Function with arbitrary small period

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Is there a function f: $mathbbR to mathbbR$ with arbitrary small period different from $f(x) = k$? ($forall epsilon >0 exists a < epsilon $ such that f(x) has a periodicity $a$)
I think the function is the Dirichlet function but I don't know how to prove it properly.










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  • 2




    F(x)=1 for numbers with a halting decimal expansion, for example $3.452$ and not $1/3$
    – Empy2
    Oct 4 at 12:00










  • @Empy2 Nice: your example also works for domain $mathbbQ$.
    – aschepler
    Oct 4 at 12:38






  • 3




    I would avoid saying “such that the period of $f(x)$ is $a$” here. The term period would usually denote the smallest $a>0$ that fulfills $f(x+a) = f(x)$. Here you should rather just say “such that $f$ has a periodicity $a$”, or “such that $f$ is $a$-periodic”, which does assert $f(x+a) = f(x)$ but does not make any statement as to whether there exists also $b in ]0,a[$ with $f(x+b)=f(x)$.
    – leftaroundabout
    Oct 4 at 13:55











  • Of course you want $0<a<epsilon$
    – MPW
    Oct 4 at 18:07










  • You can construct a lot of such functions. The recipe: 1) Take any sequence $a_k$ which has infinite number of values in any neighbourhood of zero; 2) Make a set consisting of all finite sums of $a_k$ with arbitrary integer coefficients; 3) The characteristic function of this set will satisfy your requirements for $f$. If $a_k=1/k$ then you're getting rationals. If $a_k=10^-k$ then you're getting Empy2's set. If $a_k=k^-pi$ then you're getting new set which is hard to imagine :-) If $a_k=sink$ then you're getting new interesting set.
    – Egor Skriptunoff
    2 days ago














up vote
10
down vote

favorite












Is there a function f: $mathbbR to mathbbR$ with arbitrary small period different from $f(x) = k$? ($forall epsilon >0 exists a < epsilon $ such that f(x) has a periodicity $a$)
I think the function is the Dirichlet function but I don't know how to prove it properly.










share|cite|improve this question









New contributor




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















  • 2




    F(x)=1 for numbers with a halting decimal expansion, for example $3.452$ and not $1/3$
    – Empy2
    Oct 4 at 12:00










  • @Empy2 Nice: your example also works for domain $mathbbQ$.
    – aschepler
    Oct 4 at 12:38






  • 3




    I would avoid saying “such that the period of $f(x)$ is $a$” here. The term period would usually denote the smallest $a>0$ that fulfills $f(x+a) = f(x)$. Here you should rather just say “such that $f$ has a periodicity $a$”, or “such that $f$ is $a$-periodic”, which does assert $f(x+a) = f(x)$ but does not make any statement as to whether there exists also $b in ]0,a[$ with $f(x+b)=f(x)$.
    – leftaroundabout
    Oct 4 at 13:55











  • Of course you want $0<a<epsilon$
    – MPW
    Oct 4 at 18:07










  • You can construct a lot of such functions. The recipe: 1) Take any sequence $a_k$ which has infinite number of values in any neighbourhood of zero; 2) Make a set consisting of all finite sums of $a_k$ with arbitrary integer coefficients; 3) The characteristic function of this set will satisfy your requirements for $f$. If $a_k=1/k$ then you're getting rationals. If $a_k=10^-k$ then you're getting Empy2's set. If $a_k=k^-pi$ then you're getting new set which is hard to imagine :-) If $a_k=sink$ then you're getting new interesting set.
    – Egor Skriptunoff
    2 days ago












up vote
10
down vote

favorite









up vote
10
down vote

favorite











Is there a function f: $mathbbR to mathbbR$ with arbitrary small period different from $f(x) = k$? ($forall epsilon >0 exists a < epsilon $ such that f(x) has a periodicity $a$)
I think the function is the Dirichlet function but I don't know how to prove it properly.










