x)>-epsilon_0$, which is absurd.



But on a second thought, I realized that my argument says that $M-m>varepsilon$ is false for any $varepsilon>0$, not just for $epsilon_0$, implying $M-m=0$. What did I do wrong?










share|cite|improve this question
















Suppose $f:[a,b]tomathbb R$ be a function such that $|f(x)-f(y)|<epsilon_0$ for all $x,yin[a,b]$. Then $M-mleepsilon_0$, where $M=sup{f(x):xin[a,b]}$ and $m=inf{f(x):xin[a,b]}$.




I went this way: suppose $M-m>epsilon_0$, hence $M>epsilon_0+m$, hence there is an $xin[a,b]$ such that$$M>f(x)>epsilon_0+m,tag1$$and similarly $M-epsilon_0>m$ implies$$M-epsilon_0>f(y)>m.tag2$$Subtracting $(1)$ from $(2)$ says $-epsilon_0>f(y)-f(x)>-epsilon_0$, which is absurd.



But on a second thought, I realized that my argument says that $M-m>varepsilon$ is false for any $varepsilon>0$, not just for $epsilon_0$, implying $M-m=0$. What did I do wrong?







real-analysis proof-verification supremum-and-infimum






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edited Dec 4 at 9:48









Glorfindel

3,38471730




3,38471730










asked Dec 4 at 8:51









Silent

2,63732050




2,63732050












  • I hope that you don't mind about the way I've edited your question.
    – José Carlos Santos
    Dec 4 at 8:58










  • Also, you only use the fact that $M$ is an upper-bound, not a supremum. Same remark for $m$. To solve you exercise, remark that for all $x,yin [a,b]$ $f(y)-varepsilon_0 <f(x)<f(y)+varepsilon _0$. The conclusion is straightforward.
    – Surb
    Dec 4 at 8:58












  • @JoséCarlosSantos, no, in fact, this seems awesome. Thank you
    – Silent
    Dec 4 at 8:59










  • An additional minor error in your proof (unrelated to the problem you noticed): You are guaranteed an $x$ such that $M ge f(x) > epsilon_0 + m$, but you are not guaranteed that $M > f(x)$. It could be that $f(x_0) = M$ and for every other $x ne x_0, f(x) < m + epsilon_0$.
    – Paul Sinclair
    Dec 4 at 17:24


















  • I hope that you don't mind about the way I've edited your question.
    – José Carlos Santos
    Dec 4 at 8:58










  • Also, you only use the fact that $M$ is an upper-bound, not a supremum. Same remark for $m$. To solve you exercise, remark that for all $x,yin [a,b]$ $f(y)-varepsilon_0 <f(x)<f(y)+varepsilon _0$. The conclusion is straightforward.
    – Surb
    Dec 4 at 8:58












  • @JoséCarlosSantos, no, in fact, this seems awesome. Thank you
    – Silent
    Dec 4 at 8:59










  • An additional minor error in your proof (unrelated to the problem you noticed): You are guaranteed an $x$ such that $M ge f(x) > epsilon_0 + m$, but you are not guaranteed that $M > f(x)$. It could be that $f(x_0) = M$ and for every other $x ne x_0, f(x) < m + epsilon_0$.
    – Paul Sinclair
    Dec 4 at 17:24
















I hope that you don't mind about the way I've edited your question.
– José Carlos Santos
Dec 4 at 8:58




I hope that you don't mind about the way I've edited your question.
– José Carlos Santos
Dec 4 at 8:58












Also, you only use the fact that $M$ is an upper-bound, not a supremum. Same remark for $m$. To solve you exercise, remark that for all $x,yin [a,b]$ $f(y)-varepsilon_0 <f(x)<f(y)+varepsilon _0$. The conclusion is straightforward.
– Surb
Dec 4 at 8:58






Also, you only use the fact that $M$ is an upper-bound, not a supremum. Same remark for $m$. To solve you exercise, remark that for all $x,yin [a,b]$ $f(y)-varepsilon_0 <f(x)<f(y)+varepsilon _0$. The conclusion is straightforward.
– Surb
Dec 4 at 8:58














@JoséCarlosSantos, no, in fact, this seems awesome. Thank you
– Silent
Dec 4 at 8:59




@JoséCarlosSantos, no, in fact, this seems awesome. Thank you
– Silent
Dec 4 at 8:59












An additional minor error in your proof (unrelated to the problem you noticed): You are guaranteed an $x$ such that $M ge f(x) > epsilon_0 + m$, but you are not guaranteed that $M > f(x)$. It could be that $f(x_0) = M$ and for every other $x ne x_0, f(x) < m + epsilon_0$.
– Paul Sinclair
Dec 4 at 17:24




An additional minor error in your proof (unrelated to the problem you noticed): You are guaranteed an $x$ such that $M ge f(x) > epsilon_0 + m$, but you are not guaranteed that $M > f(x)$. It could be that $f(x_0) = M$ and for every other $x ne x_0, f(x) < m + epsilon_0$.
– Paul Sinclair
Dec 4 at 17:24










3 Answers
3






active

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up vote
8
down vote



accepted










Your mistake is thinking you can just subtract inequalities. You can't do that. For example,



$$1>0$$



is true, and $$2>0$$ is also true, but subtracting these two inequalities, I get $$-1>0$$ which is not true.





