Cardinality of the set of all real functions of real variable












35














How does one compute the cardinality of the set of functions $f:mathbb{R} to mathbb{R}$ (not necessarily continuous)?










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  • You have used the assumption that the Cardinality of power set of R is equal to the Cardinality of functions from real to {0,1}. How to we prove that as well?
    – user39246
    Sep 2 '12 at 10:59










  • @krishnanem: Please ask new questions in a new thread, rather than as an answer to a previous question. Furthermore, this question was asked and answered several times before.
    – Asaf Karagila
    Sep 2 '12 at 11:17












  • @AsafKaragila: I have moved this non-answer to a comment so that krishnanem can pose it as a new question and can read your comment.
    – robjohn
    Sep 2 '12 at 20:39










  • @robjohn: I think Michael's comment would have been a more suitable choice over mine.
    – Asaf Karagila
    Sep 2 '12 at 21:44










  • @AsafKaragila: The point is to have krishnanem post the comment as a new question. Your comment solely addresses that point. Michael's also addresses the misplaced question and might encourage a reply.
    – robjohn
    Sep 3 '12 at 3:18
















35














How does one compute the cardinality of the set of functions $f:mathbb{R} to mathbb{R}$ (not necessarily continuous)?










share|cite|improve this question
























  • You have used the assumption that the Cardinality of power set of R is equal to the Cardinality of functions from real to {0,1}. How to we prove that as well?
    – user39246
    Sep 2 '12 at 10:59










  • @krishnanem: Please ask new questions in a new thread, rather than as an answer to a previous question. Furthermore, this question was asked and answered several times before.
    – Asaf Karagila
    Sep 2 '12 at 11:17












  • @AsafKaragila: I have moved this non-answer to a comment so that krishnanem can pose it as a new question and can read your comment.
    – robjohn
    Sep 2 '12 at 20:39










  • @robjohn: I think Michael's comment would have been a more suitable choice over mine.
    – Asaf Karagila
    Sep 2 '12 at 21:44










  • @AsafKaragila: The point is to have krishnanem post the comment as a new question. Your comment solely addresses that point. Michael's also addresses the misplaced question and might encourage a reply.
    – robjohn
    Sep 3 '12 at 3:18














35












35








35


15





How does one compute the cardinality of the set of functions $f:mathbb{R} to mathbb{R}$ (not necessarily continuous)?










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How does one compute the cardinality of the set of functions $f:mathbb{R} to mathbb{R}$ (not necessarily continuous)?







elementary-set-theory cardinals






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edited Sep 2 '12 at 11:05









Asaf Karagila

301k32422755




301k32422755










asked Jan 17 '11 at 23:26









Benji

2,12631823




2,12631823












  • You have used the assumption that the Cardinality of power set of R is equal to the Cardinality of functions from real to {0,1}. How to we prove that as well?
    – user39246
    Sep 2 '12 at 10:59










  • @krishnanem: Please ask new questions in a new thread, rather than as an answer to a previous question. Furthermore, this question was asked and answered several times before.
    – Asaf Karagila
    Sep 2 '12 at 11:17












  • @AsafKaragila: I have moved this non-answer to a comment so that krishnanem can pose it as a new question and can read your comment.
    – robjohn
    Sep 2 '12 at 20:39










  • @robjohn: I think Michael's comment would have been a more suitable choice over mine.
    – Asaf Karagila
    Sep 2 '12 at 21:44










  • @AsafKaragila: The point is to have krishnanem post the comment as a new question. Your comment solely addresses that point. Michael's also addresses the misplaced question and might encourage a reply.
    – robjohn
    Sep 3 '12 at 3:18


















  • You have used the assumption that the Cardinality of power set of R is equal to the Cardinality of functions from real to {0,1}. How to we prove that as well?
    – user39246
    Sep 2 '12 at 10:59










  • @krishnanem: Please ask new questions in a new thread, rather than as an answer to a previous question. Furthermore, this question was asked and answered several times before.
    – Asaf Karagila
    Sep 2 '12 at 11:17












  • @AsafKaragila: I have moved this non-answer to a comment so that krishnanem can pose it as a new question and can read your comment.
    – robjohn
    Sep 2 '12 at 20:39










  • @robjohn: I think Michael's comment would have been a more suitable choice over mine.
    – Asaf Karagila
    Sep 2 '12 at 21:44










  • @AsafKaragila: The point is to have krishnanem post the comment as a new question. Your comment solely addresses that point. Michael's also addresses the misplaced question and might encourage a reply.
    – robjohn
    Sep 3 '12 at 3:18
















You have used the assumption that the Cardinality of power set of R is equal to the Cardinality of functions from real to {0,1}. How to we prove that as well?
– user39246
Sep 2 '12 at 10:59




You have used the assumption that the Cardinality of power set of R is equal to the Cardinality of functions from real to {0,1}. How to we prove that as well?
– user39246
Sep 2 '12 at 10:59












@krishnanem: Please ask new questions in a new thread, rather than as an answer to a previous question. Furthermore, this question was asked and answered several times before.
– Asaf Karagila
Sep 2 '12 at 11:17






@krishnanem: Please ask new questions in a new thread, rather than as an answer to a previous question. Furthermore, this question was asked and answered several times before.
– Asaf Karagila
Sep 2 '12 at 11:17














@AsafKaragila: I have moved this non-answer to a comment so that krishnanem can pose it as a new question and can read your comment.
– robjohn
Sep 2 '12 at 20:39




@AsafKaragila: I have moved this non-answer to a comment so that krishnanem can pose it as a new question and can read your comment.
– robjohn
Sep 2 '12 at 20:39












@robjohn: I think Michael's comment would have been a more suitable choice over mine.
– Asaf Karagila
Sep 2 '12 at 21:44




@robjohn: I think Michael's comment would have been a more suitable choice over mine.
– Asaf Karagila
Sep 2 '12 at 21:44












@AsafKaragila: The point is to have krishnanem post the comment as a new question. Your comment solely addresses that point. Michael's also addresses the misplaced question and might encourage a reply.
– robjohn
Sep 3 '12 at 3:18




@AsafKaragila: The point is to have krishnanem post the comment as a new question. Your comment solely addresses that point. Michael's also addresses the misplaced question and might encourage a reply.
– robjohn
Sep 3 '12 at 3:18










4 Answers
4






active

oldest

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38














All you need is a few basics of cardinal arithmetic: if $kappa$ and $lambda$ are cardinals, none of them zero, and at least one of them is infinite, then $kappa+lambda = kappalambda = max{kappa,lambda}$. And cardinal exponentiation satisfies some of the same laws as regular exponentiation; in particular, $(kappa^{lambda})^{nu} = kappa^{lambdanu}$.