share|cite|improve this question









New contributor




Lance is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Is there a function f: $mathbbR to mathbbR$ with arbitrary small period different from $f(x) = k$? ($forall epsilon >0 exists a < epsilon $ such that f(x) has a periodicity $a$)
I think the function is the Dirichlet function but I don't know how to prove it properly.







periodic-functions






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edited Oct 4 at 19:06





















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asked Oct 4 at 11:45









Lance

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  • 2




    F(x)=1 for numbers with a halting decimal expansion, for example $3.452$ and not $1/3$
    – Empy2
    Oct 4 at 12:00










  • @Empy2 Nice: your example also works for domain $mathbbQ$.
    – aschepler
    Oct 4 at 12:38






  • 3




    I would avoid saying “such that the period of $f(x)$ is $a$” here. The term period would usually denote the smallest $a>0$ that fulfills $f(x+a) = f(x)$. Here you should rather just say “such that $f$ has a periodicity $a$”, or “such that $f$ is $a$-periodic”, which does assert $f(x+a) = f(x)$ but does not make any statement as to whether there exists also $b in ]0,a[$ with $f(x+b)=f(x)$.
    – leftaroundabout
    Oct 4 at 13:55











  • Of course you want $0<a<epsilon$
    – MPW
    Oct 4 at 18:07










  • You can construct a lot of such functions. The recipe: 1) Take any sequence $a_k$ which has infinite number of values in any neighbourhood of zero; 2) Make a set consisting of all finite sums of $a_k$ with arbitrary integer coefficients; 3) The characteristic function of this set will satisfy your requirements for $f$. If $a_k=1/k$ then you're getting rationals. If $a_k=10^-k$ then you're getting Empy2's set. If $a_k=k^-pi$ then you're getting new set which is hard to imagine :-) If $a_k=sink$ then you're getting new interesting set.
    – Egor Skriptunoff
    2 days ago












  • 2




    F(x)=1 for numbers with a halting decimal expansion, for example $3.452$ and not $1/3$
    – Empy2
    Oct 4 at 12:00










  • @Empy2 Nice: your example also works for domain $mathbbQ$.
    – aschepler
    Oct 4 at 12:38






  • 3




    I would avoid saying “such that the period of $f(x)$ is $a$” here. The term period would usually denote the smallest $a>0$ that fulfills $f(x+a) = f(x)$. Here you should rather just say “such that $f$ has a periodicity $a$”, or “such that $f$ is $a$-periodic”, which does assert $f(x+a) = f(x)$ but does not make any statement as to whether there exists also $b in ]0,a[$ with $f(x+b)=f(x)$.
    – leftaroundabout
    Oct 4 at 13:55











  • Of course you want $0<a<epsilon$
    – MPW
    Oct 4 at 18:07










  • You can construct a lot of such functions. The recipe: 1) Take any sequence $a_k$ which has infinite number of values in any neighbourhood of zero; 2) Make a set consisting of all finite sums of $a_k$ with arbitrary integer coefficients; 3) The characteristic function of this set will satisfy your requirements for $f$. If $a_k=1/k$ then you're getting rationals. If $a_k=10^-k$ then you're getting Empy2's set. If $a_k=k^-pi$ then you're getting new set which is hard to imagine :-) If $a_k=sink$ then you're getting new interesting set.
    – Egor Skriptunoff
    2 days ago







2




2




F(x)=1 for numbers with a halting decimal expansion, for example $3.452$ and not $1/3$
– Empy2
Oct 4 at 12:00




F(x)=1 for numbers with a halting decimal expansion, for example $3.452$ and not $1/3$
– Empy2
Oct 4 at 12:00












@Empy2 Nice: your example also works for domain $mathbbQ$.
– aschepler
Oct 4 at 12:38




@Empy2 Nice: your example also works for domain $mathbbQ$.
– aschepler
Oct 4 at 12:38




3




3




I would avoid saying “such that the period of $f(x)$ is $a$” here. The term period would usually denote the smallest $a>0$ that fulfills $f(x+a) = f(x)$. Here you should rather just say “such that $f$ has a periodicity $a$”, or “such that $f$ is $a$-periodic”, which does assert $f(x+a) = f(x)$ but does not make any statement as to whether there exists also $b in ]0,a[$ with $f(x+b)=f(x)$.
– leftaroundabout
Oct 4 at 13:55





