Cardinality of ${f:mathbb{R}rightarrow mathbb{R}: (forall x in mathbb{R} setminus mathbb{Q})(f(x)-x in...











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Let $S$ be ${f:mathbb{R}rightarrow mathbb{R}: (forall x in mathbb{R} setminus mathbb{Q})(f(x)-x in mathbb{N})}$. Determine the cardinality of S!



My attempt:



I honestly have no clue, I know that $forall x in mathbb{R} setminus mathbb{Q} ,f(x) = x+n, n in mathbb{N}$ but other than that no useful conclusions. I can't even seem to intuitively guess this. Of course, $k(S) leq k(mathbb{R}^{mathbb{R}})$, but other than that I'm lost here and would appreciate any hints!



I've tried to build different injections from $P(mathbb R)$,$mathbb{R}^{mathbb{R}}$ and $P(mathbb{R} setminus mathbb{Q})$ to $S$ but have failed miserably!










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    up vote
    0
    down vote

    favorite












    Let $S$ be ${f:mathbb{R}rightarrow mathbb{R}: (forall x in mathbb{R} setminus mathbb{Q})(f(x)-x in mathbb{N})}$. Determine the cardinality of S!



    My attempt:



    I honestly have no clue, I know that $forall x in mathbb{R} setminus mathbb{Q} ,f(x) = x+n, n in mathbb{N}$ but other than that no useful conclusions. I can't even seem to intuitively guess this. Of course, $k(S) leq k(mathbb{R}^{mathbb{R}})$, but other than that I'm lost here and would appreciate any hints!



    I've tried to build different injections from $P(mathbb R)$,$mathbb{R}^{mathbb{R}}$ and $P(mathbb{R} setminus mathbb{Q})$ to $S$ but have failed miserably!










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $S$ be ${f:mathbb{R}rightarrow mathbb{R}: (forall x in mathbb{R} setminus mathbb{Q})(f(x)-x in mathbb{N})}$. Determine the cardinality of S!



      My attempt:



      I honestly have no clue, I know that $forall x in mathbb{R} setminus mathbb{Q} ,f(x) = x+n, n in mathbb{N}$ but other than that no useful conclusions. I can't even seem to intuitively guess this. Of course, $k(S) leq k(mathbb{R}^{mathbb{R}})$, but other than that I'm lost here and would appreciate any hints!



      I've tried to build different injections from $P(mathbb R)$,$mathbb{R}^{mathbb{R}}$ and $P(mathbb{R} setminus mathbb{Q})$ to $S$ but have failed miserably!










      share|cite|improve this question















      Let $S$ be ${f:mathbb{R}rightarrow mathbb{R}: (forall x in mathbb{R} setminus mathbb{Q})(f(x)-x in mathbb{N})}$. Determine the cardinality of S!



      My attempt:



      I honestly have no clue, I know that $forall x in mathbb{R} setminus mathbb{Q} ,f(x) = x+n, n in mathbb{N}$ but other than that no useful conclusions. I can't even seem to intuitively guess this. Of course, $k(S) leq k(mathbb{R}^{mathbb{R}})$, but other than that I'm lost here and would appreciate any hints!



      I've tried to build different injections from $P(mathbb R)$,$mathbb{R}^{mathbb{R}}$ and $P(mathbb{R} setminus mathbb{Q})$ to $S$ but have failed miserably!







      elementary-set-theory






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      edited Nov 22 at 16:17









      Asaf Karagila

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      asked Nov 22 at 16:14









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          If $fin S$, define $g_f$ to be $g_f(x)=f(x)-x$. Then $g_finBbb{N^{Rsetminus Q}}$, define $h_f=frestrictionBbb Q$. Note, moreover, that that $fmapsto (g_f,h_f)inBbb{N^{Rsetminus Q}times R^Q}$ is a bijection.



          What is the cardinality of $Bbb{N^{Rsetminus Q}times R^Q}$?






          share|cite|improve this answer





















          • The cardinality of that set is $2^c$ ? (where $c=k(mathbb{R})$?
            – Collapse
            Nov 22 at 16:23












          • Yes. It is.${}{}$
            – Asaf Karagila
            Nov 22 at 16:24













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          1 Answer
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          active

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          1 Answer
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          active

          oldest

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          active

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          active

          oldest

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          up vote
          1
          down vote



          accepted










          If $fin S$, define $g_f$ to be $g_f(x)=f(x)-x$. Then $g_finBbb{N^{Rsetminus Q}}$, define $h_f=frestrictionBbb Q$. Note, moreover, that that $fmapsto (g_f,h_f)inBbb{N^{Rsetminus Q}times R^Q}$ is a bijection.



          What is the cardinality of $Bbb{N^{Rsetminus Q}times R^Q}$?






          share|cite|improve this answer





















          • The cardinality of that set is $2^c$ ? (where $c=k(mathbb{R})$?
            – Collapse
            Nov 22 at 16:23












          • Yes. It is.${}{}$
            – Asaf Karagila
            Nov 22 at 16:24

















          up vote
          1
          down vote



          accepted










          If $fin S$, define $g_f$ to be $g_f(x)=f(x)-x$. Then $g_finBbb{N^{Rsetminus Q}}$, define $h_f=frestrictionBbb Q$. Note, moreover, that that $fmapsto (g_f,h_f)inBbb{N^{Rsetminus Q}times R^Q}$ is a bijection.



          What is the cardinality of $Bbb{N^{Rsetminus Q}times R^Q}$?






          share|cite|improve this answer





















          • The cardinality of that set is $2^c$ ? (where $c=k(mathbb{R})$?
            – Collapse
            Nov 22 at 16:23












          • Yes. It is.${}{}$
            – Asaf Karagila
            Nov 22 at 16:24















          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          If $fin S$, define $g_f$ to be $g_f(x)=f(x)-x$. Then $g_finBbb{N^{Rsetminus Q}}$, define $h_f=frestrictionBbb Q$. Note, moreover, that that $fmapsto (g_f,h_f)inBbb{N^{Rsetminus Q}times R^Q}$ is a bijection.



          What is the cardinality of $Bbb{N^{Rsetminus Q}times R^Q}$?






          share|cite|improve this answer












          If $fin S$, define $g_f$ to be $g_f(x)=f(x)-x$. Then $g_finBbb{N^{Rsetminus Q}}$, define $h_f=frestrictionBbb Q$. Note, moreover, that that $fmapsto (g_f,h_f)inBbb{N^{Rsetminus Q}times R^Q}$ is a bijection.



          What is the cardinality of $Bbb{N^{Rsetminus Q}times R^Q}$?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 16:17









          Asaf Karagila

          301k32422753




          301k32422753












          • The cardinality of that set is $2^c$ ? (where $c=k(mathbb{R})$?
            – Collapse
            Nov 22 at 16:23












          • Yes. It is.${}{}$
            – Asaf Karagila
            Nov 22 at 16:24




















          • The cardinality of that set is $2^c$ ? (where $c=k(mathbb{R})$?
            – Collapse
            Nov 22 at 16:23












          • Yes. It is.${}{}$
            – Asaf Karagila
            Nov 22 at 16:24


















          The cardinality of that set is $2^c$ ? (where $c=k(mathbb{R})$?
          – Collapse
          Nov 22 at 16:23






          The cardinality of that set is $2^c$ ? (where $c=k(mathbb{R})$?
          – Collapse
          Nov 22 at 16:23














          Yes. It is.${}{}$
          – Asaf Karagila
          Nov 22 at 16:24






          Yes. It is.${}{}$
          – Asaf Karagila
          Nov 22 at 16:24




















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