Problem: proving that function is constant [on hold]











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Let $f colon mathbb{R} to mathbb{R} $ be a continuous function so that $ f(x) in mathbb I = mathbb Rsetminus mathbb Q, forall x in mathbb{R}$. Prove that $f$ is constant.



I tried assuming it's not a constant but I can't get to a contradiction with continuity.






functions continuity







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put on hold as off-topic by user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R 9 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 2




    What's $mathbb{I}$?
    – Dante Grevino
    23 hours ago










  • @DanteGrevino it's set of irrational numbers
    – user560461
    23 hours ago






  • 5




    Hint: assume it is not and then use the Intermediate Value Theorem to show that $f$ must take on a rational value.
    – John Douma
    23 hours ago










  • Basically, rationals are everywhere. You cannot move an inch, a millimeter or an angstrom without stepping on a rational, or, for that matter, an infinity of rationals.
    – Eric Duminil
    20 hours ago















up vote
4
down vote

favorite
1












Let $f colon mathbb{R} to mathbb{R} $ be a continuous function so that $ f(x) in mathbb I = mathbb Rsetminus mathbb Q, forall x in mathbb{R}$. Prove that $f$ is constant.



I tried assuming it's not a constant but I can't get to a contradiction with continuity.






functions continuity







share|cite|improve this question















put on hold as off-topic by user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R 9 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 2




    What's $mathbb{I}$?
    – Dante Grevino
    23 hours ago










  • @DanteGrevino it's set of irrational numbers
    – user560461
    23 hours ago






  • 5




    Hint: assume it is not and then use the Intermediate Value Theorem to show that $f$ must take on a rational value.
    – John Douma
    23 hours ago










  • Basically, rationals are everywhere. You cannot move an inch, a millimeter or an angstrom without stepping on a rational, or, for that matter, an infinity of rationals.
    – Eric Duminil
    20 hours ago













up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





Let $f colon mathbb{R} to mathbb{R} $ be a continuous function so that $ f(x) in mathbb I = mathbb Rsetminus mathbb Q, forall x in mathbb{R}$. Prove that $f$ is constant.



I tried assuming it's not a constant but I can't get to a contradiction with continuity.






functions continuity







share|cite|improve this question















Let $f colon mathbb{R} to mathbb{R} $ be a continuous function so that $ f(x) in mathbb I = mathbb Rsetminus mathbb Q, forall x in mathbb{R}$. Prove that $f$ is constant.



I tried assuming it's not a constant but I can't get to a contradiction with continuity.






functions continuity




functions continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited 11 hours ago









Surb

37k94375




37k94375










asked 23 hours ago









user560461

525




525




put on hold as off-topic by user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R 9 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R 9 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj R

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    What's $mathbb{I}$?
    – Dante Grevino
    23 hours ago










  • @DanteGrevino it's set of irrational numbers
    – user560461
    23 hours ago






  • 5




    Hint: assume it is not and then use the Intermediate Value Theorem to show that $f$ must take on a rational value.
    – John Douma
    23 hours ago










  • Basically, rationals are everywhere. You cannot move an inch, a millimeter or an angstrom without stepping on a rational, or, for that matter, an infinity of rationals.
    – Eric Duminil
    20 hours ago














  • 2




    What's $mathbb{I}$?
    – Dante Grevino
    23 hours ago










  • @DanteGrevino it's set of irrational numbers
    – user560461
    23 hours ago






  • 5




    Hint: assume it is not and then use the Intermediate Value Theorem to show that $f$ must take on a rational value.
    – John Douma
    23 hours ago










  • Basically, rationals are everywhere. You cannot move an inch, a millimeter or an angstrom without stepping on a rational, or, for that matter, an infinity of rationals.
    – Eric Duminil
    20 hours ago