I would avoid saying “such that the period of $f(x)$ is $a$” here. The term period would usually denote the smallest $a>0$ that fulfills $f(x+a) = f(x)$. Here you should rather just say “such that $f$ has a periodicity $a$”, or “such that $f$ is $a$-periodic”, which does assert $f(x+a) = f(x)$ but does not make any statement as to whether there exists also $b in ]0,a[$ with $f(x+b)=f(x)$.
– leftaroundabout
Oct 4 at 13:55













Of course you want $0<a<epsilon$
– MPW
Oct 4 at 18:07




Of course you want $0<a<epsilon$
– MPW
Oct 4 at 18:07












You can construct a lot of such functions. The recipe: 1) Take any sequence $a_k$ which has infinite number of values in any neighbourhood of zero; 2) Make a set consisting of all finite sums of $a_k$ with arbitrary integer coefficients; 3) The characteristic function of this set will satisfy your requirements for $f$. If $a_k=1/k$ then you're getting rationals. If $a_k=10^-k$ then you're getting Empy2's set. If $a_k=k^-pi$ then you're getting new set which is hard to imagine :-) If $a_k=sink$ then you're getting new interesting set.
– Egor Skriptunoff
2 days ago




You can construct a lot of such functions. The recipe: 1) Take any sequence $a_k$ which has infinite number of values in any neighbourhood of zero; 2) Make a set consisting of all finite sums of $a_k$ with arbitrary integer coefficients; 3) The characteristic function of this set will satisfy your requirements for $f$. If $a_k=1/k$ then you're getting rationals. If $a_k=10^-k$ then you're getting Empy2's set. If $a_k=k^-pi$ then you're getting new set which is hard to imagine :-) If $a_k=sink$ then you're getting new interesting set.
– Egor Skriptunoff
2 days ago










5 Answers
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up vote
17
down vote













You're right. The characteristic function of the rationals is periodic of period $1/n$ for all $n in mathbb N$ because $x$ is rational iff $x+1/n$ is rational.






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    up vote
    7
    down vote













    You're correct.



    Let $epsilon$ be arbitrarily small. You need to prove that there exists some $0<p<epsilon$ such that $D(x)=D(x+p)$ for all $xin mathbbR$.



    We know that $epsilon$ is some positive real number, so there exists some $pin mathbbQ$ such that $0<p<epsilon$. Let's look at an arbitrary $xin mathbbR$ and see if our property is satisfied or not:



    If $x$ is rational, then $D(x)=1$. Since the sum of two rationals is rational, then $D(x+p)=1$ too.



    If $x$ is irrational, then $D(x)=0$. Since the sum of a rational and an irrational is irrational, then $D(x+p)=0$ too.



    So if we choose our period to be $p$, our property is satisfied.






    share|cite|improve this answer



























      up vote
      3
      down vote













      The main question has been answered in other answers well enough, but I would like to address a few natural follow-up questions. What about continuous functions $f$ with this property?



      It turns out that in this case there are no nontrivial solutions - every such function is constant. Here's a topological proof: Let $K=xinBbb Rmid forall y, f(y)=f(x+y)$ be the set of periods of $f$. If $f$ is continuous, then this is an intersection of the sets $xinBbb Rmid f(y)=f(x+y)$, which is closed (it is the preimage of $0$ under the function $g(x)=f(y)-f(x+y)$), so $K$ itself is closed. $K$ is also dense in $Bbb R$, because it is an additive group with arbitrarily small elements, so $K=Bbb R$ and hence $f(x)=f(y)$ for all $x,yin Bbb R$.



      If we consider discontinuous functions again, then we know $K$ is a dense additive subgroup of $Bbb R$. Does every dense additive subgroup generate such a function? Yes, we can just take the characteristic function of $K$. For a fixed $K$, the space of such functions is just all functions $Bbb R/Kto Bbb R$. This is another way to get at the constancy result, since as a topological group, $Bbb R/K$ has the indiscrete topology, because any open set will cover $Bbb R$ if copied around with translations by $K$.



