Continuous function on $mathbb{Q}$












6












$begingroup$


Let $f:mathbb{Q}tomathbb{R}$ be a function defined as:

$$f(x) =
begin{cases}
0 & x^2 < 2\
1 & x^2 geq 2
end{cases}
$$ Is this function continuous? How can we check the continuity around $sqrt{2}$ since it's not in $mathbb{Q}$?






calculus real-analysis continuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    Note (perhaps relevantly) that this function isn't just $mathbb{Q}mapstomathbb{R}$; it's also $mathbb{Q}mapstomathbb{Q}$ (which may be relevant for the next thing you learn using this function!).
    $endgroup$
    – Steven Stadnicki
    Feb 13 '14 at 16:32
















6












$begingroup$


Let $f:mathbb{Q}tomathbb{R}$ be a function defined as:

$$f(x) =
begin{cases}
0 & x^2 < 2\
1 & x^2 geq 2
end{cases}
$$ Is this function continuous? How can we check the continuity around $sqrt{2}$ since it's not in $mathbb{Q}$?






calculus real-analysis continuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    Note (perhaps relevantly) that this function isn't just $mathbb{Q}mapstomathbb{R}$; it's also $mathbb{Q}mapstomathbb{Q}$ (which may be relevant for the next thing you learn using this function!).
    $endgroup$
    – Steven Stadnicki
    Feb 13 '14 at 16:32














6












6








6


0



$begingroup$


Let $f:mathbb{Q}tomathbb{R}$ be a function defined as:

$$f(x) =
begin{cases}
0 & x^2 < 2\
1 & x^2 geq 2
end{cases}
$$ Is this function continuous? How can we check the continuity around $sqrt{2}$ since it's not in $mathbb{Q}$?






calculus real-analysis continuity







share|cite|improve this question











$endgroup$




Let $f:mathbb{Q}tomathbb{R}$ be a function defined as:

$$f(x) =
begin{cases}
0 & x^2 < 2\
1 & x^2 geq 2
end{cases}
$$ Is this function continuous? How can we check the continuity around $sqrt{2}$ since it's not in $mathbb{Q}$?






calculus real-analysis continuity




calculus real-analysis continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 13 '14 at 8:20









Brian Fitzpatrick

21.1k42958




21.1k42958










asked Feb 13 '14 at 8:16









SpockSpock

950719




950719












  • $begingroup$
    Note (perhaps relevantly) that this function isn't just $mathbb{Q}mapstomathbb{R}$; it's also $mathbb{Q}mapstomathbb{Q}$ (which may be relevant for the next thing you learn using this function!).
    $endgroup$
    – Steven Stadnicki
    Feb 13 '14 at 16:32


















  • $begingroup$
    Note (perhaps relevantly) that this function isn't just $mathbb{Q}mapstomathbb{R}$; it's also $mathbb{Q}mapstomathbb{Q}$ (which may be relevant for the next thing you learn using this function!).
    $endgroup$
    – Steven Stadnicki
    Feb 13 '14 at 16:32
















$begingroup$
Note (perhaps relevantly) that this function isn't just $mathbb{Q}mapstomathbb{R}$; it's also $mathbb{Q}mapstomathbb{Q}$ (which may be relevant for the next thing you learn using this function!).
$endgroup$
– Steven Stadnicki
Feb 13 '14 at 16:32




$begingroup$
Note (perhaps relevantly) that this function isn't just $mathbb{Q}mapstomathbb{R}$; it's also $mathbb{Q}mapstomathbb{Q}$ (which may be relevant for the next thing you learn using this function!).
$endgroup$
– Steven Stadnicki
Feb 13 '14 at 16:32










5 Answers
5






active

oldest

votes


















7












$begingroup$

Because $sqrt{2}$ is not in $mathbb{Q}$, you don't have to check continuity at $sqrt{2}$: it's completely irrelevant!






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    Hint: You are right, you need only check continuity around (i.e. nearby) $sqrt 2$, not at $sqrt 2$.






    share|cite|improve this answer









    $endgroup$





















      4












      $begingroup$

      $f$ is continuous if it is continuous in every point in $mathbb Q$. You don't have to check for continuity around a point that does not exist in the domain.



      Hint: for all other points, you can see that there exists a neighborhood on which $f$ is constant.






      share|cite|improve this answer









      $endgroup$





















        2












        $begingroup$

        Since the image of $f$ is discrete, we need only check that the fibers of $f$ are open. But
        $$
        f^{-1}(0)=(-infty,sqrt{2})capmathbb Q
        quadtext{and}quad
        f^{-1}(1)=(sqrt{2},infty)capmathbb Q
        $$
        so $f$ is continuous.






        share|cite|improve this answer











        $endgroup$





















          0












          $begingroup$

          Try to apply the definition of the continuity. Here's my solution.



          Let $epsilon > 0$ and let $x_0 in mathbb{Q}$. Now set $delta$ := |$sqrt{2}-x_0$|. Then, clearly $delta neq 0$ and $delta >0$.



