Counterexample for $f$ is strictly increasing ,$ g$ and $fcirc g$ is continous but f is not continous












0












$begingroup$


I wanted to find counterexample



Counterexample for $f$ is strictly increasing,$ g$ and $fcirc g$ is continuous but $f$ is not continuous



Where f and g are function form $[0,1]to [0,1]$



How to approach to find such example



Any Help will be appreciated






real-analysis examples-counterexamples







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    0












    $begingroup$


    I wanted to find counterexample



    Counterexample for $f$ is strictly increasing,$ g$ and $fcirc g$ is continuous but $f$ is not continuous



    Where f and g are function form $[0,1]to [0,1]$



    How to approach to find such example



    Any Help will be appreciated






    real-analysis examples-counterexamples







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I wanted to find counterexample



      Counterexample for $f$ is strictly increasing,$ g$ and $fcirc g$ is continuous but $f$ is not continuous



      Where f and g are function form $[0,1]to [0,1]$



      How to approach to find such example



      Any Help will be appreciated






      real-analysis examples-counterexamples







      share|cite|improve this question









      $endgroup$




      I wanted to find counterexample



      Counterexample for $f$ is strictly increasing,$ g$ and $fcirc g$ is continuous but $f$ is not continuous



      Where f and g are function form $[0,1]to [0,1]$



      How to approach to find such example



      Any Help will be appreciated






      real-analysis examples-counterexamples




      real-analysis examples-counterexamples






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 2 '18 at 9:24









      MathLoverMathLover

      49710




      49710






















          1 Answer
          1






          active

          oldest

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          3












          $begingroup$

          Hint: Let $g(x)=0{}{}{}{}{}{}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            great Sir...I am always missing such examples...
            $endgroup$
            – MathLover
            Dec 2 '18 at 9:29










          • $begingroup$
            But Sir If We want both map as surjective map then is it possible to such construct map?
            $endgroup$
            – MathLover
            Dec 2 '18 at 9:31








          • 2




            $begingroup$
            @MathLover No, if $g$ is surjective and continuous, and $f$ is not continuous, then for some $x_0$, we will have $g(x_0)$ be some point of discontinuity of $f$, which means that $f(g(x))$ must be discontinuous at that point. I've glossed over a few details here, but that's the general argument.
            $endgroup$
            – Arthur
            Dec 2 '18 at 9:36













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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Hint: Let $g(x)=0{}{}{}{}{}{}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            great Sir...I am always missing such examples...
            $endgroup$
            – MathLover
            Dec 2 '18 at 9:29










          • $begingroup$
            But Sir If We want both map as surjective map then is it possible to such construct map?
            $endgroup$
            – MathLover
            Dec 2 '18 at 9:31








          • 2




            $begingroup$
            @MathLover No, if $g$ is surjective and continuous, and $f$ is not continuous, then for some $x_0$, we will have $g(x_0)$ be some point of discontinuity of $f$, which means that $f(g(x))$ must be discontinuous at that point. I've glossed over a few details here, but that's the general argument.
            $endgroup$
            – Arthur
            Dec 2 '18 at 9:36


















          3












          $begingroup$

          Hint: Let $g(x)=0{}{}{}{}{}{}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            great Sir...I am always missing such examples...
            $endgroup$
            – MathLover
            Dec 2 '18 at 9:29










          • $begingroup$
            But Sir If We want both map as surjective map then is it possible to such construct map?
            $endgroup$
            – MathLover
            Dec 2 '18 at 9:31








          • 2




            $begingroup$
            @MathLover No, if $g$ is surjective and continuous, and $f$ is not continuous, then for some $x_0$, we will have $g(x_0)$ be some point of discontinuity of $f$, which means that $f(g(x))$ must be discontinuous at that point. I've glossed over a few details here, but that's the general argument.
            $endgroup$
            – Arthur
            Dec 2 '18 at 9:36
















          3












          3








          3





          $begingroup$

          Hint: Let $g(x)=0{}{}{}{}{}{}$.






          share|cite|improve this answer









          $endgroup$



          Hint: Let $g(x)=0{}{}{}{}{}{}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 9:28









          ArthurArthur

          113k7110193




          113k7110193












          • $begingroup$
            great Sir...I am always missing such examples...
            $endgroup$
            – MathLover
            Dec 2 '18 at 9:29










          • $begingroup$
            But Sir If We want both map as surjective map then is it possible to such construct map?
            $endgroup$
            – MathLover
            Dec 2 '18 at 9:31








          • 2




            $begingroup$
            @MathLover No, if $g$ is surjective and continuous, and $f$ is not continuous, then for some $x_0$, we will have $g(x_0)$ be some point of discontinuity of $f$, which means that $f(g(x))$ must be discontinuous at that point. I've glossed over a few details here, but that's the general argument.
            $endgroup$
            – Arthur
            Dec 2 '18 at 9:36




















          • $begingroup$
            great Sir...I am always missing such examples...
            $endgroup$
            – MathLover
            Dec 2 '18 at 9:29










          • $begingroup$
            But Sir If We want both map as surjective map then is it possible to such construct map?
            $endgroup$
            – MathLover
            Dec 2 '18 at 9:31








          • 2




            $begingroup$
            @MathLover No, if $g$ is surjective and continuous, and $f$ is not continuous, then for some $x_0$, we will have $g(x_0)$ be some point of discontinuity of $f$, which means that $f(g(x))$ must be discontinuous at that point. I've glossed over a few details here, but that's the general argument.
            $endgroup$
            – Arthur
            Dec 2 '18 at 9:36


















          $begingroup$
          great Sir...I am always missing such examples...
          $endgroup$
          – MathLover
          Dec 2 '18 at 9:29




          $begingroup$
          great Sir...I am always missing such examples...
          $endgroup$
          – MathLover
          Dec 2 '18 at 9:29












          $begingroup$
          But Sir If We want both map as surjective map then is it possible to such construct map?
          $endgroup$
          – MathLover
          Dec 2 '18 at 9:31






          $begingroup$
          But Sir If We want both map as surjective map then is it possible to such construct map?
          $endgroup$
          – MathLover
          Dec 2 '18 at 9:31






          2




          2




          $begingroup$
          @MathLover No, if $g$ is surjective and continuous, and $f$ is not continuous, then for some $x_0$, we will have $g(x_0)$ be some point of discontinuity of $f$, which means that $f(g(x))$ must be discontinuous at that point. I've glossed over a few details here, but that's the general argument.
          $endgroup$
          – Arthur
          Dec 2 '18 at 9:36






          $begingroup$
          @MathLover No, if $g$ is surjective and continuous, and $f$ is not continuous, then for some $x_0$, we will have $g(x_0)$ be some point of discontinuity of $f$, which means that $f(g(x))$ must be discontinuous at that point. I've glossed over a few details here, but that's the general argument.
          $endgroup$
          – Arthur
          Dec 2 '18 at 9:36




















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