      Of course $Bbb R/K$ can be uncountable, for example if $K=Bbb Q$ or any other countable subgroup. Can it be countable or finite? It can be countable assuming some choice, as observed in TomGrubb's answer. If we consider a Hamel basis $B$ of $Bbb R$ over $Bbb Q$, then the set of all real numbers with zero first projection is a subgroup $K$ of $Bbb R$ for which $Bbb R/Ksimeq Bbb Q$.



      But it can't be finite (unless it is trivial). In other words, there is no coherent way to talk about real numbers being partitioned into the "even" and "odd" ones. If $Bbb R/K$ has $n>1$ elements, then that means that every number which is a multiple of $n$ is in $K$; but every real number is a multiple of $n$, to wit, $x=n(x/n)$.






      share|cite|improve this answer



























        up vote
        3
        down vote













        Another way to look at it is to think about the set of periods, i.e.



        $$P = f(x + p) = f(x) text for all x in mathbbR $$



        Zero is clearly a member of this set no matter what f is. P is closed under addition and negation. So clearly P is a group over addition. So, if you want to find a function that has arbitrarily small periods, you want to find a subgroup of $mathbbR$ that has arbitrarily small values. $mathbbQ$ is the obvious choice, so the characteristic function for $mathbbQ$ works, as stated in another answer.






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          up vote
          0
          down vote













          Here's a different function with the same property (which relies on a fair bit of choice). Choose a Hamel basis for $mathbbR$ over $mathbbQ$ and pick a basis vector $v$. Let $f$ be the function which projects onto the $v$ coordinate. Then for any other basis vector $u$ and any integer $n$,
          $$
          f(x+u/n)=f(x).
          $$

          More can be found on these functions in the article "Discontinuous additive functions" by Bernardi.






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          • Hope you don't mind my edit, since it's not that much choice needed for what you want. =)
            – user21820
            Oct 4 at 17:10










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          5 Answers
          5






          active

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          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          17
          down vote













          You're right. The characteristic function of the rationals is periodic of period $1/n$ for all $n in mathbb N$ because $x$ is rational iff $x+1/n$ is rational.






          share|cite|improve this answer
























            up vote
            17
            down vote













            You're right. The characteristic function of the rationals is periodic of period $1/n$ for all $n in mathbb N$ because $x$ is rational iff $x+1/n$ is rational.






            share|cite|improve this answer






















              up vote
              17
              down vote










              up vote
              17
              down vote









              You're right. The characteristic function of the rationals is periodic of period $1/n$ for all $n in mathbb N$ because $x$ is rational iff $x+1/n$ is rational.






              share|cite|improve this answer












              You're right. The characteristic function of the rationals is periodic of period $1/n$ for all $n in mathbb N$ because $x$ is rational iff $x+1/n$ is rational.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Oct 4 at 11:50









              lhf

              159k9161376




              159k9161376




















                  up vote
                  7
                  down vote













                  You're correct.



                  Let $epsilon$ be arbitrarily small. You need to prove that there exists some $0<p<epsilon$ such that $D(x)=D(x+p)$ for all $xin mathbbR$.



                  We know that $epsilon$ is some positive real number, so there exists some $pin mathbbQ$ such that $0<p<epsilon$. Let's look at an arbitrary $xin mathbbR$ and see if our property is satisfied or not:



                  If $x$ is rational, then $D(x)=1$. Since the sum of two rationals is rational, then $D(x+p)=1$ too.



                  If $x$ is irrational, then $D(x)=0$. Since the sum of a rational and an irrational is irrational, then $D(x+p)=0$ too.



                  So if we choose our period to be $p$, our property is satisfied.






                  share|cite|improve this answer
























                    up vote
                    7
                    down vote













                    You're correct.



                    Let $epsilon$ be arbitrarily small. You need to prove that there exists some $0<p<epsilon$ such that $D(x)=D(x+p)$ for all $xin mathbbR$.



