Real Analysis - Continuity











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a. Give an example of a function defined everywhere on the interval $[0,1]$, which does not achieve its maximum.



b. Give an example of a function defined on $mathbb{R}$, that is nowhere continuous.



c. Give an example of a continuous function defined on a bounded set, which is not uniformly continuous.



My answers are




  1. $f(x) = x^2$


  2. $f(x) = {1$, if $x$ is rational; $-1$, if $x$ is irrational$}$

  3. $f(x) = 1/x$


My workings are attached. Please help verify if the working is correct.Solutions










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  • 4




    2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
    – Ted Shifrin
    Dec 1 at 22:41















up vote
3
down vote

favorite












a. Give an example of a function defined everywhere on the interval $[0,1]$, which does not achieve its maximum.



b. Give an example of a function defined on $mathbb{R}$, that is nowhere continuous.



c. Give an example of a continuous function defined on a bounded set, which is not uniformly continuous.



My answers are




  1. $f(x) = x^2$


  2. $f(x) = {1$, if $x$ is rational; $-1$, if $x$ is irrational$}$

  3. $f(x) = 1/x$


My workings are attached. Please help verify if the working is correct.Solutions










share|cite|improve this question







New contributor




Ty Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 4




    2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
    – Ted Shifrin
    Dec 1 at 22:41













up vote
3
down vote

favorite









up vote
3
down vote

favorite











a. Give an example of a function defined everywhere on the interval $[0,1]$, which does not achieve its maximum.



b. Give an example of a function defined on $mathbb{R}$, that is nowhere continuous.



c. Give an example of a continuous function defined on a bounded set, which is not uniformly continuous.



My answers are




  1. $f(x) = x^2$


  2. $f(x) = {1$, if $x$ is rational; $-1$, if $x$ is irrational$}$

  3. $f(x) = 1/x$


My workings are attached. Please help verify if the working is correct.Solutions










share|cite|improve this question







New contributor




Ty Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











a. Give an example of a function defined everywhere on the interval $[0,1]$, which does not achieve its maximum.



b. Give an example of a function defined on $mathbb{R}$, that is nowhere continuous.



c. Give an example of a continuous function defined on a bounded set, which is not uniformly continuous.



My answers are




  1. $f(x) = x^2$


  2. $f(x) = {1$, if $x$ is rational; $-1$, if $x$ is irrational$}$

  3. $f(x) = 1/x$


My workings are attached. Please help verify if the working is correct.Solutions







real-analysis continuity maxima-minima uniform-continuity






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asked Dec 1 at 22:36









Ty Johnson

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Ty Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 4




    2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
    – Ted Shifrin
    Dec 1 at 22:41














  • 4




    2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
    – Ted Shifrin
    Dec 1 at 22:41








4




4




2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
– Ted Shifrin
Dec 1 at 22:41




2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
– Ted Shifrin
Dec 1 at 22:41










3 Answers
3






active

oldest

votes

















up vote
6
down vote



accepted










$a)$ Consider $f(x)$ on $[0,1]$ such that $f(0) = 0, f(1) = 0, f(x) = dfrac{1}{x}, 0 <x < 1$.



$b)$ The function you had is a good one.



$c)$ Consider $f(x)$ on $(0,1)$ and $f(x) = dfrac{1}{x}$.






share|cite|improve this answer




























    up vote
    4
    down vote














    1. Wrong: the maximum is $1$, which is $f(1)$.

    2. Right.

    3. Your answer is incomplete, at best, since you did not state what is the domain of your function.






    share|cite|improve this answer




























      up vote
      2
      down vote













      1)



      Extreme value theorem says that every continuous function over a closed interval has a maximum.



      You need a function that is not continuous.



      $f(x) = begin{cases} f(x) = x & x<1\f(x) = 0 & x=1 end{cases}$






      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
        6
        down vote



        accepted










        $a)$ Consider $f(x)$ on $[0,1]$ such that $f(0) = 0, f(1) = 0, f(x) = dfrac{1}{x}, 0 <x < 1$.



