How to prove $mathbb{R}^n < mathbb{N}^mathbb{R}$












2












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I know that the cardinality of $mathbb{R}^n$ is equal to $mathbb{R}$. I also know that the cardinality of $mathbb{N}^n$ is equal to $mathbb{N}$, but how do I prove that the cardinality of $mathbb{R}^mathbb{N}$ is smaller than the cardinality of $mathbb{N}^mathbb{R}$.






elementary-set-theory cardinals







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  • 1




    $begingroup$
    You can start first by showing $mathbb{R}^{mathbb N}$ is actually $mathbb{R}$.
    $endgroup$
    – Quang Hoang
    Jan 28 at 17:10










  • $begingroup$
    Your choice of fonts is very inconsistent.
    $endgroup$
    – Asaf Karagila♦
    Jan 28 at 17:10










  • $begingroup$
    $|mathbb{R}^{mathbb{N}}| = |mathbb{R}|^{|mathbb{N}|} = (2^{aleph_0})^{aleph_0} = 2^{aleph_0aleph_0} = 2^{aleph_0}$. $|mathbb{N}^{mathbb{R}}| = |mathbb{N}|^{|mathbb{R}|} = (aleph_0)^{2^{aleph_0}} geq 2^{2^{aleph_0}}$.
    $endgroup$
    – Arturo Magidin
    Jan 28 at 18:17


















2












$begingroup$


I know that the cardinality of $mathbb{R}^n$ is equal to $mathbb{R}$. I also know that the cardinality of $mathbb{N}^n$ is equal to $mathbb{N}$, but how do I prove that the cardinality of $mathbb{R}^mathbb{N}$ is smaller than the cardinality of $mathbb{N}^mathbb{R}$.






elementary-set-theory cardinals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You can start first by showing $mathbb{R}^{mathbb N}$ is actually $mathbb{R}$.
    $endgroup$
    – Quang Hoang
    Jan 28 at 17:10










  • $begingroup$
    Your choice of fonts is very inconsistent.
    $endgroup$
    – Asaf Karagila♦
    Jan 28 at 17:10










  • $begingroup$
    $|mathbb{R}^{mathbb{N}}| = |mathbb{R}|^{|mathbb{N}|} = (2^{aleph_0})^{aleph_0} = 2^{aleph_0aleph_0} = 2^{aleph_0}$. $|mathbb{N}^{mathbb{R}}| = |mathbb{N}|^{|mathbb{R}|} = (aleph_0)^{2^{aleph_0}} geq 2^{2^{aleph_0}}$.
    $endgroup$
    – Arturo Magidin
    Jan 28 at 18:17
















2












2








2





$begingroup$


I know that the cardinality of $mathbb{R}^n$ is equal to $mathbb{R}$. I also know that the cardinality of $mathbb{N}^n$ is equal to $mathbb{N}$, but how do I prove that the cardinality of $mathbb{R}^mathbb{N}$ is smaller than the cardinality of $mathbb{N}^mathbb{R}$.






elementary-set-theory cardinals







share|cite|improve this question











$endgroup$




I know that the cardinality of $mathbb{R}^n$ is equal to $mathbb{R}$. I also know that the cardinality of $mathbb{N}^n$ is equal to $mathbb{N}$, but how do I prove that the cardinality of $mathbb{R}^mathbb{N}$ is smaller than the cardinality of $mathbb{N}^mathbb{R}$.






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 Jan 28 at 18:06









Björn Friedrich

2,69461831




2,69461831










asked Jan 28 at 17:08









Jan ZavecJan Zavec

161




161








  • 1




    $begingroup$
    You can start first by showing $mathbb{R}^{mathbb N}$ is actually $mathbb{R}$.
    $endgroup$
    – Quang Hoang
    Jan 28 at 17:10










  • $begingroup$
    Your choice of fonts is very inconsistent.
    $endgroup$
    – Asaf Karagila♦
    Jan 28 at 17:10