                    We know that $epsilon$ is some positive real number, so there exists some $pin mathbbQ$ such that $0<p<epsilon$. Let's look at an arbitrary $xin mathbbR$ and see if our property is satisfied or not:



                    If $x$ is rational, then $D(x)=1$. Since the sum of two rationals is rational, then $D(x+p)=1$ too.



                    If $x$ is irrational, then $D(x)=0$. Since the sum of a rational and an irrational is irrational, then $D(x+p)=0$ too.



                    So if we choose our period to be $p$, our property is satisfied.






                    share|cite|improve this answer






















                      up vote
                      7
                      down vote










                      up vote
                      7
                      down vote









                      You're correct.



                      Let $epsilon$ be arbitrarily small. You need to prove that there exists some $0<p<epsilon$ such that $D(x)=D(x+p)$ for all $xin mathbbR$.



                      We know that $epsilon$ is some positive real number, so there exists some $pin mathbbQ$ such that $0<p<epsilon$. Let's look at an arbitrary $xin mathbbR$ and see if our property is satisfied or not:



                      If $x$ is rational, then $D(x)=1$. Since the sum of two rationals is rational, then $D(x+p)=1$ too.



                      If $x$ is irrational, then $D(x)=0$. Since the sum of a rational and an irrational is irrational, then $D(x+p)=0$ too.



                      So if we choose our period to be $p$, our property is satisfied.






                      share|cite|improve this answer












                      You're correct.



                      Let $epsilon$ be arbitrarily small. You need to prove that there exists some $0<p<epsilon$ such that $D(x)=D(x+p)$ for all $xin mathbbR$.



                      We know that $epsilon$ is some positive real number, so there exists some $pin mathbbQ$ such that $0<p<epsilon$. Let's look at an arbitrary $xin mathbbR$ and see if our property is satisfied or not:



                      If $x$ is rational, then $D(x)=1$. Since the sum of two rationals is rational, then $D(x+p)=1$ too.



                      If $x$ is irrational, then $D(x)=0$. Since the sum of a rational and an irrational is irrational, then $D(x+p)=0$ too.



                      So if we choose our period to be $p$, our property is satisfied.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Oct 4 at 11:57









                      GSofer

                      469211




                      469211




















                          up vote
                          3
                          down vote













                          The main question has been answered in other answers well enough, but I would like to address a few natural follow-up questions. What about continuous functions $f$ with this property?



                          It turns out that in this case there are no nontrivial solutions - every such function is constant. Here's a topological proof: Let $K=xinBbb Rmid forall y, f(y)=f(x+y)$ be the set of periods of $f$. If $f$ is continuous, then this is an intersection of the sets $xinBbb Rmid f(y)=f(x+y)$, which is closed (it is the preimage of $0$ under the function $g(x)=f(y)-f(x+y)$), so $K$ itself is closed. $K$ is also dense in $Bbb R$, because it is an additive group with arbitrarily small elements, so $K=Bbb R$ and hence $f(x)=f(y)$ for all $x,yin Bbb R$.



                          If we consider discontinuous functions again, then we know $K$ is a dense additive subgroup of $Bbb R$. Does every dense additive subgroup generate such a function? Yes, we can just take the characteristic function of $K$. For a fixed $K$, the space of such functions is just all functions $Bbb R/Kto Bbb R$. This is another way to get at the constancy result, since as a topological group, $Bbb R/K$ has the indiscrete topology, because any open set will cover $Bbb R$ if copied around with translations by $K$.



                          Of course $Bbb R/K$ can be uncountable, for example if $K=Bbb Q$ or any other countable subgroup. Can it be countable or finite? It can be countable assuming some choice, as observed in TomGrubb's answer. If we consider a Hamel basis $B$ of $Bbb R$ over $Bbb Q$, then the set of all real numbers with zero first projection is a subgroup $K$ of $Bbb R$ for which $Bbb R/Ksimeq Bbb Q$.



                          But it can't be finite (unless it is trivial). In other words, there is no coherent way to talk about real numbers being partitioned into the "even" and "odd" ones. If $Bbb R/K$ has $n>1$ elements, then that means that every number which is a multiple of $n$ is in $K$; but every real number is a multiple of $n$, to wit, $x=n(x/n)$.