        $b)$ The function you had is a good one.



        $c)$ Consider $f(x)$ on $(0,1)$ and $f(x) = dfrac{1}{x}$.






        share|cite|improve this answer

























          up vote
          6
          down vote



          accepted










          $a)$ Consider $f(x)$ on $[0,1]$ such that $f(0) = 0, f(1) = 0, f(x) = dfrac{1}{x}, 0 <x < 1$.



          $b)$ The function you had is a good one.



          $c)$ Consider $f(x)$ on $(0,1)$ and $f(x) = dfrac{1}{x}$.






          share|cite|improve this answer























            up vote
            6
            down vote



            accepted







            up vote
            6
            down vote



            accepted






            $a)$ Consider $f(x)$ on $[0,1]$ such that $f(0) = 0, f(1) = 0, f(x) = dfrac{1}{x}, 0 <x < 1$.



            $b)$ The function you had is a good one.



            $c)$ Consider $f(x)$ on $(0,1)$ and $f(x) = dfrac{1}{x}$.






            share|cite|improve this answer












            $a)$ Consider $f(x)$ on $[0,1]$ such that $f(0) = 0, f(1) = 0, f(x) = dfrac{1}{x}, 0 <x < 1$.



            $b)$ The function you had is a good one.



            $c)$ Consider $f(x)$ on $(0,1)$ and $f(x) = dfrac{1}{x}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 1 at 22:50









            DeepSea

            70.6k54487




            70.6k54487






















                up vote
                4
                down vote














                1. Wrong: the maximum is $1$, which is $f(1)$.

                2. Right.

                3. Your answer is incomplete, at best, since you did not state what is the domain of your function.






                share|cite|improve this answer

























                  up vote
                  4
                  down vote














                  1. Wrong: the maximum is $1$, which is $f(1)$.

                  2. Right.

                  3. Your answer is incomplete, at best, since you did not state what is the domain of your function.






                  share|cite|improve this answer























                    up vote
                    4
                    down vote










                    up vote
                    4
                    down vote










                    1. Wrong: the maximum is $1$, which is $f(1)$.

                    2. Right.

                    3. Your answer is incomplete, at best, since you did not state what is the domain of your function.






                    share|cite|improve this answer













                    1. Wrong: the maximum is $1$, which is $f(1)$.

                    2. Right.

                    3. Your answer is incomplete, at best, since you did not state what is the domain of your function.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 1 at 22:44









                    José Carlos Santos

                    144k20113214




                    144k20113214






















                        up vote
                        2
                        down vote













                        1)



                        Extreme value theorem says that every continuous function over a closed interval has a maximum.



                        You need a function that is not continuous.



                        $f(x) = begin{cases} f(x) = x & x<1\f(x) = 0 & x=1 end{cases}$






                        share|cite|improve this answer

























                          up vote
                          2
                          down vote













                          1)



                          Extreme value theorem says that every continuous function over a closed interval has a maximum.



                          You need a function that is not continuous.



                          $f(x) = begin{cases} f(x) = x & x<1\f(x) = 0 & x=1 end{cases}$






                          share|cite|improve this answer























                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            1)



                            Extreme value theorem says that every continuous function over a closed interval has a maximum.



                            You need a function that is not continuous.



                            $f(x) = begin{cases} f(x) = x & x<1\f(x) = 0 & x=1 end{cases}$






                            share|cite|improve this answer












                            1)



                            Extreme value theorem says that every continuous function over a closed interval has a maximum.



                            You need a function that is not continuous.



                            $f(x) = begin{cases} f(x) = x & x<1\f(x) = 0 & x=1 end{cases}$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 1 at 23:27









                            Doug M

                            43k31753




                            43k31753






















                                Ty Johnson is a new contributor. Be nice, and check out our Code of Conduct.










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