The cardinality of the set of all real functions is then
$$|mathbb{R}|^{|mathbb{R}|} =mathfrak{c}^{mathfrak{c}} = (2^{aleph_0})^{2^{aleph_0}} = 2^{aleph_02^{aleph_0}} = 2^{2^{aleph_0}} = 2^{mathfrak{c}}.$$
In other words, it is equal to the cardinality of the power set of $mathbb{R}$.



With a few extra facts, you can get more. In general, if $kappa$ is an infinite cardinal, and $2leqlambdaleqkappa$, then $lambda^{kappa}=2^{kappa}$. This follows because:
$$2^{kappa} leq lambda^{kappa} leq (2^{lambda})^{kappa} = 2^{lambdakappa} = 2^{kappa},$$
so you get equality throughout. The extra information you need for this is to know that if $kappa$, $lambda$, and $nu$ are nonzero cardinals, $kappaleqlambda$, then $kappa^{nu}leq lambda^{nu}$.



In particular, for any infinite cardinal $kappa$ you have $kappa^{kappa} = 2^{kappa}$.






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    29














    I guess that you know that $|mathbb{N}| = |mathbb{N}timesmathbb{N}|$ and thus $|mathbb{R}| = |2^{mathbb{N}}| = |2^{mathbb{N}timesmathbb{N}}| = |2^mathbb{N}times 2^mathbb{N}| = |mathbb{R}timesmathbb{R}|$



    This means that $|P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})|$. Since $fcolonmathbb{R}tomathbb{R}$ is an element of $P(mathbb{R}timesmathbb{R})$ you have that $mathbb{R}^mathbb{R}$ (all the functions from $mathbb{R}$ to itself) is of cardinality less or equal to the one of $P(mathbb{R}timesmathbb{R})$ which in turn means that $|mathbb{R}^mathbb{R}|le |P(mathbb{R})|$.



    Now, since $|P(mathbb{R})| = |2^mathbb{R}|$ which is the set of all functions from $mathbb{R}$ to ${0,1}$, and clearly every function from $mathbb{R}$ into ${0,1}$ is in particular a function from $mathbb{R}$ into itself, we have:
    $$|P(mathbb{R})| = |2^mathbb{R}| le |mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})| = |P(mathbb{R})|$$



    So all in all we have that $|mathbb{R}^mathbb{R}| = |P(mathbb{R})| = |2^mathbb{R}|$.






    share|cite|improve this answer



















    • 2




      Did you mean $|P(mathbb{R})|$ at the end of your second paragraph?
      – Arturo Magidin
      Jan 17 '11 at 23:51






    • 2




      @AsafKaraglia: Could you please detail how and why $|mathbb{R}| = |mathbb{R}timesmathbb{R}|$ and $ mathbb{R} neq mathbb{R}timesmathbb{R} $ $Longrightarrow |P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})|$? I made an incidental edits which I hope will help and referenced math.stackexchange.com/questions/29366/….
      – Greek - Area 51 Proposal
      Nov 7 '13 at 1:10








    • 1




      @LePressentiment: Don't add color to my posts. Thank you.
      – Asaf Karagila
      Nov 7 '13 at 5:36






    • 2




      @AsafKaragila: No problem at all. I'll post my edition separately below. In your previous version (rollback) above, you write that $mathbb{R}^mathbb{R}$ = all the functions from $mathbb{R}$ to itself. Should this be the set of all such functions? Also, will you please to let me know of my previous comment, preceding your comment?
      – Greek - Area 51 Proposal
      Nov 7 '13 at 9:48










    • @LePressentiment: I don't know what you mean by that. $|A|=|B|$ implies that $|mathcal P(A)|=|mathcal P(B)|$. That's a simple exercise in the definition of cardinalities. And yes $A^B$ is the set of all functions from $B$ into $A$, although it is sometimes denoted by ${}^BA$.
      – Asaf Karagila
      Nov 7 '13 at 10:02





















    3














    This answer is based on, but differs slightly from, user Asaf Karaglia's above.





    First, observe that by definition, ${text{all real functions of real variable}}:= {f: ; f: mathbb{R}tomathbb{R}} := mathbb{R}^mathbb{R}$.



    The question is about $|{text{all real functions of real variable}}|$, so examine an arbitrary real function of real variable: $f,colon,mathbb{R}tomathbb{R}.$

    By inspection, $f,colon,mathbb{R}tomathbb{R} := {(r, f(r)) : r in mathbb{R}} quad subseteq quad P(mathbb{R} times mathbb{R})$.

    Thus, $color{green}{|mathbb{R}^{mathbb{R}}| le |P(mathbb{R}timesmathbb{R})|}$.



    Before continuing, let's try to simplify $|P(mathbb{R}timesmathbb{R})|$. Observe that $|mathbb{R}| = |mathbb{R}^k| , forall , k in mathbb{N}$. Its proof by mathematical induction requires the induction hypothesis of $|mathbb{R}| = |mathbb{R}^2|$, one proof of which is : $|mathbb{N}| = |mathbb{N}timesmathbb{N}| implies |mathbb{R}| = |2^{mathbb{N}}| = |2^{mathbb{N}timesmathbb{N}}| = |2^mathbb{N}times 2^mathbb{N}| = |mathbb{R}timesmathbb{R}|$.



    Verily, $mathbb{R} neq mathbb{R}^2$. Howbeit, for infinite sets $A,B$: $|A| = |B| Longrightarrow require{cancel} cancel{Longleftarrow} |P(A)| = |P(B)|$.

    (The converse is discussed here.)



    Thus, $|P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})| implies color{green}{|mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})|} = |P(mathbb{R})|$. Now scrutinise $|P(mathbb{R})|$:



    ● $color{#A9057D}{|P(mathbb{R})| = |2^{mathbb{R}}|}$, where $2^{mathbb{R}} := {f : ; f: mathbb{R} to {0,1}}$,

    ● Every $f: mathbb{R} to {0,1}$ is a particular case of a function from $mathbb{R}$ to $mathbb{R}$, thus $color{#EC5021}{2^{mathbb{R}} subsetneq mathbb{R}^mathbb{R}}$.



