Countability of set of all functions with domain $mathbb{Z}$ and codomain $mathbb{R}$












1












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enter image description here



I’m not sure how to attempt this problem and would appreciate some guidance. I have the following, but I’m sure it’s incorrect:



enter image description here






elementary-set-theory







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  • $begingroup$
    A set is infinite if its cardinality is equal to $|Z^+|$, still countable but infinite.
    $endgroup$
    – gd1035
    Nov 29 '18 at 23:25
















1












$begingroup$


enter image description here



I’m not sure how to attempt this problem and would appreciate some guidance. I have the following, but I’m sure it’s incorrect:



enter image description here






elementary-set-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    A set is infinite if its cardinality is equal to $|Z^+|$, still countable but infinite.
    $endgroup$
    – gd1035
    Nov 29 '18 at 23:25














1












1








1





$begingroup$


enter image description here



I’m not sure how to attempt this problem and would appreciate some guidance. I have the following, but I’m sure it’s incorrect:



enter image description here






elementary-set-theory







share|cite|improve this question









$endgroup$




enter image description here



I’m not sure how to attempt this problem and would appreciate some guidance. I have the following, but I’m sure it’s incorrect:



enter image description here






elementary-set-theory




elementary-set-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 '18 at 23:22









darylnakdarylnak

167111




167111












  • $begingroup$
    A set is infinite if its cardinality is equal to $|Z^+|$, still countable but infinite.
    $endgroup$
    – gd1035
    Nov 29 '18 at 23:25


















  • $begingroup$
    A set is infinite if its cardinality is equal to $|Z^+|$, still countable but infinite.
    $endgroup$
    – gd1035
    Nov 29 '18 at 23:25
















$begingroup$
A set is infinite if its cardinality is equal to $|Z^+|$, still countable but infinite.
$endgroup$
– gd1035
Nov 29 '18 at 23:25




$begingroup$
A set is infinite if its cardinality is equal to $|Z^+|$, still countable but infinite.
$endgroup$
– gd1035
Nov 29 '18 at 23:25










2 Answers
2






active

oldest

votes


















1












$begingroup$

Because $mathbb{N}$ and $mathbb{Z}$ have the same cardinality, the space of functions from $mathbb{Z}$ to $mathbb{R}$ has the same size as the space of functions from $mathbb{N}$ to $mathbb{R}$ (this is the space of all real-valued-sequences).



Cantor's diagonality argument gives us that the space of all ${0,1}$-sequences is uncountable. Therefore the space of all real-valued-sequences is uncountable. Therefore the collection in question is uncountable.






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  • 1




    $begingroup$
    You might not even need to go this far. The set of all constant functions $mathbb{Z} to mathbb{R}$ is already uncountable.
    $endgroup$
    – platty
    Nov 29 '18 at 23:31






  • 1




    $begingroup$
    I agree. It is because $mathbb{R}$ is uncountable. This is typically proven via Cantor's diagonality argument. Both have merits because of the common lemma with Cantor's diagonality argument.
    $endgroup$
    – Alberto Takase
    Nov 29 '18 at 23:35



















0












$begingroup$

Your proof would be complete if you actually show that the set has cardinality $ge|mathbb{R}|$. This is essentially obvious: for every $rinmathbb{R}$, define $bar{r}colonmathbb{Z}tomathbb{R}$ to be the constant function $bar{r}(x)=r$.



Then $rmapstobar{r}$ is an injective map $mathbb{R}to mathbb{R}^{mathbb{Z}}$, proving that
$$
|mathbb{R}|le|mathbb{R}^{mathbb{Z}}|
$$



More generally, $|X|le|X^Y|$ for every nonempty set $Y$, the argument is the same (where $X^Y$ denotes the set of all maps $Yto X$).



Actually, also the converse inequality holds, in your case:
$$
|mathbb{R}^{mathbb{Z}}|=(2^{aleph_0})^{aleph_0}=2^{aleph_0aleph_0}=
2^{aleph_0}=|mathbb{R}|
$$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Because $mathbb{N}$ and $mathbb{Z}$ have the same cardinality, the space of functions from $mathbb{Z}$ to $mathbb{R}$ has the same size as the space of functions from $mathbb{N}$ to $mathbb{R}$ (this is the space of all real-valued-sequences).



