Understanding power of $left{ f | f: mathbb R rightarrow mathbb R right}$












1












$begingroup$


I have problem with understanding power of $left{ f | f: mathbb R rightarrow mathbb R right}$

Generally, from lecture I know that power of set
$$left{ f | f: A rightarrow B right}$$
is $ |B|^{|A|} $ - ok, so it seems that power of
$$left{ f | f: mathbb R rightarrow mathbb R right}$$
should be $ |mathbb R|^{|mathbb R|} = mathfrak{c}^{mathfrak{c}} = 2^{mathfrak{c}} $

Ok, but from the other hand, I can interpret that function as infinity set of pairs $ (x,y) $

So for each $ x $ I choose $y$. I can do this on $mathfrak{c}$ ways. After that I repeat this process $mathfrak{c} $ times so I have
$$mathfrak{c} cdot mathfrak{c} cdot mathfrak{c} cdot ... $$
But from lecture I know that $$mathfrak{c} cdot mathfrak{c} = mathfrak{c} $$
So I reduce it to $mathfrak{c}$.

I know that somewhere I have failed, but I have some doubts about that :(






cardinals







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  • $begingroup$
    You have infinitely many (uncountable many) $mathfrak c$ factors, whereas with your argument you can only remove finitely many.
    $endgroup$
    – Dog_69
    Jan 20 at 16:09












  • $begingroup$
    so should I write that as $mathfrak{c}^{mathfrak{c}} $?
    $endgroup$
    – VirtualUser
    Jan 20 at 16:29
















1












$begingroup$


I have problem with understanding power of $left{ f | f: mathbb R rightarrow mathbb R right}$

Generally, from lecture I know that power of set
$$left{ f | f: A rightarrow B right}$$
is $ |B|^{|A|} $ - ok, so it seems that power of
$$left{ f | f: mathbb R rightarrow mathbb R right}$$
should be $ |mathbb R|^{|mathbb R|} = mathfrak{c}^{mathfrak{c}} = 2^{mathfrak{c}} $

Ok, but from the other hand, I can interpret that function as infinity set of pairs $ (x,y) $

So for each $ x $ I choose $y$. I can do this on $mathfrak{c}$ ways. After that I repeat this process $mathfrak{c} $ times so I have
$$mathfrak{c} cdot mathfrak{c} cdot mathfrak{c} cdot ... $$
But from lecture I know that $$mathfrak{c} cdot mathfrak{c} = mathfrak{c} $$
So I reduce it to $mathfrak{c}$.

I know that somewhere I have failed, but I have some doubts about that :(






cardinals







share|cite|improve this question









$endgroup$












  • $begingroup$
    You have infinitely many (uncountable many) $mathfrak c$ factors, whereas with your argument you can only remove finitely many.
    $endgroup$
    – Dog_69
    Jan 20 at 16:09












  • $begingroup$
    so should I write that as $mathfrak{c}^{mathfrak{c}} $?
    $endgroup$
    – VirtualUser
    Jan 20 at 16:29














1












1








1





$begingroup$


I have problem with understanding power of $left{ f | f: mathbb R rightarrow mathbb R right}$

Generally, from lecture I know that power of set
$$left{ f | f: A rightarrow B right}$$
is $ |B|^{|A|} $ - ok, so it seems that power of
$$left{ f | f: mathbb R rightarrow mathbb R right}$$
should be $ |mathbb R|^{|mathbb R|} = mathfrak{c}^{mathfrak{c}} = 2^{mathfrak{c}} $

Ok, but from the other hand, I can interpret that function as infinity set of pairs $ (x,y) $

So for each $ x $ I choose $y$. I can do this on $mathfrak{c}$ ways. After that I repeat this process $mathfrak{c} $ times so I have
$$mathfrak{c} cdot mathfrak{c} cdot mathfrak{c} cdot ... $$
But from lecture I know that $$mathfrak{c} cdot mathfrak{c} = mathfrak{c} $$
So I reduce it to $mathfrak{c}$.

I know that somewhere I have failed, but I have some doubts about that :(






cardinals







share|cite|improve this question









$endgroup$




I have problem with understanding power of $left{ f | f: mathbb R rightarrow mathbb R right}$

Generally, from lecture I know that power of set
$$left{ f | f: A rightarrow B right}$$
is $ |B|^{|A|} $ - ok, so it seems that power of
$$left{ f | f: mathbb R rightarrow mathbb R right}$$
should be $ |mathbb R|^{|mathbb R|} = mathfrak{c}^{mathfrak{c}} = 2^{mathfrak{c}} $

Ok, but from the other hand, I can interpret that function as infinity set of pairs $ (x,y) $

So for each $ x $ I choose $y$. I can do this on $mathfrak{c}$ ways. After that I repeat this process $mathfrak{c} $ times so I have
$$mathfrak{c} cdot mathfrak{c} cdot mathfrak{c} cdot ... $$
But from lecture I know that $$mathfrak{c} cdot mathfrak{c} = mathfrak{c} $$
So I reduce it to $mathfrak{c}$.