    Altogether, $color{#A9057D}{|P(mathbb{R})| =} color{#EC5021}{|2^mathbb{R}| le} color{green}{|mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})|} = |P(mathbb{R})|$



    $implies |P(mathbb{R})| qquad qquad quad leq |mathbb{R}^mathbb{R}| leq |P(mathbb{R})| implies color{#A9057D}{underbrace{|P(mathbb{R})|}_{= |2^mathbb{R}|}} = |mathbb{R}^mathbb{R}| $.






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    • Downvoters, pursuant to my edit, please let me know of further sugggestions which would be more instructive than a mere downvote.
      – Greek - Area 51 Proposal
      Nov 8 '13 at 7:35






    • 4




      Use less colors, so people with disabilities could read this without getting a headache.
      – Asaf Karagila
      Nov 8 '13 at 8:31



















    2














    This is irrelevent here, still it is 'relevent'. The cardinality of set of all continuous function from $mathbb{R}$ to $mathbb{R}$ $(C(mathbb{R},mathbb{R}))$ is $2 ^ mathbb{N_0} = mathfrak{c}$ because any such function is determined by its value on rationals. hence #$(C(mathbb{R},mathbb{R}))$ = # $mathbb{R}^mathbb{Q}$ which has cardinality $2^mathbb{N_0}$.






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    • Yes, it's relevant, because it sets a lower bound on the cardinality of the question, i.e., $|mathbb{R}^{mathbb{R}}| geq |mathbb{R}| = mathfrak{c}$. Moreover, it is a nice, easy-to-understand, result :)
      – loved.by.Jesus
      Apr 29 '17 at 21:48













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    4 Answers
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    4 Answers
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    38














    All you need is a few basics of cardinal arithmetic: if $kappa$ and $lambda$ are cardinals, none of them zero, and at least one of them is infinite, then $kappa+lambda = kappalambda = max{kappa,lambda}$. And cardinal exponentiation satisfies some of the same laws as regular exponentiation; in particular, $(kappa^{lambda})^{nu} = kappa^{lambdanu}$.



    The cardinality of the set of all real functions is then
    $$|mathbb{R}|^{|mathbb{R}|} =mathfrak{c}^{mathfrak{c}} = (2^{aleph_0})^{2^{aleph_0}} = 2^{aleph_02^{aleph_0}} = 2^{2^{aleph_0}} = 2^{mathfrak{c}}.$$
    In other words, it is equal to the cardinality of the power set of $mathbb{R}$.



    With a few extra facts, you can get more. In general, if $kappa$ is an infinite cardinal, and $2leqlambdaleqkappa$, then $lambda^{kappa}=2^{kappa}$. This follows because:
    $$2^{kappa} leq lambda^{kappa} leq (2^{lambda})^{kappa} = 2^{lambdakappa} = 2^{kappa},$$
    so you get equality throughout. The extra information you need for this is to know that if $kappa$, $lambda$, and $nu$ are nonzero cardinals, $kappaleqlambda$, then $kappa^{nu}leq lambda^{nu}$.



    In particular, for any infinite cardinal $kappa$ you have $kappa^{kappa} = 2^{kappa}$.






    share|cite|improve this answer




























      38














      All you need is a few basics of cardinal arithmetic: if $kappa$ and $lambda$ are cardinals, none of them zero, and at least one of them is infinite, then $kappa+lambda = kappalambda = max{kappa,lambda}$. And cardinal exponentiation satisfies some of the same laws as regular exponentiation; in particular, $(kappa^{lambda})^{nu} = kappa^{lambdanu}$.



      The cardinality of the set of all real functions is then
      $$|mathbb{R}|^{|mathbb{R}|} =mathfrak{c}^{mathfrak{c}} = (2^{aleph_0})^{2^{aleph_0}} = 2^{aleph_02^{aleph_0}} = 2^{2^{aleph_0}} = 2^{mathfrak{c}}.$$
      In other words, it is equal to the cardinality of the power set of $mathbb{R}$.



      With a few extra facts, you can get more. In general, if $kappa$ is an infinite cardinal, and $2leqlambdaleqkappa$, then $lambda^{kappa}=2^{kappa}$. This follows because:
      $$2^{kappa} leq lambda^{kappa} leq (2^{lambda})^{kappa} = 2^{lambdakappa} = 2^{kappa},$$
      so you get equality throughout. The extra information you need for this is to know that if $kappa$, $lambda$, and $nu$ are nonzero cardinals, $kappaleqlambda$, then $kappa^{nu}leq lambda^{nu}$.



      In particular, for any infinite cardinal $kappa$ you have $kappa^{kappa} = 2^{kappa}$.






      share|cite|improve this answer


























        38












        38








        38






        All you need is a few basics of cardinal arithmetic: if $kappa$ and $lambda$ are cardinals, none of them zero, and at least one of them is infinite, then $kappa+lambda = kappalambda = max{kappa,lambda}$. And cardinal exponentiation satisfies some of the same laws as regular exponentiation; in particular, $(kappa^{lambda})^{nu} = kappa^{lambdanu}$.



        The cardinality of the set of all real functions is then
        $$|mathbb{R}|^{|mathbb{R}|} =mathfrak{c}^{mathfrak{c}} = (2^{aleph_0})^{2^{aleph_0}} = 2^{aleph_02^{aleph_0}} = 2^{2^{aleph_0}} = 2^{mathfrak{c}}.$$
        In other words, it is equal to the cardinality of the power set of $mathbb{R}$.



        With a few extra facts, you can get more. In general, if $kappa$ is an infinite cardinal, and $2leqlambdaleqkappa$, then $lambda^{kappa}=2^{kappa}$. This follows because:
        $$2^{kappa} leq lambda^{kappa} leq (2^{lambda})^{kappa} = 2^{lambdakappa} = 2^{kappa},$$
        so you get equality throughout. The extra information you need for this is to know that if $kappa$, $lambda$, and $nu$ are nonzero cardinals, $kappaleqlambda$, then $kappa^{nu}leq lambda^{nu}$.



        In particular, for any infinite cardinal $kappa$ you have $kappa^{kappa} = 2^{kappa}$.






        share|cite|improve this answer














        All you need is a few basics of cardinal arithmetic: if $kappa$ and $lambda$ are cardinals, none of them zero, and at least one of them is infinite, then $kappa+lambda = kappalambda = max{kappa,lambda}$. And cardinal exponentiation satisfies some of the same laws as regular exponentiation; in particular, $(kappa^{lambda})^{nu} = kappa^{lambdanu}$.