2




2




What's $mathbb{I}$?
– Dante Grevino
23 hours ago




What's $mathbb{I}$?
– Dante Grevino
23 hours ago












@DanteGrevino it's set of irrational numbers
– user560461
23 hours ago




@DanteGrevino it's set of irrational numbers
– user560461
23 hours ago




5




5




Hint: assume it is not and then use the Intermediate Value Theorem to show that $f$ must take on a rational value.
– John Douma
23 hours ago




Hint: assume it is not and then use the Intermediate Value Theorem to show that $f$ must take on a rational value.
– John Douma
23 hours ago












Basically, rationals are everywhere. You cannot move an inch, a millimeter or an angstrom without stepping on a rational, or, for that matter, an infinity of rationals.
– Eric Duminil
20 hours ago




Basically, rationals are everywhere. You cannot move an inch, a millimeter or an angstrom without stepping on a rational, or, for that matter, an infinity of rationals.
– Eric Duminil
20 hours ago










4 Answers
4






active

oldest

votes

















up vote
10
down vote













Assuming we know 1. The Intermediate Value Theorem and 2. that between any two irrationals there is a rational, then the proof by contradiction should work fine.



If $f(a)<f(b)$ then there has to be a rational number $r$ with $f(a)<r<f(b).$ By IVT, there is a $c$ such that $f(c)=r$, contradiction.






share|cite|improve this answer




























    up vote
    9
    down vote













    The image $f(mathbb{R})$ is path-connected and the only non-empty path-connected subspaces of $mathbb{I}$ are the points. So $f$ is constant.






    share|cite|improve this answer








    New contributor




    Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      up vote
      2
      down vote













      The continuous image of a connected set is connected.



      The only connected sets in $mathbb Rsetminus mathbb Q$ are single points.



      Done.






      share|cite|improve this answer




























        up vote
        2
        down vote













        Suppose that $f$ is not constant. Then there are $i_1, i_2 in mathbb I$ such that $i_1 ne i_2$ and $i_1,i_2 in f(mathbb R)$. WLOG we can assume that $i_1 <i_2$. Now pick some rational number in $(i_1, i_2)$. The intermediate value theorem gives $r in f(mathbb R)$, a contradiction.






        share|cite|improve this answer






























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          10
          down vote













          Assuming we know 1. The Intermediate Value Theorem and 2. that between any two irrationals there is a rational, then the proof by contradiction should work fine.



          If $f(a)<f(b)$ then there has to be a rational number $r$ with $f(a)<r<f(b).$ By IVT, there is a $c$ such that $f(c)=r$, contradiction.






          share|cite|improve this answer

























            up vote
            10
            down vote













            Assuming we know 1. The Intermediate Value Theorem and 2. that between any two irrationals there is a rational, then the proof by contradiction should work fine.



            If $f(a)<f(b)$ then there has to be a rational number $r$ with $f(a)<r<f(b).$ By IVT, there is a $c$ such that $f(c)=r$, contradiction.






            share|cite|improve this answer























              up vote
              10
              down vote










              up vote
              10
              down vote









              Assuming we know 1. The Intermediate Value Theorem and 2. that between any two irrationals there is a rational, then the proof by contradiction should work fine.



              If $f(a)<f(b)$ then there has to be a rational number $r$ with $f(a)<r<f(b).$ By IVT, there is a $c$ such that $f(c)=r$, contradiction.






              share|cite|improve this answer












              Assuming we know 1. The Intermediate Value Theorem and 2. that between any two irrationals there is a rational, then the proof by contradiction should work fine.