                          share|cite|improve this answer
























                            up vote
                            3
                            down vote













                            The main question has been answered in other answers well enough, but I would like to address a few natural follow-up questions. What about continuous functions $f$ with this property?



                            It turns out that in this case there are no nontrivial solutions - every such function is constant. Here's a topological proof: Let $K=xinBbb Rmid forall y, f(y)=f(x+y)$ be the set of periods of $f$. If $f$ is continuous, then this is an intersection of the sets $xinBbb Rmid f(y)=f(x+y)$, which is closed (it is the preimage of $0$ under the function $g(x)=f(y)-f(x+y)$), so $K$ itself is closed. $K$ is also dense in $Bbb R$, because it is an additive group with arbitrarily small elements, so $K=Bbb R$ and hence $f(x)=f(y)$ for all $x,yin Bbb R$.



                            If we consider discontinuous functions again, then we know $K$ is a dense additive subgroup of $Bbb R$. Does every dense additive subgroup generate such a function? Yes, we can just take the characteristic function of $K$. For a fixed $K$, the space of such functions is just all functions $Bbb R/Kto Bbb R$. This is another way to get at the constancy result, since as a topological group, $Bbb R/K$ has the indiscrete topology, because any open set will cover $Bbb R$ if copied around with translations by $K$.



                            Of course $Bbb R/K$ can be uncountable, for example if $K=Bbb Q$ or any other countable subgroup. Can it be countable or finite? It can be countable assuming some choice, as observed in TomGrubb's answer. If we consider a Hamel basis $B$ of $Bbb R$ over $Bbb Q$, then the set of all real numbers with zero first projection is a subgroup $K$ of $Bbb R$ for which $Bbb R/Ksimeq Bbb Q$.



                            But it can't be finite (unless it is trivial). In other words, there is no coherent way to talk about real numbers being partitioned into the "even" and "odd" ones. If $Bbb R/K$ has $n>1$ elements, then that means that every number which is a multiple of $n$ is in $K$; but every real number is a multiple of $n$, to wit, $x=n(x/n)$.






                            share|cite|improve this answer






















                              up vote
                              3
                              down vote










                              up vote
                              3
                              down vote









                              The main question has been answered in other answers well enough, but I would like to address a few natural follow-up questions. What about continuous functions $f$ with this property?



                              It turns out that in this case there are no nontrivial solutions - every such function is constant. Here's a topological proof: Let $K=xinBbb Rmid forall y, f(y)=f(x+y)$ be the set of periods of $f$. If $f$ is continuous, then this is an intersection of the sets $xinBbb Rmid f(y)=f(x+y)$, which is closed (it is the preimage of $0$ under the function $g(x)=f(y)-f(x+y)$), so $K$ itself is closed. $K$ is also dense in $Bbb R$, because it is an additive group with arbitrarily small elements, so $K=Bbb R$ and hence $f(x)=f(y)$ for all $x,yin Bbb R$.



                              If we consider discontinuous functions again, then we know $K$ is a dense additive subgroup of $Bbb R$. Does every dense additive subgroup generate such a function? Yes, we can just take the characteristic function of $K$. For a fixed $K$, the space of such functions is just all functions $Bbb R/Kto Bbb R$. This is another way to get at the constancy result, since as a topological group, $Bbb R/K$ has the indiscrete topology, because any open set will cover $Bbb R$ if copied around with translations by $K$.



                              Of course $Bbb R/K$ can be uncountable, for example if $K=Bbb Q$ or any other countable subgroup. Can it be countable or finite? It can be countable assuming some choice, as observed in TomGrubb's answer. If we consider a Hamel basis $B$ of $Bbb R$ over $Bbb Q$, then the set of all real numbers with zero first projection is a subgroup $K$ of $Bbb R$ for which $Bbb R/Ksimeq Bbb Q$.