        The cardinality of the set of all real functions is then
        $$|mathbb{R}|^{|mathbb{R}|} =mathfrak{c}^{mathfrak{c}} = (2^{aleph_0})^{2^{aleph_0}} = 2^{aleph_02^{aleph_0}} = 2^{2^{aleph_0}} = 2^{mathfrak{c}}.$$
        In other words, it is equal to the cardinality of the power set of $mathbb{R}$.



        With a few extra facts, you can get more. In general, if $kappa$ is an infinite cardinal, and $2leqlambdaleqkappa$, then $lambda^{kappa}=2^{kappa}$. This follows because:
        $$2^{kappa} leq lambda^{kappa} leq (2^{lambda})^{kappa} = 2^{lambdakappa} = 2^{kappa},$$
        so you get equality throughout. The extra information you need for this is to know that if $kappa$, $lambda$, and $nu$ are nonzero cardinals, $kappaleqlambda$, then $kappa^{nu}leq lambda^{nu}$.



        In particular, for any infinite cardinal $kappa$ you have $kappa^{kappa} = 2^{kappa}$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 17 '11 at 23:43

























        answered Jan 17 '11 at 23:32









        Arturo Magidin

        260k32584904




        260k32584904























            29














            I guess that you know that $|mathbb{N}| = |mathbb{N}timesmathbb{N}|$ and thus $|mathbb{R}| = |2^{mathbb{N}}| = |2^{mathbb{N}timesmathbb{N}}| = |2^mathbb{N}times 2^mathbb{N}| = |mathbb{R}timesmathbb{R}|$



            This means that $|P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})|$. Since $fcolonmathbb{R}tomathbb{R}$ is an element of $P(mathbb{R}timesmathbb{R})$ you have that $mathbb{R}^mathbb{R}$ (all the functions from $mathbb{R}$ to itself) is of cardinality less or equal to the one of $P(mathbb{R}timesmathbb{R})$ which in turn means that $|mathbb{R}^mathbb{R}|le |P(mathbb{R})|$.



            Now, since $|P(mathbb{R})| = |2^mathbb{R}|$ which is the set of all functions from $mathbb{R}$ to ${0,1}$, and clearly every function from $mathbb{R}$ into ${0,1}$ is in particular a function from $mathbb{R}$ into itself, we have:
            $$|P(mathbb{R})| = |2^mathbb{R}| le |mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})| = |P(mathbb{R})|$$



            So all in all we have that $|mathbb{R}^mathbb{R}| = |P(mathbb{R})| = |2^mathbb{R}|$.






            share|cite|improve this answer



















            • 2




              Did you mean $|P(mathbb{R})|$ at the end of your second paragraph?
              – Arturo Magidin
              Jan 17 '11 at 23:51






            • 2




              @AsafKaraglia: Could you please detail how and why $|mathbb{R}| = |mathbb{R}timesmathbb{R}|$ and $ mathbb{R} neq mathbb{R}timesmathbb{R} $ $Longrightarrow |P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})|$? I made an incidental edits which I hope will help and referenced math.stackexchange.com/questions/29366/….
              – Greek - Area 51 Proposal
              Nov 7 '13 at 1:10








            • 1




              @LePressentiment: Don't add color to my posts. Thank you.
              – Asaf Karagila
              Nov 7 '13 at 5:36






            • 2




              @AsafKaragila: No problem at all. I'll post my edition separately below. In your previous version (rollback) above, you write that $mathbb{R}^mathbb{R}$ = all the functions from $mathbb{R}$ to itself. Should this be the set of all such functions? Also, will you please to let me know of my previous comment, preceding your comment?
              – Greek - Area 51 Proposal
              Nov 7 '13 at 9:48










            • @LePressentiment: I don't know what you mean by that. $|A|=|B|$ implies that $|mathcal P(A)|=|mathcal P(B)|$. That's a simple exercise in the definition of cardinalities. And yes $A^B$ is the set of all functions from $B$ into $A$, although it is sometimes denoted by ${}^BA$.
              – Asaf Karagila
              Nov 7 '13 at 10:02


















            29














            I guess that you know that $|mathbb{N}| = |mathbb{N}timesmathbb{N}|$ and thus $|mathbb{R}| = |2^{mathbb{N}}| = |2^{mathbb{N}timesmathbb{N}}| = |2^mathbb{N}times 2^mathbb{N}| = |mathbb{R}timesmathbb{R}|$



            This means that $|P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})|$. Since $fcolonmathbb{R}tomathbb{R}$ is an element of $P(mathbb{R}timesmathbb{R})$ you have that $mathbb{R}^mathbb{R}$ (all the functions from $mathbb{R}$ to itself) is of cardinality less or equal to the one of $P(mathbb{R}timesmathbb{R})$ which in turn means that $|mathbb{R}^mathbb{R}|le |P(mathbb{R})|$.



            Now, since $|P(mathbb{R})| = |2^mathbb{R}|$ which is the set of all functions from $mathbb{R}$ to ${0,1}$, and clearly every function from $mathbb{R}$ into ${0,1}$ is in particular a function from $mathbb{R}$ into itself, we have:
            $$|P(mathbb{R})| = |2^mathbb{R}| le |mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})| = |P(mathbb{R})|$$



            So all in all we have that $|mathbb{R}^mathbb{R}| = |P(mathbb{R})| = |2^mathbb{R}|$.






            share|cite|improve this answer



















            • 2




              Did you mean $|P(mathbb{R})|$ at the end of your second paragraph?
              – Arturo Magidin
              Jan 17 '11 at 23:51






            • 2




              @AsafKaraglia: Could you please detail how and why $|mathbb{R}| = |mathbb{R}timesmathbb{R}|$ and $ mathbb{R} neq mathbb{R}timesmathbb{R} $ $Longrightarrow |P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})|$? I made an incidental edits which I hope will help and referenced math.stackexchange.com/questions/29366/….
              – Greek - Area 51 Proposal
              Nov 7 '13 at 1:10








            • 1




              @LePressentiment: Don't add color to my posts. Thank you.
              – Asaf Karagila
              Nov 7 '13 at 5:36






            • 2




              @AsafKaragila: No problem at all. I'll post my edition separately below. In your previous version (rollback) above, you write that $mathbb{R}^mathbb{R}$ = all the functions from $mathbb{R}$ to itself. Should this be the set of all such functions? Also, will you please to let me know of my previous comment, preceding your comment?
              – Greek - Area 51 Proposal
              Nov 7 '13 at 9:48










            • @LePressentiment: I don't know what you mean by that. $|A|=|B|$ implies that $|mathcal P(A)|=|mathcal P(B)|$. That's a simple exercise in the definition of cardinalities. And yes $A^B$ is the set of all functions from $B$ into $A$, although it is sometimes denoted by ${}^BA$.
              – Asaf Karagila
              Nov 7 '13 at 10:02
















            29












            29








            29






            I guess that you know that $|mathbb{N}| = |mathbb{N}timesmathbb{N}|$ and thus $|mathbb{R}| = |2^{mathbb{N}}| = |2^{mathbb{N}timesmathbb{N}}| = |2^mathbb{N}times 2^mathbb{N}| = |mathbb{R}timesmathbb{R}|$



            This means that $|P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})|$. Since $fcolonmathbb{R}tomathbb{R}$ is an element of $P(mathbb{R}timesmathbb{R})$ you have that $mathbb{R}^mathbb{R}$ (all the functions from $mathbb{R}$ to itself) is of cardinality less or equal to the one of $P(mathbb{R}timesmathbb{R})$ which in turn means that $|mathbb{R}^mathbb{R}|le |P(mathbb{R})|$.



