What is the cardinality of the following set?












0












$begingroup$


What is the cardinality of the following set: ${f: mathbb{R} rightarrow {a+bsqrt[3]4 | a,b in mathbb{Q}}}$?



Let $S = {f: mathbb{R} rightarrow {a+bsqrt[3]4 | a,b in mathbb{Q}}}$.



I already show that ${f:mathbb{R}rightarrow{0,1}} subset S$, then $|S| ge |{f:mathbb{R}rightarrow{0,1}}| = 2^c$, and I need to show the rest of the part i.e.: $|S| le 2^c$, then to conclude that the cardinality of the set $S$ is $2^c$. (please correct me if this part is wrong)



But I get stuck, I don't know how to show that $|S| le 2^c$. Can you please help me out? Thanks in advance!






abstract-algebra elementary-set-theory cardinals







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  • $begingroup$
    I assume $R$ and $Q$ are supposed to be $mathbb{R}$ and $mathbb{Q}$?
    $endgroup$
    – GenericMathematician
    Dec 9 '18 at 21:57






  • 2




    $begingroup$
    Would you be more comfortable if ${,a+bsqrt[3]4mid a,binBbb Q,}$ were replaced with $Bbb N$?
    $endgroup$
    – Hagen von Eitzen
    Dec 9 '18 at 21:57


















0












$begingroup$


What is the cardinality of the following set: ${f: mathbb{R} rightarrow {a+bsqrt[3]4 | a,b in mathbb{Q}}}$?



Let $S = {f: mathbb{R} rightarrow {a+bsqrt[3]4 | a,b in mathbb{Q}}}$.



I already show that ${f:mathbb{R}rightarrow{0,1}} subset S$, then $|S| ge |{f:mathbb{R}rightarrow{0,1}}| = 2^c$, and I need to show the rest of the part i.e.: $|S| le 2^c$, then to conclude that the cardinality of the set $S$ is $2^c$. (please correct me if this part is wrong)



But I get stuck, I don't know how to show that $|S| le 2^c$. Can you please help me out? Thanks in advance!






abstract-algebra elementary-set-theory cardinals







share|cite|improve this question











$endgroup$












  • $begingroup$
    I assume $R$ and $Q$ are supposed to be $mathbb{R}$ and $mathbb{Q}$?
    $endgroup$
    – GenericMathematician
    Dec 9 '18 at 21:57






  • 2




    $begingroup$
    Would you be more comfortable if ${,a+bsqrt[3]4mid a,binBbb Q,}$ were replaced with $Bbb N$?
    $endgroup$
    – Hagen von Eitzen
    Dec 9 '18 at 21:57
















0












0








0


1



$begingroup$


What is the cardinality of the following set: ${f: mathbb{R} rightarrow {a+bsqrt[3]4 | a,b in mathbb{Q}}}$?



Let $S = {f: mathbb{R} rightarrow {a+bsqrt[3]4 | a,b in mathbb{Q}}}$.



I already show that ${f:mathbb{R}rightarrow{0,1}} subset S$, then $|S| ge |{f:mathbb{R}rightarrow{0,1}}| = 2^c$, and I need to show the rest of the part i.e.: $|S| le 2^c$, then to conclude that the cardinality of the set $S$ is $2^c$. (please correct me if this part is wrong)



But I get stuck, I don't know how to show that $|S| le 2^c$. Can you please help me out? Thanks in advance!






abstract-algebra elementary-set-theory cardinals







share|cite|improve this question











$endgroup$




What is the cardinality of the following set: ${f: mathbb{R} rightarrow {a+bsqrt[3]4 | a,b in mathbb{Q}}}$?



Let $S = {f: mathbb{R} rightarrow {a+bsqrt[3]4 | a,b in mathbb{Q}}}$.



I already show that ${f:mathbb{R}rightarrow{0,1}} subset S$, then $|S| ge |{f:mathbb{R}rightarrow{0,1}}| = 2^c$, and I need to show the rest of the part i.e.: $|S| le 2^c$, then to conclude that the cardinality of the set $S$ is $2^c$. (please correct me if this part is wrong)



But I get stuck, I don't know how to show that $|S| le 2^c$. Can you please help me out? Thanks in advance!






abstract-algebra elementary-set-theory cardinals




abstract-algebra 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 9 '18 at 22:08









ODF

1,486510




1,486510










asked Dec 9 '18 at 21:50









EdwardEdward

132




132












  • $begingroup$
    I assume $R$ and $Q$ are supposed to be $mathbb{R}$ and $mathbb{Q}$?
    $endgroup$
    – GenericMathematician
    Dec 9 '18 at 21:57