The problem with subtracting inequalities is that when you subtract equations, you actually multiply one of them by $(-1)$ and then add them. With inequalities, you cannot do that because multiplication by a negative number reverses the inequality.





For an actual proof, a sketch of it would be this:




  • Find some $x$ for which $f(x)$ is "near" $M$

  • Find some $y$ for which $f(y)$ is "near" $m$

  • Use the fact that $|f(x)-f(y)|<epsilon_0$ and the fact that $$|f(x)-f(y)| = |f(x)-M+M-f(y)+m-m|leq |f(x)-M| + |f(y)-m| + |M-m|$$ to reach a conclusion.


Naturally, the "near" in this sketch must, in the final proof, be a more rigorous statement. Good luck!






share|cite|improve this answer






























    up vote
    3
    down vote













    Last step is wrong. $aleq bleq c$ and $a'leq b'leq c'$ do not imply $a-a'leq b-b'leq c-c'$.






    share|cite|improve this answer




























      up vote
      1
      down vote













      You may also proceed as follows:




      • Note that $|x|$ is continuous.

      • Choose sequences $(x_n), (y_n)$ with $lim_{n to infty} f(x_n) = M$ and $lim_{n to infty} f(y_n) = m$.


      It follows:
      $$|f(x_n) - f(y_n)| stackrel{n to infty}{longrightarrow} M-m$$
      Now, as $|f(x_n) - f(y_n)| < epsilon_0 Rightarrow M-m leq epsilon_0$.






      share|cite|improve this answer





















        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        8
        down vote



        accepted










        Your mistake is thinking you can just subtract inequalities. You can't do that. For example,



        $$1>0$$



        is true, and $$2>0$$ is also true, but subtracting these two inequalities, I get $$-1>0$$ which is not true.





        The problem with subtracting inequalities is that when you subtract equations, you actually multiply one of them by $(-1)$ and then add them. With inequalities, you cannot do that because multiplication by a negative number reverses the inequality.





        For an actual proof, a sketch of it would be this:




        • Find some $x$ for which $f(x)$ is "near" $M$

        • Find some $y$ for which $f(y)$ is "near" $m$

        • Use the fact that $|f(x)-f(y)|<epsilon_0$ and the fact that $$|f(x)-f(y)| = |f(x)-M+M-f(y)+m-m|leq |f(x)-M| + |f(y)-m| + |M-m|$$ to reach a conclusion.


        Naturally, the "near" in this sketch must, in the final proof, be a more rigorous statement. Good luck!






        share|cite|improve this answer



























          up vote
          8
          down vote



          accepted










          Your mistake is thinking you can just subtract inequalities. You can't do that. For example,



          $$1>0$$



          is true, and $$2>0$$ is also true, but subtracting these two inequalities, I get $$-1>0$$ which is not true.





          The problem with subtracting inequalities is that when you subtract equations, you actually multiply one of them by $(-1)$ and then add them. With inequalities, you cannot do that because multiplication by a negative number reverses the inequality.





          For an actual proof, a sketch of it would be this:




          • Find some $x$ for which $f(x)$ is "near" $M$

          • Find some $y$ for which $f(y)$ is "near" $m$

          • Use the fact that $|f(x)-f(y)|<epsilon_0$ and the fact that $$|f(x)-f(y)| = |f(x)-M+M-f(y)+m-m|leq |f(x)-M| + |f(y)-m| + |M-m|$$ to reach a conclusion.


          Naturally, the "near" in this sketch must, in the final proof, be a more rigorous statement. Good luck!






          share|cite|improve this answer

























            up vote
            8
            down vote



            accepted







            up vote
            8
            down vote



            accepted






            Your mistake is thinking you can just subtract inequalities. You can't do that. For example,



            $$1>0$$



            is true, and $$2>0$$ is also true, but subtracting these two inequalities, I get $$-1>0$$ which is not true.





            The problem with subtracting inequalities is that when you subtract equations, you actually multiply one of them by $(-1)$ and then add them. With inequalities, you cannot do that because multiplication by a negative number reverses the inequality.