            Now, since $|P(mathbb{R})| = |2^mathbb{R}|$ which is the set of all functions from $mathbb{R}$ to ${0,1}$, and clearly every function from $mathbb{R}$ into ${0,1}$ is in particular a function from $mathbb{R}$ into itself, we have:
            $$|P(mathbb{R})| = |2^mathbb{R}| le |mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})| = |P(mathbb{R})|$$



            So all in all we have that $|mathbb{R}^mathbb{R}| = |P(mathbb{R})| = |2^mathbb{R}|$.






            share|cite|improve this answer














            I guess that you know that $|mathbb{N}| = |mathbb{N}timesmathbb{N}|$ and thus $|mathbb{R}| = |2^{mathbb{N}}| = |2^{mathbb{N}timesmathbb{N}}| = |2^mathbb{N}times 2^mathbb{N}| = |mathbb{R}timesmathbb{R}|$



            This means that $|P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})|$. Since $fcolonmathbb{R}tomathbb{R}$ is an element of $P(mathbb{R}timesmathbb{R})$ you have that $mathbb{R}^mathbb{R}$ (all the functions from $mathbb{R}$ to itself) is of cardinality less or equal to the one of $P(mathbb{R}timesmathbb{R})$ which in turn means that $|mathbb{R}^mathbb{R}|le |P(mathbb{R})|$.



            Now, since $|P(mathbb{R})| = |2^mathbb{R}|$ which is the set of all functions from $mathbb{R}$ to ${0,1}$, and clearly every function from $mathbb{R}$ into ${0,1}$ is in particular a function from $mathbb{R}$ into itself, we have:
            $$|P(mathbb{R})| = |2^mathbb{R}| le |mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})| = |P(mathbb{R})|$$



            So all in all we have that $|mathbb{R}^mathbb{R}| = |P(mathbb{R})| = |2^mathbb{R}|$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 7 '13 at 5:35

























            answered Jan 17 '11 at 23:45









            Asaf Karagila

            301k32422755




            301k32422755








            • 2




              Did you mean $|P(mathbb{R})|$ at the end of your second paragraph?
              – Arturo Magidin
              Jan 17 '11 at 23:51






            • 2




              @AsafKaraglia: Could you please detail how and why $|mathbb{R}| = |mathbb{R}timesmathbb{R}|$ and $ mathbb{R} neq mathbb{R}timesmathbb{R} $ $Longrightarrow |P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})|$? I made an incidental edits which I hope will help and referenced math.stackexchange.com/questions/29366/….
              – Greek - Area 51 Proposal
              Nov 7 '13 at 1:10








            • 1




              @LePressentiment: Don't add color to my posts. Thank you.
              – Asaf Karagila
              Nov 7 '13 at 5:36






            • 2




              @AsafKaragila: No problem at all. I'll post my edition separately below. In your previous version (rollback) above, you write that $mathbb{R}^mathbb{R}$ = all the functions from $mathbb{R}$ to itself. Should this be the set of all such functions? Also, will you please to let me know of my previous comment, preceding your comment?
              – Greek - Area 51 Proposal
              Nov 7 '13 at 9:48










            • @LePressentiment: I don't know what you mean by that. $|A|=|B|$ implies that $|mathcal P(A)|=|mathcal P(B)|$. That's a simple exercise in the definition of cardinalities. And yes $A^B$ is the set of all functions from $B$ into $A$, although it is sometimes denoted by ${}^BA$.
              – Asaf Karagila
              Nov 7 '13 at 10:02
















            • 2




              Did you mean $|P(mathbb{R})|$ at the end of your second paragraph?
              – Arturo Magidin
              Jan 17 '11 at 23:51






            • 2




              @AsafKaraglia: Could you please detail how and why $|mathbb{R}| = |mathbb{R}timesmathbb{R}|$ and $ mathbb{R} neq mathbb{R}timesmathbb{R} $ $Longrightarrow |P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})|$? I made an incidental edits which I hope will help and referenced math.stackexchange.com/questions/29366/….
              – Greek - Area 51 Proposal
              Nov 7 '13 at 1:10








            • 1




              @LePressentiment: Don't add color to my posts. Thank you.
              – Asaf Karagila
              Nov 7 '13 at 5:36






            • 2




              @AsafKaragila: No problem at all. I'll post my edition separately below. In your previous version (rollback) above, you write that $mathbb{R}^mathbb{R}$ = all the functions from $mathbb{R}$ to itself. Should this be the set of all such functions? Also, will you please to let me know of my previous comment, preceding your comment?
              – Greek - Area 51 Proposal
              Nov 7 '13 at 9:48










            • @LePressentiment: I don't know what you mean by that. $|A|=|B|$ implies that $|mathcal P(A)|=|mathcal P(B)|$. That's a simple exercise in the definition of cardinalities. And yes $A^B$ is the set of all functions from $B$ into $A$, although it is sometimes denoted by ${}^BA$.
              – Asaf Karagila
              Nov 7 '13 at 10:02










            2




            2




            Did you mean $|P(mathbb{R})|$ at the end of your second paragraph?
            – Arturo Magidin
            Jan 17 '11 at 23:51




            Did you mean $|P(mathbb{R})|$ at the end of your second paragraph?
            – Arturo Magidin
            Jan 17 '11 at 23:51




            2




            2




            @AsafKaraglia: Could you please detail how and why $|mathbb{R}| = |mathbb{R}timesmathbb{R}|$ and $ mathbb{R} neq mathbb{R}timesmathbb{R} $ $Longrightarrow |P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})|$? I made an incidental edits which I hope will help and referenced math.stackexchange.com/questions/29366/….
            – Greek - Area 51 Proposal
            Nov 7 '13 at 1:10