          Let $x in mathbb{Q}$ such that |$x-x_0$| < $delta$



          then if $x_0 > sqrt{2}$ we have $x > sqrt{2}$ and so $f(x)=1=f(x_0)$



          and if $x_0 < sqrt{2}$ we have $x < sqrt{2} $ and so $f(x)=0=f(x_0)$



          So we have | $f(x)- f(x_0)$ | $= 0 < epsilon$



          Hence $f$ is continuous in $x_0 in mathbb{Q}$ and since this calculation holds for any such $x_0 in mathbb{Q}$, we have that $f$ is continuous on $mathbb{Q}$.






          share|cite|improve this answer









          $endgroup$













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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            7












            $begingroup$

            Because $sqrt{2}$ is not in $mathbb{Q}$, you don't have to check continuity at $sqrt{2}$: it's completely irrelevant!






            share|cite|improve this answer









            $endgroup$


















              7












              $begingroup$

              Because $sqrt{2}$ is not in $mathbb{Q}$, you don't have to check continuity at $sqrt{2}$: it's completely irrelevant!






              share|cite|improve this answer









              $endgroup$
















                7












                7








                7





                $begingroup$

                Because $sqrt{2}$ is not in $mathbb{Q}$, you don't have to check continuity at $sqrt{2}$: it's completely irrelevant!






                share|cite|improve this answer









                $endgroup$



                Because $sqrt{2}$ is not in $mathbb{Q}$, you don't have to check continuity at $sqrt{2}$: it's completely irrelevant!







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 13 '14 at 8:20









                HurkylHurkyl

                111k9118262




                111k9118262























                    4












                    $begingroup$

                    Hint: You are right, you need only check continuity around (i.e. nearby) $sqrt 2$, not at $sqrt 2$.






                    share|cite|improve this answer









                    $endgroup$


















                      4












                      $begingroup$

                      Hint: You are right, you need only check continuity around (i.e. nearby) $sqrt 2$, not at $sqrt 2$.






                      share|cite|improve this answer









                      $endgroup$
















                        4












                        4








                        4





                        $begingroup$

                        Hint: You are right, you need only check continuity around (i.e. nearby) $sqrt 2$, not at $sqrt 2$.






                        share|cite|improve this answer









                        $endgroup$



                        Hint: You are right, you need only check continuity around (i.e. nearby) $sqrt 2$, not at $sqrt 2$.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Feb 13 '14 at 8:20









                        Hagen von EitzenHagen von Eitzen

                        278k22269497




                        278k22269497























                            4












                            $begingroup$

                            $f$ is continuous if it is continuous in every point in $mathbb Q$. You don't have to check for continuity around a point that does not exist in the domain.



                            Hint: for all other points, you can see that there exists a neighborhood on which $f$ is constant.






                            share|cite|improve this answer









                            $endgroup$


















                              4












                              $begingroup$

                              $f$ is continuous if it is continuous in every point in $mathbb Q$. You don't have to check for continuity around a point that does not exist in the domain.



                              Hint: for all other points, you can see that there exists a neighborhood on which $f$ is constant.






                              share|cite|improve this answer









                              $endgroup$
















                                4












                                4








                                4





                                $begingroup$

                                $f$ is continuous if it is continuous in every point in $mathbb Q$. You don't have to check for continuity around a point that does not exist in the domain.



                                Hint: for all other points, you can see that there exists a neighborhood on which $f$ is constant.






                                share|cite|improve this answer









                                $endgroup$



                                $f$ is continuous if it is continuous in every point in $mathbb Q$. You don't have to check for continuity around a point that does not exist in the domain.



                                Hint: for all other points, you can see that there exists a neighborhood on which $f$ is constant.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Feb 13 '14 at 8:21









                                5xum5xum

                                90.2k394161




                                90.2k394161























                                    2












                                    $begingroup$

                                    Since the image of $f$ is discrete, we need only check that the fibers of $f$ are open. But
                                    $$
                                    f^{-1}(0)=(-infty,sqrt{2})capmathbb Q
                                    quadtext{and}quad
                                    f^{-1}(1)=(sqrt{2},infty)capmathbb Q
                                    $$
                                    so $f$ is continuous.






                                    share|cite|improve this answer











                                    $endgroup$


















                                      2












                                      $begingroup$

                                      Since the image of $f$ is discrete, we need only check that the fibers of $f$ are open. But
                                      $$
                                      f^{-1}(0)=(-infty,sqrt{2})capmathbb Q
                                      quadtext{and}quad
                                      f^{-1}(1)=(sqrt{2},infty)capmathbb Q
                                      $$
                                      so $f$ is continuous.