  • $begingroup$
    $|mathbb{R}^{mathbb{N}}| = |mathbb{R}|^{|mathbb{N}|} = (2^{aleph_0})^{aleph_0} = 2^{aleph_0aleph_0} = 2^{aleph_0}$. $|mathbb{N}^{mathbb{R}}| = |mathbb{N}|^{|mathbb{R}|} = (aleph_0)^{2^{aleph_0}} geq 2^{2^{aleph_0}}$.
    $endgroup$
    – Arturo Magidin
    Jan 28 at 18:17
















  • 1




    $begingroup$
    You can start first by showing $mathbb{R}^{mathbb N}$ is actually $mathbb{R}$.
    $endgroup$
    – Quang Hoang
    Jan 28 at 17:10










  • $begingroup$
    Your choice of fonts is very inconsistent.
    $endgroup$
    – Asaf Karagila♦
    Jan 28 at 17:10










  • $begingroup$
    $|mathbb{R}^{mathbb{N}}| = |mathbb{R}|^{|mathbb{N}|} = (2^{aleph_0})^{aleph_0} = 2^{aleph_0aleph_0} = 2^{aleph_0}$. $|mathbb{N}^{mathbb{R}}| = |mathbb{N}|^{|mathbb{R}|} = (aleph_0)^{2^{aleph_0}} geq 2^{2^{aleph_0}}$.
    $endgroup$
    – Arturo Magidin
    Jan 28 at 18:17










1




1




$begingroup$
You can start first by showing $mathbb{R}^{mathbb N}$ is actually $mathbb{R}$.
$endgroup$
– Quang Hoang
Jan 28 at 17:10




$begingroup$
You can start first by showing $mathbb{R}^{mathbb N}$ is actually $mathbb{R}$.
$endgroup$
– Quang Hoang
Jan 28 at 17:10












$begingroup$
Your choice of fonts is very inconsistent.
$endgroup$
– Asaf Karagila♦
Jan 28 at 17:10




$begingroup$
Your choice of fonts is very inconsistent.
$endgroup$
– Asaf Karagila♦
Jan 28 at 17:10












$begingroup$
$|mathbb{R}^{mathbb{N}}| = |mathbb{R}|^{|mathbb{N}|} = (2^{aleph_0})^{aleph_0} = 2^{aleph_0aleph_0} = 2^{aleph_0}$. $|mathbb{N}^{mathbb{R}}| = |mathbb{N}|^{|mathbb{R}|} = (aleph_0)^{2^{aleph_0}} geq 2^{2^{aleph_0}}$.
$endgroup$
– Arturo Magidin
Jan 28 at 18:17






$begingroup$
$|mathbb{R}^{mathbb{N}}| = |mathbb{R}|^{|mathbb{N}|} = (2^{aleph_0})^{aleph_0} = 2^{aleph_0aleph_0} = 2^{aleph_0}$. $|mathbb{N}^{mathbb{R}}| = |mathbb{N}|^{|mathbb{R}|} = (aleph_0)^{2^{aleph_0}} geq 2^{2^{aleph_0}}$.
$endgroup$
– Arturo Magidin
Jan 28 at 18:17












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$begingroup$

$$|mathbb{N}^mathbb{R}|ge |2^mathbb{R}|>|mathbb{R}|=|mathbb{R}^n|(=|mathbb{R}^mathbb{N}|).$$






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    4












    $begingroup$

    $$|mathbb{N}^mathbb{R}|ge |2^mathbb{R}|>|mathbb{R}|=|mathbb{R}^n|(=|mathbb{R}^mathbb{N}|).$$






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      $$|mathbb{N}^mathbb{R}|ge |2^mathbb{R}|>|mathbb{R}|=|mathbb{R}^n|(=|mathbb{R}^mathbb{N}|).$$






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        $$|mathbb{N}^mathbb{R}|ge |2^mathbb{R}|>|mathbb{R}|=|mathbb{R}^n|(=|mathbb{R}^mathbb{N}|).$$






        share|cite|improve this answer











        $endgroup$



        $$|mathbb{N}^mathbb{R}|ge |2^mathbb{R}|>|mathbb{R}|=|mathbb{R}^n|(=|mathbb{R}^mathbb{N}|).$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 29 at 7:00









        Henno Brandsma

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        114k348124










        answered Jan 28 at 17:10









        Eclipse SunEclipse Sun

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        7,9151438






























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