            @AsafKaraglia: Could you please detail how and why $|mathbb{R}| = |mathbb{R}timesmathbb{R}|$ and $ mathbb{R} neq mathbb{R}timesmathbb{R} $ $Longrightarrow |P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})|$? I made an incidental edits which I hope will help and referenced math.stackexchange.com/questions/29366/….
            – Greek - Area 51 Proposal
            Nov 7 '13 at 1:10






            1




            1




            @LePressentiment: Don't add color to my posts. Thank you.
            – Asaf Karagila
            Nov 7 '13 at 5:36




            @LePressentiment: Don't add color to my posts. Thank you.
            – Asaf Karagila
            Nov 7 '13 at 5:36




            2




            2




            @AsafKaragila: No problem at all. I'll post my edition separately below. In your previous version (rollback) above, you write that $mathbb{R}^mathbb{R}$ = all the functions from $mathbb{R}$ to itself. Should this be the set of all such functions? Also, will you please to let me know of my previous comment, preceding your comment?
            – Greek - Area 51 Proposal
            Nov 7 '13 at 9:48




            @AsafKaragila: No problem at all. I'll post my edition separately below. In your previous version (rollback) above, you write that $mathbb{R}^mathbb{R}$ = all the functions from $mathbb{R}$ to itself. Should this be the set of all such functions? Also, will you please to let me know of my previous comment, preceding your comment?
            – Greek - Area 51 Proposal
            Nov 7 '13 at 9:48












            @LePressentiment: I don't know what you mean by that. $|A|=|B|$ implies that $|mathcal P(A)|=|mathcal P(B)|$. That's a simple exercise in the definition of cardinalities. And yes $A^B$ is the set of all functions from $B$ into $A$, although it is sometimes denoted by ${}^BA$.
            – Asaf Karagila
            Nov 7 '13 at 10:02






            @LePressentiment: I don't know what you mean by that. $|A|=|B|$ implies that $|mathcal P(A)|=|mathcal P(B)|$. That's a simple exercise in the definition of cardinalities. And yes $A^B$ is the set of all functions from $B$ into $A$, although it is sometimes denoted by ${}^BA$.
            – Asaf Karagila
            Nov 7 '13 at 10:02













            3














            This answer is based on, but differs slightly from, user Asaf Karaglia's above.





            First, observe that by definition, ${text{all real functions of real variable}}:= {f: ; f: mathbb{R}tomathbb{R}} := mathbb{R}^mathbb{R}$.



            The question is about $|{text{all real functions of real variable}}|$, so examine an arbitrary real function of real variable: $f,colon,mathbb{R}tomathbb{R}.$

            By inspection, $f,colon,mathbb{R}tomathbb{R} := {(r, f(r)) : r in mathbb{R}} quad subseteq quad P(mathbb{R} times mathbb{R})$.

            Thus, $color{green}{|mathbb{R}^{mathbb{R}}| le |P(mathbb{R}timesmathbb{R})|}$.



            Before continuing, let's try to simplify $|P(mathbb{R}timesmathbb{R})|$. Observe that $|mathbb{R}| = |mathbb{R}^k| , forall , k in mathbb{N}$. Its proof by mathematical induction requires the induction hypothesis of $|mathbb{R}| = |mathbb{R}^2|$, one proof of which is : $|mathbb{N}| = |mathbb{N}timesmathbb{N}| implies |mathbb{R}| = |2^{mathbb{N}}| = |2^{mathbb{N}timesmathbb{N}}| = |2^mathbb{N}times 2^mathbb{N}| = |mathbb{R}timesmathbb{R}|$.



            Verily, $mathbb{R} neq mathbb{R}^2$. Howbeit, for infinite sets $A,B$: $|A| = |B| Longrightarrow require{cancel} cancel{Longleftarrow} |P(A)| = |P(B)|$.

            (The converse is discussed here.)



            Thus, $|P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})| implies color{green}{|mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})|} = |P(mathbb{R})|$. Now scrutinise $|P(mathbb{R})|$:



            ● $color{#A9057D}{|P(mathbb{R})| = |2^{mathbb{R}}|}$, where $2^{mathbb{R}} := {f : ; f: mathbb{R} to {0,1}}$,

            ● Every $f: mathbb{R} to {0,1}$ is a particular case of a function from $mathbb{R}$ to $mathbb{R}$, thus $color{#EC5021}{2^{mathbb{R}} subsetneq mathbb{R}^mathbb{R}}$.



            Altogether, $color{#A9057D}{|P(mathbb{R})| =} color{#EC5021}{|2^mathbb{R}| le} color{green}{|mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})|} = |P(mathbb{R})|$



            $implies |P(mathbb{R})| qquad qquad quad leq |mathbb{R}^mathbb{R}| leq |P(mathbb{R})| implies color{#A9057D}{underbrace{|P(mathbb{R})|}_{= |2^mathbb{R}|}} = |mathbb{R}^mathbb{R}| $.






            share|cite|improve this answer























            • Downvoters, pursuant to my edit, please let me know of further sugggestions which would be more instructive than a mere downvote.
              – Greek - Area 51 Proposal
              Nov 8 '13 at 7:35






            • 4




              Use less colors, so people with disabilities could read this without getting a headache.
              – Asaf Karagila
              Nov 8 '13 at 8:31
















            3














            This answer is based on, but differs slightly from, user Asaf Karaglia's above.





            First, observe that by definition, ${text{all real functions of real variable}}:= {f: ; f: mathbb{R}tomathbb{R}} := mathbb{R}^mathbb{R}$.



            The question is about $|{text{all real functions of real variable}}|$, so examine an arbitrary real function of real variable: $f,colon,mathbb{R}tomathbb{R}.$

            By inspection, $f,colon,mathbb{R}tomathbb{R} := {(r, f(r)) : r in mathbb{R}} quad subseteq quad P(mathbb{R} times mathbb{R})$.

            Thus, $color{green}{|mathbb{R}^{mathbb{R}}| le |P(mathbb{R}timesmathbb{R})|}$.



            Before continuing, let's try to simplify $|P(mathbb{R}timesmathbb{R})|$. Observe that $|mathbb{R}| = |mathbb{R}^k| , forall , k in mathbb{N}$. Its proof by mathematical induction requires the induction hypothesis of $|mathbb{R}| = |mathbb{R}^2|$, one proof of which is : $|mathbb{N}| = |mathbb{N}timesmathbb{N}| implies |mathbb{R}| = |2^{mathbb{N}}| = |2^{mathbb{N}timesmathbb{N}}| = |2^mathbb{N}times 2^mathbb{N}| = |mathbb{R}timesmathbb{R}|$.