                                      share|cite|improve this answer











                                      $endgroup$
















                                        2












                                        2








                                        2





                                        $begingroup$

                                        Since the image of $f$ is discrete, we need only check that the fibers of $f$ are open. But
                                        $$
                                        f^{-1}(0)=(-infty,sqrt{2})capmathbb Q
                                        quadtext{and}quad
                                        f^{-1}(1)=(sqrt{2},infty)capmathbb Q
                                        $$
                                        so $f$ is continuous.






                                        share|cite|improve this answer











                                        $endgroup$



                                        Since the image of $f$ is discrete, we need only check that the fibers of $f$ are open. But
                                        $$
                                        f^{-1}(0)=(-infty,sqrt{2})capmathbb Q
                                        quadtext{and}quad
                                        f^{-1}(1)=(sqrt{2},infty)capmathbb Q
                                        $$
                                        so $f$ is continuous.







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Feb 13 '14 at 16:30

























                                        answered Feb 13 '14 at 8:31









                                        Brian FitzpatrickBrian Fitzpatrick

                                        21.1k42958




                                        21.1k42958























                                            0












                                            $begingroup$

                                            Try to apply the definition of the continuity. Here's my solution.



                                            Let $epsilon > 0$ and let $x_0 in mathbb{Q}$. Now set $delta$ := |$sqrt{2}-x_0$|. Then, clearly $delta neq 0$ and $delta >0$.



                                            Let $x in mathbb{Q}$ such that |$x-x_0$| < $delta$



                                            then if $x_0 > sqrt{2}$ we have $x > sqrt{2}$ and so $f(x)=1=f(x_0)$



                                            and if $x_0 < sqrt{2}$ we have $x < sqrt{2} $ and so $f(x)=0=f(x_0)$



                                            So we have | $f(x)- f(x_0)$ | $= 0 < epsilon$



                                            Hence $f$ is continuous in $x_0 in mathbb{Q}$ and since this calculation holds for any such $x_0 in mathbb{Q}$, we have that $f$ is continuous on $mathbb{Q}$.






                                            share|cite|improve this answer









                                            $endgroup$


















                                              0












                                              $begingroup$

                                              Try to apply the definition of the continuity. Here's my solution.



                                              Let $epsilon > 0$ and let $x_0 in mathbb{Q}$. Now set $delta$ := |$sqrt{2}-x_0$|. Then, clearly $delta neq 0$ and $delta >0$.



                                              Let $x in mathbb{Q}$ such that |$x-x_0$| < $delta$



                                              then if $x_0 > sqrt{2}$ we have $x > sqrt{2}$ and so $f(x)=1=f(x_0)$



                                              and if $x_0 < sqrt{2}$ we have $x < sqrt{2} $ and so $f(x)=0=f(x_0)$



                                              So we have | $f(x)- f(x_0)$ | $= 0 < epsilon$



                                              Hence $f$ is continuous in $x_0 in mathbb{Q}$ and since this calculation holds for any such $x_0 in mathbb{Q}$, we have that $f$ is continuous on $mathbb{Q}$.






                                              share|cite|improve this answer









                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                Try to apply the definition of the continuity. Here's my solution.



                                                Let $epsilon > 0$ and let $x_0 in mathbb{Q}$. Now set $delta$ := |$sqrt{2}-x_0$|. Then, clearly $delta neq 0$ and $delta >0$.



                                                Let $x in mathbb{Q}$ such that |$x-x_0$| < $delta$



                                                then if $x_0 > sqrt{2}$ we have $x > sqrt{2}$ and so $f(x)=1=f(x_0)$



                                                and if $x_0 < sqrt{2}$ we have $x < sqrt{2} $ and so $f(x)=0=f(x_0)$



                                                So we have | $f(x)- f(x_0)$ | $= 0 < epsilon$



                                                Hence $f$ is continuous in $x_0 in mathbb{Q}$ and since this calculation holds for any such $x_0 in mathbb{Q}$, we have that $f$ is continuous on $mathbb{Q}$.






                                                share|cite|improve this answer









                                                $endgroup$



                                                Try to apply the definition of the continuity. Here's my solution.



                                                Let $epsilon > 0$ and let $x_0 in mathbb{Q}$. Now set $delta$ := |$sqrt{2}-x_0$|. Then, clearly $delta neq 0$ and $delta >0$.



                                                Let $x in mathbb{Q}$ such that |$x-x_0$| < $delta$



                                                then if $x_0 > sqrt{2}$ we have $x > sqrt{2}$ and so $f(x)=1=f(x_0)$



                                                and if $x_0 < sqrt{2}$ we have $x < sqrt{2} $ and so $f(x)=0=f(x_0)$



                                                So we have | $f(x)- f(x_0)$ | $= 0 < epsilon$



                                                Hence $f$ is continuous in $x_0 in mathbb{Q}$ and since this calculation holds for any such $x_0 in mathbb{Q}$, we have that $f$ is continuous on $mathbb{Q}$.







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Dec 6 '18 at 13:01









                                                Alain.KlbtrAlain.Klbtr

                                                827




                                                827






























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