  • 2




    $begingroup$
    Would you be more comfortable if ${,a+bsqrt[3]4mid a,binBbb Q,}$ were replaced with $Bbb N$?
    $endgroup$
    – Hagen von Eitzen
    Dec 9 '18 at 21:57




















  • $begingroup$
    I assume $R$ and $Q$ are supposed to be $mathbb{R}$ and $mathbb{Q}$?
    $endgroup$
    – GenericMathematician
    Dec 9 '18 at 21:57






  • 2




    $begingroup$
    Would you be more comfortable if ${,a+bsqrt[3]4mid a,binBbb Q,}$ were replaced with $Bbb N$?
    $endgroup$
    – Hagen von Eitzen
    Dec 9 '18 at 21:57


















$begingroup$
I assume $R$ and $Q$ are supposed to be $mathbb{R}$ and $mathbb{Q}$?
$endgroup$
– GenericMathematician
Dec 9 '18 at 21:57




$begingroup$
I assume $R$ and $Q$ are supposed to be $mathbb{R}$ and $mathbb{Q}$?
$endgroup$
– GenericMathematician
Dec 9 '18 at 21:57




2




2




$begingroup$
Would you be more comfortable if ${,a+bsqrt[3]4mid a,binBbb Q,}$ were replaced with $Bbb N$?
$endgroup$
– Hagen von Eitzen
Dec 9 '18 at 21:57






$begingroup$
Would you be more comfortable if ${,a+bsqrt[3]4mid a,binBbb Q,}$ were replaced with $Bbb N$?
$endgroup$
– Hagen von Eitzen
Dec 9 '18 at 21:57












1 Answer
1






active

oldest

votes


















0












$begingroup$

You're looking at the set of functins from $mathbb{R}$ which has size $2^{aleph_0}= mathfrak{c}$ to a countable set (as $|mathbb{Q}^2| = aleph_0$) which by definition has size $(aleph_0)^{2^{aleph_0}} = 2^{mathfrak{c}}$ by standard cardinal arithmetic : $$ 2^mathfrak{c} le aleph_0^{mathfrak{c}} le (2^mathfrak{c})^mathfrak{c} = 2^mathfrak{c}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you find an one to one function map to show that the cardinality of the set S is less than or equal to $2^c$. Since we haven't learned cardinal arithmetic yet...
    $endgroup$
    – Edward
    Dec 9 '18 at 22:11












  • $begingroup$
    You'll need axiom of choice for that, in order to use a bijection between $mathfrak c times mathfrak c$ and $mathfrak c$.
    $endgroup$
    – mathcounterexamples.net
    Dec 10 '18 at 16:52












  • $begingroup$
    @mathcounterexamples.net well for $mathfrak{c}$ we can show it without AC: use the representation as sequences of natural numbers that can be interleaved.
    $endgroup$
    – Henno Brandsma
    Dec 10 '18 at 17:38











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1 Answer
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active

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1 Answer
1






active

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active

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active

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0












$begingroup$

You're looking at the set of functins from $mathbb{R}$ which has size $2^{aleph_0}= mathfrak{c}$ to a countable set (as $|mathbb{Q}^2| = aleph_0$) which by definition has size $(aleph_0)^{2^{aleph_0}} = 2^{mathfrak{c}}$ by standard cardinal arithmetic : $$ 2^mathfrak{c} le aleph_0^{mathfrak{c}} le (2^mathfrak{c})^mathfrak{c} = 2^mathfrak{c}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you find an one to one function map to show that the cardinality of the set S is less than or equal to $2^c$. Since we haven't learned cardinal arithmetic yet...
    $endgroup$
    – Edward
    Dec 9 '18 at 22:11












  • $begingroup$
    You'll need axiom of choice for that, in order to use a bijection between $mathfrak c times mathfrak c$ and $mathfrak c$.
    $endgroup$
    – mathcounterexamples.net
    Dec 10 '18 at 16:52












  • $begingroup$
    @mathcounterexamples.net well for $mathfrak{c}$ we can show it without AC: use the representation as sequences of natural numbers that can be interleaved.
    $endgroup$
    – Henno Brandsma
    Dec 10 '18 at 17:38
















0












$begingroup$

You're looking at the set of functins from $mathbb{R}$ which has size $2^{aleph_0}= mathfrak{c}$ to a countable set (as $|mathbb{Q}^2| = aleph_0$) which by definition has size $(aleph_0)^{2^{aleph_0}} = 2^{mathfrak{c}}$ by standard cardinal arithmetic : $$ 2^mathfrak{c} le aleph_0^{mathfrak{c}} le (2^mathfrak{c})^mathfrak{c} = 2^mathfrak{c}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you find an one to one function map to show that the cardinality of the set S is less than or equal to $2^c$. Since we haven't learned cardinal arithmetic yet...
    $endgroup$
    – Edward
    Dec 9 '18 at 22:11