            Verily, $mathbb{R} neq mathbb{R}^2$. Howbeit, for infinite sets $A,B$: $|A| = |B| Longrightarrow require{cancel} cancel{Longleftarrow} |P(A)| = |P(B)|$.

            (The converse is discussed here.)



            Thus, $|P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})| implies color{green}{|mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})|} = |P(mathbb{R})|$. Now scrutinise $|P(mathbb{R})|$:



            ● $color{#A9057D}{|P(mathbb{R})| = |2^{mathbb{R}}|}$, where $2^{mathbb{R}} := {f : ; f: mathbb{R} to {0,1}}$,

            ● Every $f: mathbb{R} to {0,1}$ is a particular case of a function from $mathbb{R}$ to $mathbb{R}$, thus $color{#EC5021}{2^{mathbb{R}} subsetneq mathbb{R}^mathbb{R}}$.



            Altogether, $color{#A9057D}{|P(mathbb{R})| =} color{#EC5021}{|2^mathbb{R}| le} color{green}{|mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})|} = |P(mathbb{R})|$



            $implies |P(mathbb{R})| qquad qquad quad leq |mathbb{R}^mathbb{R}| leq |P(mathbb{R})| implies color{#A9057D}{underbrace{|P(mathbb{R})|}_{= |2^mathbb{R}|}} = |mathbb{R}^mathbb{R}| $.






            share|cite|improve this answer























            • Downvoters, pursuant to my edit, please let me know of further sugggestions which would be more instructive than a mere downvote.
              – Greek - Area 51 Proposal
              Nov 8 '13 at 7:35






            • 4




              Use less colors, so people with disabilities could read this without getting a headache.
              – Asaf Karagila
              Nov 8 '13 at 8:31














            3












            3








            3






            This answer is based on, but differs slightly from, user Asaf Karaglia's above.





            First, observe that by definition, ${text{all real functions of real variable}}:= {f: ; f: mathbb{R}tomathbb{R}} := mathbb{R}^mathbb{R}$.



            The question is about $|{text{all real functions of real variable}}|$, so examine an arbitrary real function of real variable: $f,colon,mathbb{R}tomathbb{R}.$

            By inspection, $f,colon,mathbb{R}tomathbb{R} := {(r, f(r)) : r in mathbb{R}} quad subseteq quad P(mathbb{R} times mathbb{R})$.

            Thus, $color{green}{|mathbb{R}^{mathbb{R}}| le |P(mathbb{R}timesmathbb{R})|}$.



            Before continuing, let's try to simplify $|P(mathbb{R}timesmathbb{R})|$. Observe that $|mathbb{R}| = |mathbb{R}^k| , forall , k in mathbb{N}$. Its proof by mathematical induction requires the induction hypothesis of $|mathbb{R}| = |mathbb{R}^2|$, one proof of which is : $|mathbb{N}| = |mathbb{N}timesmathbb{N}| implies |mathbb{R}| = |2^{mathbb{N}}| = |2^{mathbb{N}timesmathbb{N}}| = |2^mathbb{N}times 2^mathbb{N}| = |mathbb{R}timesmathbb{R}|$.



            Verily, $mathbb{R} neq mathbb{R}^2$. Howbeit, for infinite sets $A,B$: $|A| = |B| Longrightarrow require{cancel} cancel{Longleftarrow} |P(A)| = |P(B)|$.

            (The converse is discussed here.)



            Thus, $|P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})| implies color{green}{|mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})|} = |P(mathbb{R})|$. Now scrutinise $|P(mathbb{R})|$:



            ● $color{#A9057D}{|P(mathbb{R})| = |2^{mathbb{R}}|}$, where $2^{mathbb{R}} := {f : ; f: mathbb{R} to {0,1}}$,

            ● Every $f: mathbb{R} to {0,1}$ is a particular case of a function from $mathbb{R}$ to $mathbb{R}$, thus $color{#EC5021}{2^{mathbb{R}} subsetneq mathbb{R}^mathbb{R}}$.



            Altogether, $color{#A9057D}{|P(mathbb{R})| =} color{#EC5021}{|2^mathbb{R}| le} color{green}{|mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})|} = |P(mathbb{R})|$



            $implies |P(mathbb{R})| qquad qquad quad leq |mathbb{R}^mathbb{R}| leq |P(mathbb{R})| implies color{#A9057D}{underbrace{|P(mathbb{R})|}_{= |2^mathbb{R}|}} = |mathbb{R}^mathbb{R}| $.






            share|cite|improve this answer














            This answer is based on, but differs slightly from, user Asaf Karaglia's above.





            First, observe that by definition, ${text{all real functions of real variable}}:= {f: ; f: mathbb{R}tomathbb{R}} := mathbb{R}^mathbb{R}$.



            The question is about $|{text{all real functions of real variable}}|$, so examine an arbitrary real function of real variable: $f,colon,mathbb{R}tomathbb{R}.$

            By inspection, $f,colon,mathbb{R}tomathbb{R} := {(r, f(r)) : r in mathbb{R}} quad subseteq quad P(mathbb{R} times mathbb{R})$.

            Thus, $color{green}{|mathbb{R}^{mathbb{R}}| le |P(mathbb{R}timesmathbb{R})|}$.



            Before continuing, let's try to simplify $|P(mathbb{R}timesmathbb{R})|$. Observe that $|mathbb{R}| = |mathbb{R}^k| , forall , k in mathbb{N}$. Its proof by mathematical induction requires the induction hypothesis of $|mathbb{R}| = |mathbb{R}^2|$, one proof of which is : $|mathbb{N}| = |mathbb{N}timesmathbb{N}| implies |mathbb{R}| = |2^{mathbb{N}}| = |2^{mathbb{N}timesmathbb{N}}| = |2^mathbb{N}times 2^mathbb{N}| = |mathbb{R}timesmathbb{R}|$.



            Verily, $mathbb{R} neq mathbb{R}^2$. Howbeit, for infinite sets $A,B$: $|A| = |B| Longrightarrow require{cancel} cancel{Longleftarrow} |P(A)| = |P(B)|$.

            (The converse is discussed here.)