    Cantor's diagonality argument gives us that the space of all ${0,1}$-sequences is uncountable. Therefore the space of all real-valued-sequences is uncountable. Therefore the collection in question is uncountable.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      You might not even need to go this far. The set of all constant functions $mathbb{Z} to mathbb{R}$ is already uncountable.
      $endgroup$
      – platty
      Nov 29 '18 at 23:31






    • 1




      $begingroup$
      I agree. It is because $mathbb{R}$ is uncountable. This is typically proven via Cantor's diagonality argument. Both have merits because of the common lemma with Cantor's diagonality argument.
      $endgroup$
      – Alberto Takase
      Nov 29 '18 at 23:35
















    1












    $begingroup$

    Because $mathbb{N}$ and $mathbb{Z}$ have the same cardinality, the space of functions from $mathbb{Z}$ to $mathbb{R}$ has the same size as the space of functions from $mathbb{N}$ to $mathbb{R}$ (this is the space of all real-valued-sequences).



    Cantor's diagonality argument gives us that the space of all ${0,1}$-sequences is uncountable. Therefore the space of all real-valued-sequences is uncountable. Therefore the collection in question is uncountable.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      You might not even need to go this far. The set of all constant functions $mathbb{Z} to mathbb{R}$ is already uncountable.
      $endgroup$
      – platty
      Nov 29 '18 at 23:31






    • 1




      $begingroup$
      I agree. It is because $mathbb{R}$ is uncountable. This is typically proven via Cantor's diagonality argument. Both have merits because of the common lemma with Cantor's diagonality argument.
      $endgroup$
      – Alberto Takase
      Nov 29 '18 at 23:35














    1












    1








    1





    $begingroup$

    Because $mathbb{N}$ and $mathbb{Z}$ have the same cardinality, the space of functions from $mathbb{Z}$ to $mathbb{R}$ has the same size as the space of functions from $mathbb{N}$ to $mathbb{R}$ (this is the space of all real-valued-sequences).



    Cantor's diagonality argument gives us that the space of all ${0,1}$-sequences is uncountable. Therefore the space of all real-valued-sequences is uncountable. Therefore the collection in question is uncountable.






    share|cite|improve this answer









    $endgroup$



    Because $mathbb{N}$ and $mathbb{Z}$ have the same cardinality, the space of functions from $mathbb{Z}$ to $mathbb{R}$ has the same size as the space of functions from $mathbb{N}$ to $mathbb{R}$ (this is the space of all real-valued-sequences).



    Cantor's diagonality argument gives us that the space of all ${0,1}$-sequences is uncountable. Therefore the space of all real-valued-sequences is uncountable. Therefore the collection in question is uncountable.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 29 '18 at 23:29









    Alberto TakaseAlberto Takase

    2,083418




    2,083418








    • 1




      $begingroup$
      You might not even need to go this far. The set of all constant functions $mathbb{Z} to mathbb{R}$ is already uncountable.
      $endgroup$
      – platty
      Nov 29 '18 at 23:31






    • 1




      $begingroup$
      I agree. It is because $mathbb{R}$ is uncountable. This is typically proven via Cantor's diagonality argument. Both have merits because of the common lemma with Cantor's diagonality argument.
      $endgroup$
      – Alberto Takase
      Nov 29 '18 at 23:35














    • 1




      $begingroup$
      You might not even need to go this far. The set of all constant functions $mathbb{Z} to mathbb{R}$ is already uncountable.
      $endgroup$
      – platty
      Nov 29 '18 at 23:31






    • 1




      $begingroup$
      I agree. It is because $mathbb{R}$ is uncountable. This is typically proven via Cantor's diagonality argument. Both have merits because of the common lemma with Cantor's diagonality argument.
      $endgroup$
      – Alberto Takase
      Nov 29 '18 at 23:35








    1




    1




    $begingroup$
    You might not even need to go this far. The set of all constant functions $mathbb{Z} to mathbb{R}$ is already uncountable.
    $endgroup$
    – platty
    Nov 29 '18 at 23:31




    $begingroup$
    You might not even need to go this far. The set of all constant functions $mathbb{Z} to mathbb{R}$ is already uncountable.
    $endgroup$
    – platty
    Nov 29 '18 at 23:31