I know that somewhere I have failed, but I have some doubts about that :(






cardinals




cardinals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 20 at 15:56









VirtualUserVirtualUser

1,003115




1,003115












  • $begingroup$
    You have infinitely many (uncountable many) $mathfrak c$ factors, whereas with your argument you can only remove finitely many.
    $endgroup$
    – Dog_69
    Jan 20 at 16:09












  • $begingroup$
    so should I write that as $mathfrak{c}^{mathfrak{c}} $?
    $endgroup$
    – VirtualUser
    Jan 20 at 16:29


















  • $begingroup$
    You have infinitely many (uncountable many) $mathfrak c$ factors, whereas with your argument you can only remove finitely many.
    $endgroup$
    – Dog_69
    Jan 20 at 16:09












  • $begingroup$
    so should I write that as $mathfrak{c}^{mathfrak{c}} $?
    $endgroup$
    – VirtualUser
    Jan 20 at 16:29
















$begingroup$
You have infinitely many (uncountable many) $mathfrak c$ factors, whereas with your argument you can only remove finitely many.
$endgroup$
– Dog_69
Jan 20 at 16:09






$begingroup$
You have infinitely many (uncountable many) $mathfrak c$ factors, whereas with your argument you can only remove finitely many.
$endgroup$
– Dog_69
Jan 20 at 16:09














$begingroup$
so should I write that as $mathfrak{c}^{mathfrak{c}} $?
$endgroup$
– VirtualUser
Jan 20 at 16:29




$begingroup$
so should I write that as $mathfrak{c}^{mathfrak{c}} $?
$endgroup$
– VirtualUser
Jan 20 at 16:29










1 Answer
1






active

oldest

votes


















2












$begingroup$

Multiply a finite number of $mathfrak c$'s together and you get $mathfrak c^n =mathfrak c.$ Even for a countably infinite number of factors, $mathfrak c^{aleph_0}=mathfrak c.$ And yet, when we multiply together $mathfrak c$ factors of $mathfrak c$, we get $mathfrak c^{mathfrak c} > mathfrak c.$



There is nothing contradictory about this. You simply need to multiply together a lot of factors of $mathfrak c$ to get something larger than $mathfrak c.$ $aleph_0$-many doesn't suffice.



All you can conclude from iterating the binary idempotence $mathfrak c^2 = mathfrak c$ is the finite case $mathfrak c^n=mathfrak c$ for arbitrary $n.$ You cannot extend this to infinite powers. Since $mathfrak c-n=mathfrak c,$ pulling out a finite number of factors does nothing: $$mathfrak c^{mathfrak c} = mathfrak c^n cdot mathfrak c^{mathfrak c} = mathfrak c cdot mathfrak c^{mathfrak c} = mathfrak c^{mathfrak c}$$ We're just going in circles.



(Note, again, $mathfrak c ^{aleph_0}=mathfrak c$ happens to be true but requires a different argument. For instance, similarly, $(aleph_0)^n= aleph_0$ for any finite $n$, but $(aleph_0)^{aleph_0} =mathfrak c > aleph_0$.)






share|cite|improve this answer











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  • $begingroup$
    Ok,thanks, now I understand my fail
    $endgroup$
    – VirtualUser
    Jan 20 at 17:01











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1 Answer
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1 Answer
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active

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active

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$begingroup$

Multiply a finite number of $mathfrak c$'s together and you get $mathfrak c^n =mathfrak c.$ Even for a countably infinite number of factors, $mathfrak c^{aleph_0}=mathfrak c.$ And yet, when we multiply together $mathfrak c$ factors of $mathfrak c$, we get $mathfrak c^{mathfrak c} > mathfrak c.$



There is nothing contradictory about this. You simply need to multiply together a lot of factors of $mathfrak c$ to get something larger than $mathfrak c.$ $aleph_0$-many doesn't suffice.



All you can conclude from iterating the binary idempotence $mathfrak c^2 = mathfrak c$ is the finite case $mathfrak c^n=mathfrak c$ for arbitrary $n.$ You cannot extend this to infinite powers. Since $mathfrak c-n=mathfrak c,$ pulling out a finite number of factors does nothing: $$mathfrak c^{mathfrak c} = mathfrak c^n cdot mathfrak c^{mathfrak c} = mathfrak c cdot mathfrak c^{mathfrak c} = mathfrak c^{mathfrak c}$$ We're just going in circles.



