Infinitely many discontinuities for a bijective function from $[0,1)$ to $(0,1)$












7












$begingroup$


Show that any bijection from $[0,1)$ to $(0,1)$ has infinitely many discontinuities.



I have thought about this question but I have no any idea. Any idea is valuable for me, thanks.






general-topology







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  • $begingroup$
    @Ian such a bijection from $[0,a)$ to $(b,c)$ can be discontinuous at $0$, right?
    $endgroup$
    – Darío G
    Sep 7 '16 at 17:08










  • $begingroup$
    See my answer here (after the "edit"): math.stackexchange.com/questions/966299/…
    $endgroup$
    – Moishe Kohan
    Sep 10 '16 at 3:46
















7












$begingroup$


Show that any bijection from $[0,1)$ to $(0,1)$ has infinitely many discontinuities.



I have thought about this question but I have no any idea. Any idea is valuable for me, thanks.






general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Ian such a bijection from $[0,a)$ to $(b,c)$ can be discontinuous at $0$, right?
    $endgroup$
    – Darío G
    Sep 7 '16 at 17:08










  • $begingroup$
    See my answer here (after the "edit"): math.stackexchange.com/questions/966299/…
    $endgroup$
    – Moishe Kohan
    Sep 10 '16 at 3:46














7












7








7


2



$begingroup$


Show that any bijection from $[0,1)$ to $(0,1)$ has infinitely many discontinuities.



I have thought about this question but I have no any idea. Any idea is valuable for me, thanks.






general-topology







share|cite|improve this question











$endgroup$




Show that any bijection from $[0,1)$ to $(0,1)$ has infinitely many discontinuities.



I have thought about this question but I have no any idea. Any idea is valuable for me, thanks.






general-topology




general-topology






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share|cite|improve this question













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edited Sep 7 '16 at 16:58









Alexis Olson

4,5481030




4,5481030










asked Sep 7 '16 at 16:52









user365user365

747




747












  • $begingroup$
    @Ian such a bijection from $[0,a)$ to $(b,c)$ can be discontinuous at $0$, right?
    $endgroup$
    – Darío G
    Sep 7 '16 at 17:08










  • $begingroup$
    See my answer here (after the "edit"): math.stackexchange.com/questions/966299/…
    $endgroup$
    – Moishe Kohan
    Sep 10 '16 at 3:46


















  • $begingroup$
    @Ian such a bijection from $[0,a)$ to $(b,c)$ can be discontinuous at $0$, right?
    $endgroup$
    – Darío G
    Sep 7 '16 at 17:08










  • $begingroup$
    See my answer here (after the "edit"): math.stackexchange.com/questions/966299/…
    $endgroup$
    – Moishe Kohan
    Sep 10 '16 at 3:46
















$begingroup$
@Ian such a bijection from $[0,a)$ to $(b,c)$ can be discontinuous at $0$, right?
$endgroup$
– Darío G
Sep 7 '16 at 17:08




$begingroup$
@Ian such a bijection from $[0,a)$ to $(b,c)$ can be discontinuous at $0$, right?
$endgroup$
– Darío G
Sep 7 '16 at 17:08












$begingroup$
See my answer here (after the "edit"): math.stackexchange.com/questions/966299/…
$endgroup$
– Moishe Kohan
Sep 10 '16 at 3:46




$begingroup$
See my answer here (after the "edit"): math.stackexchange.com/questions/966299/…
$endgroup$
– Moishe Kohan
Sep 10 '16 at 3:46










2 Answers
2






active

oldest

votes


















4












$begingroup$

HINT: Suppose that $f:[0,1)to(0,1)$ is a bijection with only finitely many discontinuities. Let $F$ be the set of points at which $f$ is not continuous. Then $[0,1)setminus F$ is a union of intervals, and $f$ is continuous on each of these intervals. Moreover, all of these intervals are open except possibly one of the form $[0,a)$, in the case in which $f$ is continuous at $0$. Let $mathscr{I}$ be the set of these intervals.




  • Use the continuity of $fupharpoonright I$ and the fact that $f$ is a bijection to show that $f[I]$ is an interval for each $Iinmathscr{I}$. Moreover, $f[I]$ is an open interval if $I$ is.


  • Show that if $0in F$, so that each $Iinmathscr{I}$ is an open interval, then on the one hand $|f[F]|=|mathscr{I}|-1$, since $(0,1)setminus f[F]$ is the union of $|mathscr{I}$ open intervals, but on the other hand $|F|=|mathscr{I}|$. Conclude that $0notin F$, so $f$ is continuous at $0$.



Let $F={a_1,ldots,a_n}$, where $0<a_1<ldots<a_n<1$. Let $I_0=[0,a_1)$, for $k=1,ldots,n-1$ let $I_k=(a_k,a_{k+1})$, and let $I_n=(a_n,1)$. Then $f$ is continuous on each of the intervals $I_k$ for $k=0,ldots,n$.