              If $f(a)<f(b)$ then there has to be a rational number $r$ with $f(a)<r<f(b).$ By IVT, there is a $c$ such that $f(c)=r$, contradiction.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 23 hours ago









              B. Goddard

              17.9k21340




              17.9k21340






















                  up vote
                  9
                  down vote













                  The image $f(mathbb{R})$ is path-connected and the only non-empty path-connected subspaces of $mathbb{I}$ are the points. So $f$ is constant.






                  share|cite|improve this answer








                  New contributor




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






















                    up vote
                    9
                    down vote













                    The image $f(mathbb{R})$ is path-connected and the only non-empty path-connected subspaces of $mathbb{I}$ are the points. So $f$ is constant.






                    share|cite|improve this answer








                    New contributor




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




















                      up vote
                      9
                      down vote










                      up vote
                      9
                      down vote









                      The image $f(mathbb{R})$ is path-connected and the only non-empty path-connected subspaces of $mathbb{I}$ are the points. So $f$ is constant.






                      share|cite|improve this answer








                      New contributor




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









                      The image $f(mathbb{R})$ is path-connected and the only non-empty path-connected subspaces of $mathbb{I}$ are the points. So $f$ is constant.







                      share|cite|improve this answer








                      New contributor




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









                      share|cite|improve this answer



                      share|cite|improve this answer






                      New contributor




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









                      answered 23 hours ago









                      Dante Grevino

                      1463




                      1463




                      New contributor




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





                      New contributor





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






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






















                          up vote
                          2
                          down vote













                          The continuous image of a connected set is connected.



                          The only connected sets in $mathbb Rsetminus mathbb Q$ are single points.



                          Done.






                          share|cite|improve this answer

























                            up vote
                            2
                            down vote













                            The continuous image of a connected set is connected.



                            The only connected sets in $mathbb Rsetminus mathbb Q$ are single points.



                            Done.






                            share|cite|improve this answer























                              up vote
                              2
                              down vote










                              up vote
                              2
                              down vote









                              The continuous image of a connected set is connected.



                              The only connected sets in $mathbb Rsetminus mathbb Q$ are single points.



                              Done.






                              share|cite|improve this answer












                              The continuous image of a connected set is connected.



                              The only connected sets in $mathbb Rsetminus mathbb Q$ are single points.



                              Done.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 17 hours ago









                              MPW

                              29.4k11856




                              29.4k11856






















                                  up vote
                                  2
                                  down vote













                                  Suppose that $f$ is not constant. Then there are $i_1, i_2 in mathbb I$ such that $i_1 ne i_2$ and $i_1,i_2 in f(mathbb R)$. WLOG we can assume that $i_1 <i_2$. Now pick some rational number in $(i_1, i_2)$. The intermediate value theorem gives $r in f(mathbb R)$, a contradiction.






                                  share|cite|improve this answer



























                                    up vote
                                    2
                                    down vote













                                    Suppose that $f$ is not constant. Then there are $i_1, i_2 in mathbb I$ such that $i_1 ne i_2$ and $i_1,i_2 in f(mathbb R)$. WLOG we can assume that $i_1 <i_2$. Now pick some rational number in $(i_1, i_2)$. The intermediate value theorem gives $r in f(mathbb R)$, a contradiction.






                                    share|cite|improve this answer

























                                      up vote
                                      2
                                      down vote










                                      up vote
                                      2
                                      down vote









                                      Suppose that $f$ is not constant. Then there are $i_1, i_2 in mathbb I$ such that $i_1 ne i_2$ and $i_1,i_2 in f(mathbb R)$. WLOG we can assume that $i_1 <i_2$. Now pick some rational number in $(i_1, i_2)$. The intermediate value theorem gives $r in f(mathbb R)$, a contradiction.






                                      share|cite|improve this answer














                                      Suppose that $f$ is not constant. Then there are $i_1, i_2 in mathbb I$ such that $i_1 ne i_2$ and $i_1,i_2 in f(mathbb R)$. WLOG we can assume that $i_1 <i_2$. Now pick some rational number in $(i_1, i_2)$. The intermediate value theorem gives $r in f(mathbb R)$, a contradiction.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited 17 hours ago









                                      MPW

                                      29.4k11856




                                      29.4k11856










                                      answered 23 hours ago









                                      Fred

                                      41.9k1642




                                      41.9k1642















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