                              But it can't be finite (unless it is trivial). In other words, there is no coherent way to talk about real numbers being partitioned into the "even" and "odd" ones. If $Bbb R/K$ has $n>1$ elements, then that means that every number which is a multiple of $n$ is in $K$; but every real number is a multiple of $n$, to wit, $x=n(x/n)$.






                              share|cite|improve this answer












                              The main question has been answered in other answers well enough, but I would like to address a few natural follow-up questions. What about continuous functions $f$ with this property?



                              It turns out that in this case there are no nontrivial solutions - every such function is constant. Here's a topological proof: Let $K=xinBbb Rmid forall y, f(y)=f(x+y)$ be the set of periods of $f$. If $f$ is continuous, then this is an intersection of the sets $xinBbb Rmid f(y)=f(x+y)$, which is closed (it is the preimage of $0$ under the function $g(x)=f(y)-f(x+y)$), so $K$ itself is closed. $K$ is also dense in $Bbb R$, because it is an additive group with arbitrarily small elements, so $K=Bbb R$ and hence $f(x)=f(y)$ for all $x,yin Bbb R$.



                              If we consider discontinuous functions again, then we know $K$ is a dense additive subgroup of $Bbb R$. Does every dense additive subgroup generate such a function? Yes, we can just take the characteristic function of $K$. For a fixed $K$, the space of such functions is just all functions $Bbb R/Kto Bbb R$. This is another way to get at the constancy result, since as a topological group, $Bbb R/K$ has the indiscrete topology, because any open set will cover $Bbb R$ if copied around with translations by $K$.



                              Of course $Bbb R/K$ can be uncountable, for example if $K=Bbb Q$ or any other countable subgroup. Can it be countable or finite? It can be countable assuming some choice, as observed in TomGrubb's answer. If we consider a Hamel basis $B$ of $Bbb R$ over $Bbb Q$, then the set of all real numbers with zero first projection is a subgroup $K$ of $Bbb R$ for which $Bbb R/Ksimeq Bbb Q$.



                              But it can't be finite (unless it is trivial). In other words, there is no coherent way to talk about real numbers being partitioned into the "even" and "odd" ones. If $Bbb R/K$ has $n>1$ elements, then that means that every number which is a multiple of $n$ is in $K$; but every real number is a multiple of $n$, to wit, $x=n(x/n)$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Oct 4 at 21:08









                              Mario Carneiro

                              18.2k33888




                              18.2k33888




















                                  up vote
                                  3
                                  down vote













                                  Another way to look at it is to think about the set of periods, i.e.



                                  $$P = f(x + p) = f(x) text for all x in mathbbR $$



                                  Zero is clearly a member of this set no matter what f is. P is closed under addition and negation. So clearly P is a group over addition. So, if you want to find a function that has arbitrarily small periods, you want to find a subgroup of $mathbbR$ that has arbitrarily small values. $mathbbQ$ is the obvious choice, so the characteristic function for $mathbbQ$ works, as stated in another answer.






                                  share|cite|improve this answer


























                                    up vote
                                    3
                                    down vote













                                    Another way to look at it is to think about the set of periods, i.e.



                                    $$P = f(x + p) = f(x) text for all x in mathbbR $$



                                    Zero is clearly a member of this set no matter what f is. P is closed under addition and negation. So clearly P is a group over addition. So, if you want to find a function that has arbitrarily small periods, you want to find a subgroup of $mathbbR$ that has arbitrarily small values. $mathbbQ$ is the obvious choice, so the characteristic function for $mathbbQ$ works, as stated in another answer.






                                    share|cite|improve this answer
























                                      up vote
                                      3
                                      down vote










                                      up vote
                                      3
                                      down vote









                                      Another way to look at it is to think about the set of periods, i.e.



                                      $$P = f(x + p) = f(x) text for all x in mathbbR $$



                                      Zero is clearly a member of this set no matter what f is. P is closed under addition and negation. So clearly P is a group over addition. So, if you want to find a function that has arbitrarily small periods, you want to find a subgroup of $mathbbR$ that has arbitrarily small values. $mathbbQ$ is the obvious choice, so the characteristic function for $mathbbQ$ works, as stated in another answer.