  • Use the fact that $f[F]=F$ to show that $f$ must map each of the intervals $I_k$ to one of the intervals $(0,a_1)$, $(a_k,a_{k+1})$ for $k=1,ldots,n-1$, or $(a_n,1)$.


  • Get a contradiction by showing that $f[I_0]$ is a half-open interval, not an open interval.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Do you actually get $f[F]=[F]$? I can get $|f[F-{0}]|=|F|$ which also shows that $0notin F$, but I don't know how you get the equality of the two sets $f[F]$ and $[F]$. Could you please explain to me?
    $endgroup$
    – Darío G
    Sep 7 '16 at 18:31












  • $begingroup$
    @Wore: You’re absolutely right; I don’t know what (or if!) I was thinking. I’ve corrected that bit now; thanks!
    $endgroup$
    – Brian M. Scott
    Sep 7 '16 at 19:09





















0












$begingroup$

Suppose that $f$ is discontinuous only at a finite set $A$, and let $B = A cup {0}$. Set $|B| = n$.



$[0,1)setminus B$ is a disjoint union of $n$ open intervals ${I_k}$ on which $f$ is continuous, so they are mapped by $f$ to a disjoint union of open intervals ${f(I_k)}$ that must line up end-to-end (as their complement is finite).



But lining up $n$ open intervals end-to-end in $(0,1)$ gives us $n-1$ endpoints, which cannot be in bijection with $B$.






share|cite|improve this answer









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    4












    $begingroup$

    HINT: Suppose that $f:[0,1)to(0,1)$ is a bijection with only finitely many discontinuities. Let $F$ be the set of points at which $f$ is not continuous. Then $[0,1)setminus F$ is a union of intervals, and $f$ is continuous on each of these intervals. Moreover, all of these intervals are open except possibly one of the form $[0,a)$, in the case in which $f$ is continuous at $0$. Let $mathscr{I}$ be the set of these intervals.




    • Use the continuity of $fupharpoonright I$ and the fact that $f$ is a bijection to show that $f[I]$ is an interval for each $Iinmathscr{I}$. Moreover, $f[I]$ is an open interval if $I$ is.


    • Show that if $0in F$, so that each $Iinmathscr{I}$ is an open interval, then on the one hand $|f[F]|=|mathscr{I}|-1$, since $(0,1)setminus f[F]$ is the union of $|mathscr{I}$ open intervals, but on the other hand $|F|=|mathscr{I}|$. Conclude that $0notin F$, so $f$ is continuous at $0$.



    Let $F={a_1,ldots,a_n}$, where $0<a_1<ldots<a_n<1$. Let $I_0=[0,a_1)$, for $k=1,ldots,n-1$ let $I_k=(a_k,a_{k+1})$, and let $I_n=(a_n,1)$. Then $f$ is continuous on each of the intervals $I_k$ for $k=0,ldots,n$.




    • Use the fact that $f[F]=F$ to show that $f$ must map each of the intervals $I_k$ to one of the intervals $(0,a_1)$, $(a_k,a_{k+1})$ for $k=1,ldots,n-1$, or $(a_n,1)$.


    • Get a contradiction by showing that $f[I_0]$ is a half-open interval, not an open interval.







    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Do you actually get $f[F]=[F]$? I can get $|f[F-{0}]|=|F|$ which also shows that $0notin F$, but I don't know how you get the equality of the two sets $f[F]$ and $[F]$. Could you please explain to me?
      $endgroup$
      – Darío G
      Sep 7 '16 at 18:31












    • $begingroup$
      @Wore: You’re absolutely right; I don’t know what (or if!) I was thinking. I’ve corrected that bit now; thanks!
      $endgroup$
      – Brian M. Scott
      Sep 7 '16 at 19:09


















    4












    $begingroup$

    HINT: Suppose that $f:[0,1)to(0,1)$ is a bijection with only finitely many discontinuities. Let $F$ be the set of points at which $f$ is not continuous. Then $[0,1)setminus F$ is a union of intervals, and $f$ is continuous on each of these intervals. Moreover, all of these intervals are open except possibly one of the form $[0,a)$, in the case in which $f$ is continuous at $0$. Let $mathscr{I}$ be the set of these intervals.




    • Use the continuity of $fupharpoonright I$ and the fact that $f$ is a bijection to show that $f[I]$ is an interval for each $Iinmathscr{I}$. Moreover, $f[I]$ is an open interval if $I$ is.


    • Show that if $0in F$, so that each $Iinmathscr{I}$ is an open interval, then on the one hand $|f[F]|=|mathscr{I}|-1$, since $(0,1)setminus f[F]$ is the union of $|mathscr{I}$ open intervals, but on the other hand $|F|=|mathscr{I}|$. Conclude that $0notin F$, so $f$ is continuous at $0$.



