Supremum Infimum argument: what did I do wrong?












2












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    I hope that you don't mind about the way I've edited your question.
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 8:58










  • $begingroup$
    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.
    $endgroup$
    – Surb
    Dec 4 '18 at 8:58












  • $begingroup$
    @JoséCarlosSantos, no, in fact, this seems awesome. Thank you
    $endgroup$
    – Silent
    Dec 4 '18 at 8:59










  • $begingroup$
    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 $x ne x_0, f(x) < m + epsilon_0$.
    $endgroup$
    – Paul Sinclair
    Dec 4 '18 at 17:24
















2












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    I hope that you don't mind about the way I've edited your question.
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 8:58










  • $begingroup$
    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.
    $endgroup$
    – Surb
    Dec 4 '18 at 8:58












  • $begingroup$
    @JoséCarlosSantos, no, in fact, this seems awesome. Thank you
    $endgroup$
    – Silent
    Dec 4 '18 at 8:59










  • $begingroup$
    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 $x ne x_0, f(x) < m + epsilon_0$.
    $endgroup$
    – Paul Sinclair
    Dec 4 '18 at 17:24














2












2








2


2



$begingroup$



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







share|cite|improve this question











$endgroup$





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




real-analysis proof-verification supremum-and-infimum






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 4 '18 at 9:48









Glorfindel

3,41981830




3,41981830










asked Dec 4 '18 at 8:51









SilentSilent

2,75832150




2,75832150












  • $begingroup$
    I hope that you don't mind about the way I've edited your question.
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 8:58










  • $begingroup$
    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.
    $endgroup$
    – Surb
    Dec 4 '18 at 8:58












  • $begingroup$
    @JoséCarlosSantos, no, in fact, this seems awesome. Thank you
    $endgroup$
    – Silent
    Dec 4 '18 at 8:59










  • $begingroup$
    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 $x ne x_0, f(x) < m + epsilon_0$.
    $endgroup$
    – Paul Sinclair
    Dec 4 '18 at 17:24


















  • $begingroup$
    I hope that you don't mind about the way I've edited your question.
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 8:58










  • $begingroup$
    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.
    $endgroup$
    – Surb
    Dec 4 '18 at 8:58












  • $begingroup$
    @JoséCarlosSantos, no, in fact, this seems awesome. Thank you
    $endgroup$
    – Silent
    Dec 4 '18 at 8:59










  • $begingroup$
    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 $x ne x_0, f(x) < m + epsilon_0$.
    $endgroup$
    – Paul Sinclair
    Dec 4 '18 at 17:24
















$begingroup$
I hope that you don't mind about the way I've edited your question.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 8:58




$begingroup$
I hope that you don't mind about the way I've edited your question.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 8:58












$begingroup$
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.
$endgroup$
– Surb
Dec 4 '18 at 8:58






$begingroup$
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.
$endgroup$
– Surb
Dec 4 '18 at 8:58














$begingroup$
@JoséCarlosSantos, no, in fact, this seems awesome. Thank you
$endgroup$
– Silent
Dec 4 '18 at 8:59




$begingroup$
@JoséCarlosSantos, no, in fact, this seems awesome. Thank you
$endgroup$
– Silent
Dec 4 '18 at 8:59












$begingroup$
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 $x ne x_0, f(x) < m + epsilon_0$.
$endgroup$
– Paul Sinclair
Dec 4 '18 at 17:24




$begingroup$
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 $x ne x_0, f(x) < m + epsilon_0$.
$endgroup$
– Paul Sinclair
Dec 4 '18 at 17:24










3 Answers
3






active

oldest

votes


















8












$begingroup$

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











$endgroup$





















    3












    $begingroup$

    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









    $endgroup$





















      1












      $begingroup$

      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









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        8












        $begingroup$

        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











        $endgroup$


















          8












          $begingroup$

          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











          $endgroup$
















            8












            8








            8





            $begingroup$

            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











            $endgroup$



            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 '18 at 9:02

























            answered Dec 4 '18 at 8:57









            5xum5xum

            90.2k393161




            90.2k393161























                3












                $begingroup$

                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









                $endgroup$


















                  3












                  $begingroup$

                  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









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    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









                    $endgroup$



                    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 '18 at 8:57









                    Kavi Rama MurthyKavi Rama Murthy

                    54.4k32055




                    54.4k32055























                        1












                        $begingroup$

                        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









                        $endgroup$


















                          1












                          $begingroup$

                          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









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            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









                            $endgroup$



                            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 '18 at 10:13









                            trancelocationtrancelocation

                            9,8201722




                            9,8201722






























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                                Are they similar matrix

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                                0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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