(Note, again, $mathfrak c ^{aleph_0}=mathfrak c$ happens to be true but requires a different argument. For instance, similarly, $(aleph_0)^n= aleph_0$ for any finite $n$, but $(aleph_0)^{aleph_0} =mathfrak c > aleph_0$.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok,thanks, now I understand my fail
    $endgroup$
    – VirtualUser
    Jan 20 at 17:01
















2












$begingroup$

Multiply a finite number of $mathfrak c$'s together and you get $mathfrak c^n =mathfrak c.$ Even for a countably infinite number of factors, $mathfrak c^{aleph_0}=mathfrak c.$ And yet, when we multiply together $mathfrak c$ factors of $mathfrak c$, we get $mathfrak c^{mathfrak c} > mathfrak c.$



There is nothing contradictory about this. You simply need to multiply together a lot of factors of $mathfrak c$ to get something larger than $mathfrak c.$ $aleph_0$-many doesn't suffice.



All you can conclude from iterating the binary idempotence $mathfrak c^2 = mathfrak c$ is the finite case $mathfrak c^n=mathfrak c$ for arbitrary $n.$ You cannot extend this to infinite powers. Since $mathfrak c-n=mathfrak c,$ pulling out a finite number of factors does nothing: $$mathfrak c^{mathfrak c} = mathfrak c^n cdot mathfrak c^{mathfrak c} = mathfrak c cdot mathfrak c^{mathfrak c} = mathfrak c^{mathfrak c}$$ We're just going in circles.



(Note, again, $mathfrak c ^{aleph_0}=mathfrak c$ happens to be true but requires a different argument. For instance, similarly, $(aleph_0)^n= aleph_0$ for any finite $n$, but $(aleph_0)^{aleph_0} =mathfrak c > aleph_0$.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok,thanks, now I understand my fail
    $endgroup$
    – VirtualUser
    Jan 20 at 17:01














2












2








2





$begingroup$

Multiply a finite number of $mathfrak c$'s together and you get $mathfrak c^n =mathfrak c.$ Even for a countably infinite number of factors, $mathfrak c^{aleph_0}=mathfrak c.$ And yet, when we multiply together $mathfrak c$ factors of $mathfrak c$, we get $mathfrak c^{mathfrak c} > mathfrak c.$



There is nothing contradictory about this. You simply need to multiply together a lot of factors of $mathfrak c$ to get something larger than $mathfrak c.$ $aleph_0$-many doesn't suffice.



All you can conclude from iterating the binary idempotence $mathfrak c^2 = mathfrak c$ is the finite case $mathfrak c^n=mathfrak c$ for arbitrary $n.$ You cannot extend this to infinite powers. Since $mathfrak c-n=mathfrak c,$ pulling out a finite number of factors does nothing: $$mathfrak c^{mathfrak c} = mathfrak c^n cdot mathfrak c^{mathfrak c} = mathfrak c cdot mathfrak c^{mathfrak c} = mathfrak c^{mathfrak c}$$ We're just going in circles.



(Note, again, $mathfrak c ^{aleph_0}=mathfrak c$ happens to be true but requires a different argument. For instance, similarly, $(aleph_0)^n= aleph_0$ for any finite $n$, but $(aleph_0)^{aleph_0} =mathfrak c > aleph_0$.)






share|cite|improve this answer











$endgroup$



Multiply a finite number of $mathfrak c$'s together and you get $mathfrak c^n =mathfrak c.$ Even for a countably infinite number of factors, $mathfrak c^{aleph_0}=mathfrak c.$ And yet, when we multiply together $mathfrak c$ factors of $mathfrak c$, we get $mathfrak c^{mathfrak c} > mathfrak c.$



There is nothing contradictory about this. You simply need to multiply together a lot of factors of $mathfrak c$ to get something larger than $mathfrak c.$ $aleph_0$-many doesn't suffice.



All you can conclude from iterating the binary idempotence $mathfrak c^2 = mathfrak c$ is the finite case $mathfrak c^n=mathfrak c$ for arbitrary $n.$ You cannot extend this to infinite powers. Since $mathfrak c-n=mathfrak c,$ pulling out a finite number of factors does nothing: $$mathfrak c^{mathfrak c} = mathfrak c^n cdot mathfrak c^{mathfrak c} = mathfrak c cdot mathfrak c^{mathfrak c} = mathfrak c^{mathfrak c}$$ We're just going in circles.



(Note, again, $mathfrak c ^{aleph_0}=mathfrak c$ happens to be true but requires a different argument. For instance, similarly, $(aleph_0)^n= aleph_0$ for any finite $n$, but $(aleph_0)^{aleph_0} =mathfrak c > aleph_0$.)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 29 at 1:31

























answered Jan 20 at 16:40









spaceisdarkgreenspaceisdarkgreen

33.4k21753




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  • $begingroup$
    Ok,thanks, now I understand my fail
    $endgroup$
    – VirtualUser
    Jan 20 at 17:01


















  • $begingroup$
    Ok,thanks, now I understand my fail
    $endgroup$
    – VirtualUser
    Jan 20 at 17:01
















$begingroup$
Ok,thanks, now I understand my fail
$endgroup$
– VirtualUser
Jan 20 at 17:01




$begingroup$
Ok,thanks, now I understand my fail
$endgroup$
– VirtualUser
Jan 20 at 17:01


















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