    Let $F={a_1,ldots,a_n}$, where $0<a_1<ldots<a_n<1$. Let $I_0=[0,a_1)$, for $k=1,ldots,n-1$ let $I_k=(a_k,a_{k+1})$, and let $I_n=(a_n,1)$. Then $f$ is continuous on each of the intervals $I_k$ for $k=0,ldots,n$.




    • Use the fact that $f[F]=F$ to show that $f$ must map each of the intervals $I_k$ to one of the intervals $(0,a_1)$, $(a_k,a_{k+1})$ for $k=1,ldots,n-1$, or $(a_n,1)$.


    • Get a contradiction by showing that $f[I_0]$ is a half-open interval, not an open interval.







    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Do you actually get $f[F]=[F]$? I can get $|f[F-{0}]|=|F|$ which also shows that $0notin F$, but I don't know how you get the equality of the two sets $f[F]$ and $[F]$. Could you please explain to me?
      $endgroup$
      – Darío G
      Sep 7 '16 at 18:31












    • $begingroup$
      @Wore: You’re absolutely right; I don’t know what (or if!) I was thinking. I’ve corrected that bit now; thanks!
      $endgroup$
      – Brian M. Scott
      Sep 7 '16 at 19:09
















    4












    4








    4





    $begingroup$

    HINT: Suppose that $f:[0,1)to(0,1)$ is a bijection with only finitely many discontinuities. Let $F$ be the set of points at which $f$ is not continuous. Then $[0,1)setminus F$ is a union of intervals, and $f$ is continuous on each of these intervals. Moreover, all of these intervals are open except possibly one of the form $[0,a)$, in the case in which $f$ is continuous at $0$. Let $mathscr{I}$ be the set of these intervals.




    • Use the continuity of $fupharpoonright I$ and the fact that $f$ is a bijection to show that $f[I]$ is an interval for each $Iinmathscr{I}$. Moreover, $f[I]$ is an open interval if $I$ is.


    • Show that if $0in F$, so that each $Iinmathscr{I}$ is an open interval, then on the one hand $|f[F]|=|mathscr{I}|-1$, since $(0,1)setminus f[F]$ is the union of $|mathscr{I}$ open intervals, but on the other hand $|F|=|mathscr{I}|$. Conclude that $0notin F$, so $f$ is continuous at $0$.



    Let $F={a_1,ldots,a_n}$, where $0<a_1<ldots<a_n<1$. Let $I_0=[0,a_1)$, for $k=1,ldots,n-1$ let $I_k=(a_k,a_{k+1})$, and let $I_n=(a_n,1)$. Then $f$ is continuous on each of the intervals $I_k$ for $k=0,ldots,n$.




    • Use the fact that $f[F]=F$ to show that $f$ must map each of the intervals $I_k$ to one of the intervals $(0,a_1)$, $(a_k,a_{k+1})$ for $k=1,ldots,n-1$, or $(a_n,1)$.


    • Get a contradiction by showing that $f[I_0]$ is a half-open interval, not an open interval.







    share|cite|improve this answer











    $endgroup$



    HINT: Suppose that $f:[0,1)to(0,1)$ is a bijection with only finitely many discontinuities. Let $F$ be the set of points at which $f$ is not continuous. Then $[0,1)setminus F$ is a union of intervals, and $f$ is continuous on each of these intervals. Moreover, all of these intervals are open except possibly one of the form $[0,a)$, in the case in which $f$ is continuous at $0$. Let $mathscr{I}$ be the set of these intervals.




    • Use the continuity of $fupharpoonright I$ and the fact that $f$ is a bijection to show that $f[I]$ is an interval for each $Iinmathscr{I}$. Moreover, $f[I]$ is an open interval if $I$ is.


    • Show that if $0in F$, so that each $Iinmathscr{I}$ is an open interval, then on the one hand $|f[F]|=|mathscr{I}|-1$, since $(0,1)setminus f[F]$ is the union of $|mathscr{I}$ open intervals, but on the other hand $|F|=|mathscr{I}|$. Conclude that $0notin F$, so $f$ is continuous at $0$.



    Let $F={a_1,ldots,a_n}$, where $0<a_1<ldots<a_n<1$. Let $I_0=[0,a_1)$, for $k=1,ldots,n-1$ let $I_k=(a_k,a_{k+1})$, and let $I_n=(a_n,1)$. Then $f$ is continuous on each of the intervals $I_k$ for $k=0,ldots,n$.




    • Use the fact that $f[F]=F$ to show that $f$ must map each of the intervals $I_k$ to one of the intervals $(0,a_1)$, $(a_k,a_{k+1})$ for $k=1,ldots,n-1$, or $(a_n,1)$.