            Thus, $|P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})| implies color{green}{|mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})|} = |P(mathbb{R})|$. Now scrutinise $|P(mathbb{R})|$:



            ● $color{#A9057D}{|P(mathbb{R})| = |2^{mathbb{R}}|}$, where $2^{mathbb{R}} := {f : ; f: mathbb{R} to {0,1}}$,

            ● Every $f: mathbb{R} to {0,1}$ is a particular case of a function from $mathbb{R}$ to $mathbb{R}$, thus $color{#EC5021}{2^{mathbb{R}} subsetneq mathbb{R}^mathbb{R}}$.



            Altogether, $color{#A9057D}{|P(mathbb{R})| =} color{#EC5021}{|2^mathbb{R}| le} color{green}{|mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})|} = |P(mathbb{R})|$



            $implies |P(mathbb{R})| qquad qquad quad leq |mathbb{R}^mathbb{R}| leq |P(mathbb{R})| implies color{#A9057D}{underbrace{|P(mathbb{R})|}_{= |2^mathbb{R}|}} = |mathbb{R}^mathbb{R}| $.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 13 '17 at 12:19









            Community

            1




            1










            answered Nov 7 '13 at 9:49









            Greek - Area 51 Proposal

            3,155669103




            3,155669103












            • Downvoters, pursuant to my edit, please let me know of further sugggestions which would be more instructive than a mere downvote.
              – Greek - Area 51 Proposal
              Nov 8 '13 at 7:35






            • 4




              Use less colors, so people with disabilities could read this without getting a headache.
              – Asaf Karagila
              Nov 8 '13 at 8:31


















            • Downvoters, pursuant to my edit, please let me know of further sugggestions which would be more instructive than a mere downvote.
              – Greek - Area 51 Proposal
              Nov 8 '13 at 7:35






            • 4




              Use less colors, so people with disabilities could read this without getting a headache.
              – Asaf Karagila
              Nov 8 '13 at 8:31
















            Downvoters, pursuant to my edit, please let me know of further sugggestions which would be more instructive than a mere downvote.
            – Greek - Area 51 Proposal
            Nov 8 '13 at 7:35




            Downvoters, pursuant to my edit, please let me know of further sugggestions which would be more instructive than a mere downvote.
            – Greek - Area 51 Proposal
            Nov 8 '13 at 7:35




            4




            4




            Use less colors, so people with disabilities could read this without getting a headache.
            – Asaf Karagila
            Nov 8 '13 at 8:31




            Use less colors, so people with disabilities could read this without getting a headache.
            – Asaf Karagila
            Nov 8 '13 at 8:31











            2














            This is irrelevent here, still it is 'relevent'. The cardinality of set of all continuous function from $mathbb{R}$ to $mathbb{R}$ $(C(mathbb{R},mathbb{R}))$ is $2 ^ mathbb{N_0} = mathfrak{c}$ because any such function is determined by its value on rationals. hence #$(C(mathbb{R},mathbb{R}))$ = # $mathbb{R}^mathbb{Q}$ which has cardinality $2^mathbb{N_0}$.






            share|cite|improve this answer























            • Yes, it's relevant, because it sets a lower bound on the cardinality of the question, i.e., $|mathbb{R}^{mathbb{R}}| geq |mathbb{R}| = mathfrak{c}$. Moreover, it is a nice, easy-to-understand, result :)
              – loved.by.Jesus
              Apr 29 '17 at 21:48


















            2














            This is irrelevent here, still it is 'relevent'. The cardinality of set of all continuous function from $mathbb{R}$ to $mathbb{R}$ $(C(mathbb{R},mathbb{R}))$ is $2 ^ mathbb{N_0} = mathfrak{c}$ because any such function is determined by its value on rationals. hence #$(C(mathbb{R},mathbb{R}))$ = # $mathbb{R}^mathbb{Q}$ which has cardinality $2^mathbb{N_0}$.






            share|cite|improve this answer























            • Yes, it's relevant, because it sets a lower bound on the cardinality of the question, i.e., $|mathbb{R}^{mathbb{R}}| geq |mathbb{R}| = mathfrak{c}$. Moreover, it is a nice, easy-to-understand, result :)
              – loved.by.Jesus
              Apr 29 '17 at 21:48
















            2












            2








            2






            This is irrelevent here, still it is 'relevent'. The cardinality of set of all continuous function from $mathbb{R}$ to $mathbb{R}$ $(C(mathbb{R},mathbb{R}))$ is $2 ^ mathbb{N_0} = mathfrak{c}$ because any such function is determined by its value on rationals. hence #$(C(mathbb{R},mathbb{R}))$ = # $mathbb{R}^mathbb{Q}$ which has cardinality $2^mathbb{N_0}$.






            share|cite|improve this answer














            This is irrelevent here, still it is 'relevent'. The cardinality of set of all continuous function from $mathbb{R}$ to $mathbb{R}$ $(C(mathbb{R},mathbb{R}))$ is $2 ^ mathbb{N_0} = mathfrak{c}$ because any such function is determined by its value on rationals. hence #$(C(mathbb{R},mathbb{R}))$ = # $mathbb{R}^mathbb{Q}$ which has cardinality $2^mathbb{N_0}$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 29 '17 at 22:03









            loved.by.Jesus

            1319




            1319










            answered Nov 7 '13 at 6:14









            GA316

            2,6381132




            2,6381132












            • Yes, it's relevant, because it sets a lower bound on the cardinality of the question, i.e., $|mathbb{R}^{mathbb{R}}| geq |mathbb{R}| = mathfrak{c}$. Moreover, it is a nice, easy-to-understand, result :)
              – loved.by.Jesus
              Apr 29 '17 at 21:48




















            • Yes, it's relevant, because it sets a lower bound on the cardinality of the question, i.e., $|mathbb{R}^{mathbb{R}}| geq |mathbb{R}| = mathfrak{c}$. Moreover, it is a nice, easy-to-understand, result :)
              – loved.by.Jesus
              Apr 29 '17 at 21:48


















            Yes, it's relevant, because it sets a lower bound on the cardinality of the question, i.e., $|mathbb{R}^{mathbb{R}}| geq |mathbb{R}| = mathfrak{c}$. Moreover, it is a nice, easy-to-understand, result :)
            – loved.by.Jesus
            Apr 29 '17 at 21:48






            Yes, it's relevant, because it sets a lower bound on the cardinality of the question, i.e., $|mathbb{R}^{mathbb{R}}| geq |mathbb{R}| = mathfrak{c}$. Moreover, it is a nice, easy-to-understand, result :)
            – loved.by.Jesus
            Apr 29 '17 at 21:48




















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