  • $begingroup$
    You'll need axiom of choice for that, in order to use a bijection between $mathfrak c times mathfrak c$ and $mathfrak c$.
    $endgroup$
    – mathcounterexamples.net
    Dec 10 '18 at 16:52












  • $begingroup$
    @mathcounterexamples.net well for $mathfrak{c}$ we can show it without AC: use the representation as sequences of natural numbers that can be interleaved.
    $endgroup$
    – Henno Brandsma
    Dec 10 '18 at 17:38














0












0








0





$begingroup$

You're looking at the set of functins from $mathbb{R}$ which has size $2^{aleph_0}= mathfrak{c}$ to a countable set (as $|mathbb{Q}^2| = aleph_0$) which by definition has size $(aleph_0)^{2^{aleph_0}} = 2^{mathfrak{c}}$ by standard cardinal arithmetic : $$ 2^mathfrak{c} le aleph_0^{mathfrak{c}} le (2^mathfrak{c})^mathfrak{c} = 2^mathfrak{c}$$






share|cite|improve this answer









$endgroup$



You're looking at the set of functins from $mathbb{R}$ which has size $2^{aleph_0}= mathfrak{c}$ to a countable set (as $|mathbb{Q}^2| = aleph_0$) which by definition has size $(aleph_0)^{2^{aleph_0}} = 2^{mathfrak{c}}$ by standard cardinal arithmetic : $$ 2^mathfrak{c} le aleph_0^{mathfrak{c}} le (2^mathfrak{c})^mathfrak{c} = 2^mathfrak{c}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 22:03









Henno BrandsmaHenno Brandsma

108k347114




108k347114












  • $begingroup$
    Can you find an one to one function map to show that the cardinality of the set S is less than or equal to $2^c$. Since we haven't learned cardinal arithmetic yet...
    $endgroup$
    – Edward
    Dec 9 '18 at 22:11












  • $begingroup$
    You'll need axiom of choice for that, in order to use a bijection between $mathfrak c times mathfrak c$ and $mathfrak c$.
    $endgroup$
    – mathcounterexamples.net
    Dec 10 '18 at 16:52












  • $begingroup$
    @mathcounterexamples.net well for $mathfrak{c}$ we can show it without AC: use the representation as sequences of natural numbers that can be interleaved.
    $endgroup$
    – Henno Brandsma
    Dec 10 '18 at 17:38


















  • $begingroup$
    Can you find an one to one function map to show that the cardinality of the set S is less than or equal to $2^c$. Since we haven't learned cardinal arithmetic yet...
    $endgroup$
    – Edward
    Dec 9 '18 at 22:11












  • $begingroup$
    You'll need axiom of choice for that, in order to use a bijection between $mathfrak c times mathfrak c$ and $mathfrak c$.
    $endgroup$
    – mathcounterexamples.net
    Dec 10 '18 at 16:52












  • $begingroup$
    @mathcounterexamples.net well for $mathfrak{c}$ we can show it without AC: use the representation as sequences of natural numbers that can be interleaved.
    $endgroup$
    – Henno Brandsma
    Dec 10 '18 at 17:38
















$begingroup$
Can you find an one to one function map to show that the cardinality of the set S is less than or equal to $2^c$. Since we haven't learned cardinal arithmetic yet...
$endgroup$
– Edward
Dec 9 '18 at 22:11






$begingroup$
Can you find an one to one function map to show that the cardinality of the set S is less than or equal to $2^c$. Since we haven't learned cardinal arithmetic yet...
$endgroup$
– Edward
Dec 9 '18 at 22:11














$begingroup$
You'll need axiom of choice for that, in order to use a bijection between $mathfrak c times mathfrak c$ and $mathfrak c$.
$endgroup$
– mathcounterexamples.net
Dec 10 '18 at 16:52






$begingroup$
You'll need axiom of choice for that, in order to use a bijection between $mathfrak c times mathfrak c$ and $mathfrak c$.
$endgroup$
– mathcounterexamples.net
Dec 10 '18 at 16:52














$begingroup$
@mathcounterexamples.net well for $mathfrak{c}$ we can show it without AC: use the representation as sequences of natural numbers that can be interleaved.
$endgroup$
– Henno Brandsma
Dec 10 '18 at 17:38




$begingroup$
@mathcounterexamples.net well for $mathfrak{c}$ we can show it without AC: use the representation as sequences of natural numbers that can be interleaved.
$endgroup$
– Henno Brandsma
Dec 10 '18 at 17:38


















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