    • Get a contradiction by showing that $f[I_0]$ is a half-open interval, not an open interval.








    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Sep 7 '16 at 19:08

























    answered Sep 7 '16 at 17:22









    Brian M. ScottBrian M. Scott

    461k40518920




    461k40518920












    • $begingroup$
      Do you actually get $f[F]=[F]$? I can get $|f[F-{0}]|=|F|$ which also shows that $0notin F$, but I don't know how you get the equality of the two sets $f[F]$ and $[F]$. Could you please explain to me?
      $endgroup$
      – Darío G
      Sep 7 '16 at 18:31












    • $begingroup$
      @Wore: You’re absolutely right; I don’t know what (or if!) I was thinking. I’ve corrected that bit now; thanks!
      $endgroup$
      – Brian M. Scott
      Sep 7 '16 at 19:09




















    • $begingroup$
      Do you actually get $f[F]=[F]$? I can get $|f[F-{0}]|=|F|$ which also shows that $0notin F$, but I don't know how you get the equality of the two sets $f[F]$ and $[F]$. Could you please explain to me?
      $endgroup$
      – Darío G
      Sep 7 '16 at 18:31












    • $begingroup$
      @Wore: You’re absolutely right; I don’t know what (or if!) I was thinking. I’ve corrected that bit now; thanks!
      $endgroup$
      – Brian M. Scott
      Sep 7 '16 at 19:09


















    $begingroup$
    Do you actually get $f[F]=[F]$? I can get $|f[F-{0}]|=|F|$ which also shows that $0notin F$, but I don't know how you get the equality of the two sets $f[F]$ and $[F]$. Could you please explain to me?
    $endgroup$
    – Darío G
    Sep 7 '16 at 18:31






    $begingroup$
    Do you actually get $f[F]=[F]$? I can get $|f[F-{0}]|=|F|$ which also shows that $0notin F$, but I don't know how you get the equality of the two sets $f[F]$ and $[F]$. Could you please explain to me?
    $endgroup$
    – Darío G
    Sep 7 '16 at 18:31














    $begingroup$
    @Wore: You’re absolutely right; I don’t know what (or if!) I was thinking. I’ve corrected that bit now; thanks!
    $endgroup$
    – Brian M. Scott
    Sep 7 '16 at 19:09






    $begingroup$
    @Wore: You’re absolutely right; I don’t know what (or if!) I was thinking. I’ve corrected that bit now; thanks!
    $endgroup$
    – Brian M. Scott
    Sep 7 '16 at 19:09













    0












    $begingroup$

    Suppose that $f$ is discontinuous only at a finite set $A$, and let $B = A cup {0}$. Set $|B| = n$.



    $[0,1)setminus B$ is a disjoint union of $n$ open intervals ${I_k}$ on which $f$ is continuous, so they are mapped by $f$ to a disjoint union of open intervals ${f(I_k)}$ that must line up end-to-end (as their complement is finite).



    But lining up $n$ open intervals end-to-end in $(0,1)$ gives us $n-1$ endpoints, which cannot be in bijection with $B$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Suppose that $f$ is discontinuous only at a finite set $A$, and let $B = A cup {0}$. Set $|B| = n$.



      $[0,1)setminus B$ is a disjoint union of $n$ open intervals ${I_k}$ on which $f$ is continuous, so they are mapped by $f$ to a disjoint union of open intervals ${f(I_k)}$ that must line up end-to-end (as their complement is finite).



      But lining up $n$ open intervals end-to-end in $(0,1)$ gives us $n-1$ endpoints, which cannot be in bijection with $B$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Suppose that $f$ is discontinuous only at a finite set $A$, and let $B = A cup {0}$. Set $|B| = n$.



        $[0,1)setminus B$ is a disjoint union of $n$ open intervals ${I_k}$ on which $f$ is continuous, so they are mapped by $f$ to a disjoint union of open intervals ${f(I_k)}$ that must line up end-to-end (as their complement is finite).



        But lining up $n$ open intervals end-to-end in $(0,1)$ gives us $n-1$ endpoints, which cannot be in bijection with $B$.






        share|cite|improve this answer









        $endgroup$



        Suppose that $f$ is discontinuous only at a finite set $A$, and let $B = A cup {0}$. Set $|B| = n$.



        $[0,1)setminus B$ is a disjoint union of $n$ open intervals ${I_k}$ on which $f$ is continuous, so they are mapped by $f$ to a disjoint union of open intervals ${f(I_k)}$ that must line up end-to-end (as their complement is finite).



        But lining up $n$ open intervals end-to-end in $(0,1)$ gives us $n-1$ endpoints, which cannot be in bijection with $B$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 23 '18 at 20:35









        SladeSlade

        25.4k12665




        25.4k12665






























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