Is it true that any injective function $f: mathbb{R} to mathbb{R}$ is strictly monotone?












1












$begingroup$



Is it true that any injective function $f: mathbb{R} to mathbb{R}$ is strictly monotone?




If yes, how do I prove it? If not, are there any examples of functions that disprove this statement.



I was thinking of $f(x) = {frac1x}$, which disproves this statement. But I'm not really sure.



Thanks for the help in advance.






real-analysis







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You need connectedness of the domain. Your counter is not defined at 0
    $endgroup$
    – PSG
    Nov 20 '18 at 4:59






  • 1




    $begingroup$
    There are non-continuous examples.
    $endgroup$
    – Lord Shark the Unknown
    Nov 20 '18 at 5:00






  • 1




    $begingroup$
    You also need that the function be continuous. Your counterexample is continuous, but the first comment applies. However, if you define $f(0)=0$ in your counterexample, it now works.
    $endgroup$
    – Jean-Claude Arbaut
    Nov 20 '18 at 5:00








  • 3




    $begingroup$
    Let $f(0)=1,f(1)=0$, and $f(x)=x$ otherwise.
    $endgroup$
    – Rahul
    Nov 20 '18 at 5:02








  • 1




    $begingroup$
    Since $f(x)=1/x$ is undefined at $0$, you can decide to give it any value you wish. Of course this is another function (it's not simply $xto1/x$ anymore). That is, the new $f$ is defined on $Bbb R$ to be $f(x)=1/x$ if $xne0$, and $f(0)=0$. This one is defined on $Bbb R$ but discontinuous, whereas your original example was defined on $Bbb Rbackslash{0}$ and continuous.
    $endgroup$
    – Jean-Claude Arbaut
    Nov 20 '18 at 5:11


















1












$begingroup$



Is it true that any injective function $f: mathbb{R} to mathbb{R}$ is strictly monotone?




If yes, how do I prove it? If not, are there any examples of functions that disprove this statement.



I was thinking of $f(x) = {frac1x}$, which disproves this statement. But I'm not really sure.



Thanks for the help in advance.






real-analysis







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You need connectedness of the domain. Your counter is not defined at 0
    $endgroup$
    – PSG
    Nov 20 '18 at 4:59






  • 1




    $begingroup$
    There are non-continuous examples.
    $endgroup$
    – Lord Shark the Unknown
    Nov 20 '18 at 5:00






  • 1




    $begingroup$
    You also need that the function be continuous. Your counterexample is continuous, but the first comment applies. However, if you define $f(0)=0$ in your counterexample, it now works.
    $endgroup$
    – Jean-Claude Arbaut
    Nov 20 '18 at 5:00








  • 3




    $begingroup$
    Let $f(0)=1,f(1)=0$, and $f(x)=x$ otherwise.
    $endgroup$
    – Rahul
    Nov 20 '18 at 5:02








  • 1




    $begingroup$
    Since $f(x)=1/x$ is undefined at $0$, you can decide to give it any value you wish. Of course this is another function (it's not simply $xto1/x$ anymore). That is, the new $f$ is defined on $Bbb R$ to be $f(x)=1/x$ if $xne0$, and $f(0)=0$. This one is defined on $Bbb R$ but discontinuous, whereas your original example was defined on $Bbb Rbackslash{0}$ and continuous.
    $endgroup$
    – Jean-Claude Arbaut
    Nov 20 '18 at 5:11
















1












1








1





$begingroup$



Is it true that any injective function $f: mathbb{R} to mathbb{R}$ is strictly monotone?




If yes, how do I prove it? If not, are there any examples of functions that disprove this statement.



I was thinking of $f(x) = {frac1x}$, which disproves this statement. But I'm not really sure.



Thanks for the help in advance.






real-analysis







share|cite|improve this question











$endgroup$





Is it true that any injective function $f: mathbb{R} to mathbb{R}$ is strictly monotone?




If yes, how do I prove it? If not, are there any examples of functions that disprove this statement.



I was thinking of $f(x) = {frac1x}$, which disproves this statement. But I'm not really sure.



Thanks for the help in advance.