    1




    1




    $begingroup$
    I agree. It is because $mathbb{R}$ is uncountable. This is typically proven via Cantor's diagonality argument. Both have merits because of the common lemma with Cantor's diagonality argument.
    $endgroup$
    – Alberto Takase
    Nov 29 '18 at 23:35




    $begingroup$
    I agree. It is because $mathbb{R}$ is uncountable. This is typically proven via Cantor's diagonality argument. Both have merits because of the common lemma with Cantor's diagonality argument.
    $endgroup$
    – Alberto Takase
    Nov 29 '18 at 23:35











    0












    $begingroup$

    Your proof would be complete if you actually show that the set has cardinality $ge|mathbb{R}|$. This is essentially obvious: for every $rinmathbb{R}$, define $bar{r}colonmathbb{Z}tomathbb{R}$ to be the constant function $bar{r}(x)=r$.



    Then $rmapstobar{r}$ is an injective map $mathbb{R}to mathbb{R}^{mathbb{Z}}$, proving that
    $$
    |mathbb{R}|le|mathbb{R}^{mathbb{Z}}|
    $$



    More generally, $|X|le|X^Y|$ for every nonempty set $Y$, the argument is the same (where $X^Y$ denotes the set of all maps $Yto X$).



    Actually, also the converse inequality holds, in your case:
    $$
    |mathbb{R}^{mathbb{Z}}|=(2^{aleph_0})^{aleph_0}=2^{aleph_0aleph_0}=
    2^{aleph_0}=|mathbb{R}|
    $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Your proof would be complete if you actually show that the set has cardinality $ge|mathbb{R}|$. This is essentially obvious: for every $rinmathbb{R}$, define $bar{r}colonmathbb{Z}tomathbb{R}$ to be the constant function $bar{r}(x)=r$.



      Then $rmapstobar{r}$ is an injective map $mathbb{R}to mathbb{R}^{mathbb{Z}}$, proving that
      $$
      |mathbb{R}|le|mathbb{R}^{mathbb{Z}}|
      $$



      More generally, $|X|le|X^Y|$ for every nonempty set $Y$, the argument is the same (where $X^Y$ denotes the set of all maps $Yto X$).



      Actually, also the converse inequality holds, in your case:
      $$
      |mathbb{R}^{mathbb{Z}}|=(2^{aleph_0})^{aleph_0}=2^{aleph_0aleph_0}=
      2^{aleph_0}=|mathbb{R}|
      $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Your proof would be complete if you actually show that the set has cardinality $ge|mathbb{R}|$. This is essentially obvious: for every $rinmathbb{R}$, define $bar{r}colonmathbb{Z}tomathbb{R}$ to be the constant function $bar{r}(x)=r$.



        Then $rmapstobar{r}$ is an injective map $mathbb{R}to mathbb{R}^{mathbb{Z}}$, proving that
        $$
        |mathbb{R}|le|mathbb{R}^{mathbb{Z}}|
        $$



        More generally, $|X|le|X^Y|$ for every nonempty set $Y$, the argument is the same (where $X^Y$ denotes the set of all maps $Yto X$).



        Actually, also the converse inequality holds, in your case:
        $$
        |mathbb{R}^{mathbb{Z}}|=(2^{aleph_0})^{aleph_0}=2^{aleph_0aleph_0}=
        2^{aleph_0}=|mathbb{R}|
        $$






        share|cite|improve this answer









        $endgroup$



        Your proof would be complete if you actually show that the set has cardinality $ge|mathbb{R}|$. This is essentially obvious: for every $rinmathbb{R}$, define $bar{r}colonmathbb{Z}tomathbb{R}$ to be the constant function $bar{r}(x)=r$.



        Then $rmapstobar{r}$ is an injective map $mathbb{R}to mathbb{R}^{mathbb{Z}}$, proving that
        $$
        |mathbb{R}|le|mathbb{R}^{mathbb{Z}}|
        $$



        More generally, $|X|le|X^Y|$ for every nonempty set $Y$, the argument is the same (where $X^Y$ denotes the set of all maps $Yto X$).



        Actually, also the converse inequality holds, in your case:
        $$
        |mathbb{R}^{mathbb{Z}}|=(2^{aleph_0})^{aleph_0}=2^{aleph_0aleph_0}=
        2^{aleph_0}=|mathbb{R}|
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 '18 at 23:32









        egregegreg

        181k1485203




        181k1485203






























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