                                      share|cite|improve this answer














                                      Another way to look at it is to think about the set of periods, i.e.



                                      $$P = f(x + p) = f(x) text for all x in mathbbR $$



                                      Zero is clearly a member of this set no matter what f is. P is closed under addition and negation. So clearly P is a group over addition. So, if you want to find a function that has arbitrarily small periods, you want to find a subgroup of $mathbbR$ that has arbitrarily small values. $mathbbQ$ is the obvious choice, so the characteristic function for $mathbbQ$ works, as stated in another answer.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited 2 days ago









                                      Yanko

                                      4,217720




                                      4,217720










                                      answered Oct 4 at 15:01









                                      David Stanley

                                      18115




                                      18115




















                                          up vote
                                          0
                                          down vote













                                          Here's a different function with the same property (which relies on a fair bit of choice). Choose a Hamel basis for $mathbbR$ over $mathbbQ$ and pick a basis vector $v$. Let $f$ be the function which projects onto the $v$ coordinate. Then for any other basis vector $u$ and any integer $n$,
                                          $$
                                          f(x+u/n)=f(x).
                                          $$

                                          More can be found on these functions in the article "Discontinuous additive functions" by Bernardi.






                                          share|cite|improve this answer






















                                          • Hope you don't mind my edit, since it's not that much choice needed for what you want. =)
                                            – user21820
                                            Oct 4 at 17:10














                                          up vote
                                          0
                                          down vote













                                          Here's a different function with the same property (which relies on a fair bit of choice). Choose a Hamel basis for $mathbbR$ over $mathbbQ$ and pick a basis vector $v$. Let $f$ be the function which projects onto the $v$ coordinate. Then for any other basis vector $u$ and any integer $n$,
                                          $$
                                          f(x+u/n)=f(x).
                                          $$

                                          More can be found on these functions in the article "Discontinuous additive functions" by Bernardi.






                                          share|cite|improve this answer






















                                          • Hope you don't mind my edit, since it's not that much choice needed for what you want. =)
                                            – user21820
                                            Oct 4 at 17:10












                                          up vote
                                          0
                                          down vote










                                          up vote
                                          0
                                          down vote









                                          Here's a different function with the same property (which relies on a fair bit of choice). Choose a Hamel basis for $mathbbR$ over $mathbbQ$ and pick a basis vector $v$. Let $f$ be the function which projects onto the $v$ coordinate. Then for any other basis vector $u$ and any integer $n$,
                                          $$
                                          f(x+u/n)=f(x).
                                          $$

                                          More can be found on these functions in the article "Discontinuous additive functions" by Bernardi.






                                          share|cite|improve this answer














                                          Here's a different function with the same property (which relies on a fair bit of choice). Choose a Hamel basis for $mathbbR$ over $mathbbQ$ and pick a basis vector $v$. Let $f$ be the function which projects onto the $v$ coordinate. Then for any other basis vector $u$ and any integer $n$,
                                          $$
                                          f(x+u/n)=f(x).
                                          $$

                                          More can be found on these functions in the article "Discontinuous additive functions" by Bernardi.







                                          share|cite|improve this answer














                                          share|cite|improve this answer



                                          share|cite|improve this answer








                                          edited Oct 4 at 17:10









                                          user21820

                                          36.8k441143




                                          36.8k441143










                                          answered Oct 4 at 13:46









                                          TomGrubb

                                          10.6k11438




                                          10.6k11438











                                          • Hope you don't mind my edit, since it's not that much choice needed for what you want. =)
                                            – user21820
                                            Oct 4 at 17:10
















                                          • Hope you don't mind my edit, since it's not that much choice needed for what you want. =)
                                            – user21820
                                            Oct 4 at 17:10















                                          Hope you don't mind my edit, since it's not that much choice needed for what you want. =)
                                          – user21820
                                          Oct 4 at 17:10




                                          Hope you don't mind my edit, since it's not that much choice needed for what you want. =)
                                          – user21820
                                          Oct 4 at 17:10










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