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 '18 at 5:08









Jean-Claude Arbaut

14.8k63464




14.8k63464










asked Nov 20 '18 at 4:57









user10634718user10634718

63




63








  • 2




    $begingroup$
    You need connectedness of the domain. Your counter is not defined at 0
    $endgroup$
    – PSG
    Nov 20 '18 at 4:59






  • 1




    $begingroup$
    There are non-continuous examples.
    $endgroup$
    – Lord Shark the Unknown
    Nov 20 '18 at 5:00






  • 1




    $begingroup$
    You also need that the function be continuous. Your counterexample is continuous, but the first comment applies. However, if you define $f(0)=0$ in your counterexample, it now works.
    $endgroup$
    – Jean-Claude Arbaut
    Nov 20 '18 at 5:00








  • 3




    $begingroup$
    Let $f(0)=1,f(1)=0$, and $f(x)=x$ otherwise.
    $endgroup$
    – Rahul
    Nov 20 '18 at 5:02








  • 1




    $begingroup$
    Since $f(x)=1/x$ is undefined at $0$, you can decide to give it any value you wish. Of course this is another function (it's not simply $xto1/x$ anymore). That is, the new $f$ is defined on $Bbb R$ to be $f(x)=1/x$ if $xne0$, and $f(0)=0$. This one is defined on $Bbb R$ but discontinuous, whereas your original example was defined on $Bbb Rbackslash{0}$ and continuous.
    $endgroup$
    – Jean-Claude Arbaut
    Nov 20 '18 at 5:11
















  • 2




    $begingroup$
    You need connectedness of the domain. Your counter is not defined at 0
    $endgroup$
    – PSG
    Nov 20 '18 at 4:59






  • 1




    $begingroup$
    There are non-continuous examples.
    $endgroup$
    – Lord Shark the Unknown
    Nov 20 '18 at 5:00






  • 1




    $begingroup$
    You also need that the function be continuous. Your counterexample is continuous, but the first comment applies. However, if you define $f(0)=0$ in your counterexample, it now works.
    $endgroup$
    – Jean-Claude Arbaut
    Nov 20 '18 at 5:00








  • 3




    $begingroup$
    Let $f(0)=1,f(1)=0$, and $f(x)=x$ otherwise.
    $endgroup$
    – Rahul
    Nov 20 '18 at 5:02








  • 1




    $begingroup$
    Since $f(x)=1/x$ is undefined at $0$, you can decide to give it any value you wish. Of course this is another function (it's not simply $xto1/x$ anymore). That is, the new $f$ is defined on $Bbb R$ to be $f(x)=1/x$ if $xne0$, and $f(0)=0$. This one is defined on $Bbb R$ but discontinuous, whereas your original example was defined on $Bbb Rbackslash{0}$ and continuous.
    $endgroup$
    – Jean-Claude Arbaut
    Nov 20 '18 at 5:11










2




2




$begingroup$
You need connectedness of the domain. Your counter is not defined at 0
$endgroup$
– PSG
Nov 20 '18 at 4:59




$begingroup$
You need connectedness of the domain. Your counter is not defined at 0
$endgroup$
– PSG
Nov 20 '18 at 4:59




1




1




$begingroup$
There are non-continuous examples.
$endgroup$
– Lord Shark the Unknown
Nov 20 '18 at 5:00




$begingroup$
There are non-continuous examples.
$endgroup$
– Lord Shark the Unknown
Nov 20 '18 at 5:00




1




1




$begingroup$
You also need that the function be continuous. Your counterexample is continuous, but the first comment applies. However, if you define $f(0)=0$ in your counterexample, it now works.
$endgroup$
– Jean-Claude Arbaut
Nov 20 '18 at 5:00






$begingroup$
You also need that the function be continuous. Your counterexample is continuous, but the first comment applies. However, if you define $f(0)=0$ in your counterexample, it now works.
$endgroup$
– Jean-Claude Arbaut
Nov 20 '18 at 5:00






3




3




$begingroup$
Let $f(0)=1,f(1)=0$, and $f(x)=x$ otherwise.
$endgroup$
– Rahul
Nov 20 '18 at 5:02






$begingroup$
Let $f(0)=1,f(1)=0$, and $f(x)=x$ otherwise.
$endgroup$
– Rahul
Nov 20 '18 at 5:02