            For an actual proof, a sketch of it would be this:




            • Find some $x$ for which $f(x)$ is "near" $M$

            • Find some $y$ for which $f(y)$ is "near" $m$

            • Use the fact that $|f(x)-f(y)|<epsilon_0$ and the fact that $$|f(x)-f(y)| = |f(x)-M+M-f(y)+m-m|leq |f(x)-M| + |f(y)-m| + |M-m|$$ to reach a conclusion.


            Naturally, the "near" in this sketch must, in the final proof, be a more rigorous statement. Good luck!






            share|cite|improve this answer














            Your mistake is thinking you can just subtract inequalities. You can't do that. For example,



            $$1>0$$



            is true, and $$2>0$$ is also true, but subtracting these two inequalities, I get $$-1>0$$ which is not true.





            The problem with subtracting inequalities is that when you subtract equations, you actually multiply one of them by $(-1)$ and then add them. With inequalities, you cannot do that because multiplication by a negative number reverses the inequality.





            For an actual proof, a sketch of it would be this:




            • Find some $x$ for which $f(x)$ is "near" $M$

            • Find some $y$ for which $f(y)$ is "near" $m$

            • Use the fact that $|f(x)-f(y)|<epsilon_0$ and the fact that $$|f(x)-f(y)| = |f(x)-M+M-f(y)+m-m|leq |f(x)-M| + |f(y)-m| + |M-m|$$ to reach a conclusion.


            Naturally, the "near" in this sketch must, in the final proof, be a more rigorous statement. Good luck!







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 4 at 9:02

























            answered Dec 4 at 8:57









            5xum

            89.1k393160




            89.1k393160






















                up vote
                3
                down vote













                Last step is wrong. $aleq bleq c$ and $a'leq b'leq c'$ do not imply $a-a'leq b-b'leq c-c'$.






                share|cite|improve this answer

























                  up vote
                  3
                  down vote













                  Last step is wrong. $aleq bleq c$ and $a'leq b'leq c'$ do not imply $a-a'leq b-b'leq c-c'$.






                  share|cite|improve this answer























                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    Last step is wrong. $aleq bleq c$ and $a'leq b'leq c'$ do not imply $a-a'leq b-b'leq c-c'$.






                    share|cite|improve this answer












                    Last step is wrong. $aleq bleq c$ and $a'leq b'leq c'$ do not imply $a-a'leq b-b'leq c-c'$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 4 at 8:57









                    Kavi Rama Murthy

                    45.9k31854




                    45.9k31854






















                        up vote
                        1
                        down vote













                        You may also proceed as follows:




                        • Note that $|x|$ is continuous.

                        • Choose sequences $(x_n), (y_n)$ with $lim_{n to infty} f(x_n) = M$ and $lim_{n to infty} f(y_n) = m$.


                        It follows:
                        $$|f(x_n) - f(y_n)| stackrel{n to infty}{longrightarrow} M-m$$
                        Now, as $|f(x_n) - f(y_n)| < epsilon_0 Rightarrow M-m leq epsilon_0$.






                        share|cite|improve this answer

























                          up vote
                          1
                          down vote













                          You may also proceed as follows:




                          • Note that $|x|$ is continuous.

                          • Choose sequences $(x_n), (y_n)$ with $lim_{n to infty} f(x_n) = M$ and $lim_{n to infty} f(y_n) = m$.


                          It follows:
                          $$|f(x_n) - f(y_n)| stackrel{n to infty}{longrightarrow} M-m$$
                          Now, as $|f(x_n) - f(y_n)| < epsilon_0 Rightarrow M-m leq epsilon_0$.






                          share|cite|improve this answer























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            You may also proceed as follows:




                            • Note that $|x|$ is continuous.

                            • Choose sequences $(x_n), (y_n)$ with $lim_{n to infty} f(x_n) = M$ and $lim_{n to infty} f(y_n) = m$.


                            It follows:
                            $$|f(x_n) - f(y_n)| stackrel{n to infty}{longrightarrow} M-m$$
                            Now, as $|f(x_n) - f(y_n)| < epsilon_0 Rightarrow M-m leq epsilon_0$.






                            share|cite|improve this answer












                            You may also proceed as follows:




                            • Note that $|x|$ is continuous.

                            • Choose sequences $(x_n), (y_n)$ with $lim_{n to infty} f(x_n) = M$ and $lim_{n to infty} f(y_n) = m$.


                            It follows:
                            $$|f(x_n) - f(y_n)| stackrel{n to infty}{longrightarrow} M-m$$
                            Now, as $|f(x_n) - f(y_n)| < epsilon_0 Rightarrow M-m leq epsilon_0$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 4 at 10:13









                            trancelocation

                            8,7571521




                            8,7571521






























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