Why is ${f,|,fcolon Atomathbb N}$ not uncountable?












0












$begingroup$


Let $S={,f,|,fcolon Atomathbb N}$, where $A={1,2}$.



I thought cardinality of $S$ is $2^{|mathbb N|}=aleph_0$. But my friend told that my answer was wrong.



Please help me where is I am wrong.






elementary-set-theory cardinals







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$endgroup$












  • $begingroup$
    Why do you think the cardinality of $S$ would be $2^{|mathbb N|}$?
    $endgroup$
    – Henrik
    Dec 8 '18 at 11:18












  • $begingroup$
    Because for any function to defined every point has 2 choices so there are 2^|N|.I thought
    $endgroup$
    – MathLover
    Dec 8 '18 at 11:19










  • $begingroup$
    Note that the set $S = { f$ a function$ : f:mathbb{N}to{1,2}}$ is uncountable. To see why, think about $mathcal{P}(mathbb{N})$ and $f^{-1}(1)$ for $fin S$.
    $endgroup$
    – Santana Afton
    Dec 8 '18 at 12:47












  • $begingroup$
    You could just list down all the functions in your set and see what the answer is.
    $endgroup$
    – Andrés E. Caicedo
    Dec 8 '18 at 14:02
















0












$begingroup$


Let $S={,f,|,fcolon Atomathbb N}$, where $A={1,2}$.



I thought cardinality of $S$ is $2^{|mathbb N|}=aleph_0$. But my friend told that my answer was wrong.



Please help me where is I am wrong.






elementary-set-theory cardinals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why do you think the cardinality of $S$ would be $2^{|mathbb N|}$?
    $endgroup$
    – Henrik
    Dec 8 '18 at 11:18












  • $begingroup$
    Because for any function to defined every point has 2 choices so there are 2^|N|.I thought
    $endgroup$
    – MathLover
    Dec 8 '18 at 11:19










  • $begingroup$
    Note that the set $S = { f$ a function$ : f:mathbb{N}to{1,2}}$ is uncountable. To see why, think about $mathcal{P}(mathbb{N})$ and $f^{-1}(1)$ for $fin S$.
    $endgroup$
    – Santana Afton
    Dec 8 '18 at 12:47












  • $begingroup$
    You could just list down all the functions in your set and see what the answer is.
    $endgroup$
    – Andrés E. Caicedo
    Dec 8 '18 at 14:02














0












0








0





$begingroup$


Let $S={,f,|,fcolon Atomathbb N}$, where $A={1,2}$.



I thought cardinality of $S$ is $2^{|mathbb N|}=aleph_0$. But my friend told that my answer was wrong.



Please help me where is I am wrong.






elementary-set-theory cardinals







share|cite|improve this question











$endgroup$




Let $S={,f,|,fcolon Atomathbb N}$, where $A={1,2}$.



I thought cardinality of $S$ is $2^{|mathbb N|}=aleph_0$. But my friend told that my answer was wrong.



Please help me where is I am wrong.






elementary-set-theory cardinals




elementary-set-theory cardinals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 12:56









Asaf Karagila♦

302k32429760




302k32429760










asked Dec 8 '18 at 11:07









MathLoverMathLover

49710




49710












  • $begingroup$
    Why do you think the cardinality of $S$ would be $2^{|mathbb N|}$?
    $endgroup$
    – Henrik
    Dec 8 '18 at 11:18












  • $begingroup$
    Because for any function to defined every point has 2 choices so there are 2^|N|.I thought
    $endgroup$
    – MathLover
    Dec 8 '18 at 11:19










  • $begingroup$
    Note that the set $S = { f$ a function$ : f:mathbb{N}to{1,2}}$ is uncountable. To see why, think about $mathcal{P}(mathbb{N})$ and $f^{-1}(1)$ for $fin S$.
    $endgroup$
    – Santana Afton
    Dec 8 '18 at 12:47












  • $begingroup$
    You could just list down all the functions in your set and see what the answer is.
    $endgroup$
    – Andrés E. Caicedo
    Dec 8 '18 at 14:02


















  • $begingroup$
    Why do you think the cardinality of $S$ would be $2^{|mathbb N|}$?
    $endgroup$
    – Henrik
    Dec 8 '18 at 11:18