1




1




$begingroup$
Since $f(x)=1/x$ is undefined at $0$, you can decide to give it any value you wish. Of course this is another function (it's not simply $xto1/x$ anymore). That is, the new $f$ is defined on $Bbb R$ to be $f(x)=1/x$ if $xne0$, and $f(0)=0$. This one is defined on $Bbb R$ but discontinuous, whereas your original example was defined on $Bbb Rbackslash{0}$ and continuous.
$endgroup$
– Jean-Claude Arbaut
Nov 20 '18 at 5:11






$begingroup$
Since $f(x)=1/x$ is undefined at $0$, you can decide to give it any value you wish. Of course this is another function (it's not simply $xto1/x$ anymore). That is, the new $f$ is defined on $Bbb R$ to be $f(x)=1/x$ if $xne0$, and $f(0)=0$. This one is defined on $Bbb R$ but discontinuous, whereas your original example was defined on $Bbb Rbackslash{0}$ and continuous.
$endgroup$
– Jean-Claude Arbaut
Nov 20 '18 at 5:11












1 Answer
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$begingroup$

Your counterexample $f(x)=1/x$ does not work as is, because it's not defined at $0$.



With the function $g:Bbb RtoBbb R$ defined by $g(0)=0$ and $g(x)=1/x$ is $xne0$, you have a valid counterexample: it's defined on $Bbb R$ and injective, but not strictly monotone.



The correct statement would be:



If $f:EtoBbb R$ is continuous and injective and $E$ is a connected subset of $Bbb R$, then $f$ is strictly monotone.



Note that here your $f$ is continuous but it's defined on $(-infty,0)cup(0,+infty)$, which is not connected. The function $g$ defined above is defined on $Bbb R$, but is not continuous.






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    $begingroup$

    Your counterexample $f(x)=1/x$ does not work as is, because it's not defined at $0$.



    With the function $g:Bbb RtoBbb R$ defined by $g(0)=0$ and $g(x)=1/x$ is $xne0$, you have a valid counterexample: it's defined on $Bbb R$ and injective, but not strictly monotone.



    The correct statement would be:



    If $f:EtoBbb R$ is continuous and injective and $E$ is a connected subset of $Bbb R$, then $f$ is strictly monotone.



    Note that here your $f$ is continuous but it's defined on $(-infty,0)cup(0,+infty)$, which is not connected. The function $g$ defined above is defined on $Bbb R$, but is not continuous.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Your counterexample $f(x)=1/x$ does not work as is, because it's not defined at $0$.



      With the function $g:Bbb RtoBbb R$ defined by $g(0)=0$ and $g(x)=1/x$ is $xne0$, you have a valid counterexample: it's defined on $Bbb R$ and injective, but not strictly monotone.



      The correct statement would be:



      If $f:EtoBbb R$ is continuous and injective and $E$ is a connected subset of $Bbb R$, then $f$ is strictly monotone.



      Note that here your $f$ is continuous but it's defined on $(-infty,0)cup(0,+infty)$, which is not connected. The function $g$ defined above is defined on $Bbb R$, but is not continuous.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Your counterexample $f(x)=1/x$ does not work as is, because it's not defined at $0$.



        With the function $g:Bbb RtoBbb R$ defined by $g(0)=0$ and $g(x)=1/x$ is $xne0$, you have a valid counterexample: it's defined on $Bbb R$ and injective, but not strictly monotone.



        The correct statement would be:



        If $f:EtoBbb R$ is continuous and injective and $E$ is a connected subset of $Bbb R$, then $f$ is strictly monotone.



        Note that here your $f$ is continuous but it's defined on $(-infty,0)cup(0,+infty)$, which is not connected. The function $g$ defined above is defined on $Bbb R$, but is not continuous.






        share|cite|improve this answer









        $endgroup$



        Your counterexample $f(x)=1/x$ does not work as is, because it's not defined at $0$.



        With the function $g:Bbb RtoBbb R$ defined by $g(0)=0$ and $g(x)=1/x$ is $xne0$, you have a valid counterexample: it's defined on $Bbb R$ and injective, but not strictly monotone.



        The correct statement would be:



        If $f:EtoBbb R$ is continuous and injective and $E$ is a connected subset of $Bbb R$, then $f$ is strictly monotone.



        Note that here your $f$ is continuous but it's defined on $(-infty,0)cup(0,+infty)$, which is not connected. The function $g$ defined above is defined on $Bbb R$, but is not continuous.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 9:48









        Jean-Claude ArbautJean-Claude Arbaut

        14.8k63464




        14.8k63464






























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