  • $begingroup$
    Because for any function to defined every point has 2 choices so there are 2^|N|.I thought
    $endgroup$
    – MathLover
    Dec 8 '18 at 11:19










  • $begingroup$
    Note that the set $S = { f$ a function$ : f:mathbb{N}to{1,2}}$ is uncountable. To see why, think about $mathcal{P}(mathbb{N})$ and $f^{-1}(1)$ for $fin S$.
    $endgroup$
    – Santana Afton
    Dec 8 '18 at 12:47












  • $begingroup$
    You could just list down all the functions in your set and see what the answer is.
    $endgroup$
    – Andrés E. Caicedo
    Dec 8 '18 at 14:02
















$begingroup$
Why do you think the cardinality of $S$ would be $2^{|mathbb N|}$?
$endgroup$
– Henrik
Dec 8 '18 at 11:18






$begingroup$
Why do you think the cardinality of $S$ would be $2^{|mathbb N|}$?
$endgroup$
– Henrik
Dec 8 '18 at 11:18














$begingroup$
Because for any function to defined every point has 2 choices so there are 2^|N|.I thought
$endgroup$
– MathLover
Dec 8 '18 at 11:19




$begingroup$
Because for any function to defined every point has 2 choices so there are 2^|N|.I thought
$endgroup$
– MathLover
Dec 8 '18 at 11:19












$begingroup$
Note that the set $S = { f$ a function$ : f:mathbb{N}to{1,2}}$ is uncountable. To see why, think about $mathcal{P}(mathbb{N})$ and $f^{-1}(1)$ for $fin S$.
$endgroup$
– Santana Afton
Dec 8 '18 at 12:47






$begingroup$
Note that the set $S = { f$ a function$ : f:mathbb{N}to{1,2}}$ is uncountable. To see why, think about $mathcal{P}(mathbb{N})$ and $f^{-1}(1)$ for $fin S$.
$endgroup$
– Santana Afton
Dec 8 '18 at 12:47














$begingroup$
You could just list down all the functions in your set and see what the answer is.
$endgroup$
– Andrés E. Caicedo
Dec 8 '18 at 14:02




$begingroup$
You could just list down all the functions in your set and see what the answer is.
$endgroup$
– Andrés E. Caicedo
Dec 8 '18 at 14:02










1 Answer
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$begingroup$

The cardinality of the set of functions from a set $A$ to a set $B$ is $|B|^{|A|}$.



In particular in your case the size of $S$ is $|mathbb{N}|^2$ (not $2^{|mathbb{N}|}$) which is the size of $mathbb{N}timesmathbb{N}$ (hence countable).






share|cite|improve this answer











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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

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    active

    oldest

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    3












    $begingroup$

    The cardinality of the set of functions from a set $A$ to a set $B$ is $|B|^{|A|}$.



    In particular in your case the size of $S$ is $|mathbb{N}|^2$ (not $2^{|mathbb{N}|}$) which is the size of $mathbb{N}timesmathbb{N}$ (hence countable).






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      The cardinality of the set of functions from a set $A$ to a set $B$ is $|B|^{|A|}$.



      In particular in your case the size of $S$ is $|mathbb{N}|^2$ (not $2^{|mathbb{N}|}$) which is the size of $mathbb{N}timesmathbb{N}$ (hence countable).






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        The cardinality of the set of functions from a set $A$ to a set $B$ is $|B|^{|A|}$.



        In particular in your case the size of $S$ is $|mathbb{N}|^2$ (not $2^{|mathbb{N}|}$) which is the size of $mathbb{N}timesmathbb{N}$ (hence countable).






        share|cite|improve this answer











        $endgroup$



        The cardinality of the set of functions from a set $A$ to a set $B$ is $|B|^{|A|}$.



        In particular in your case the size of $S$ is $|mathbb{N}|^2$ (not $2^{|mathbb{N}|}$) which is the size of $mathbb{N}timesmathbb{N}$ (hence countable).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 8 '18 at 13:35

























        answered Dec 8 '18 at 11:14









        YankoYanko

        6,3531528




        6,3531528






























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