Mixing
Cardinal numbers: if $a>2^omega$, then $a^omega=a$?
Clash Royale CLAN TAG#URR8PPP
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Here is my proof, given that $|A|=a$ and $omega=|mathbbN|$(also commonly denoted as $aleph_0$):
Let $F$ be the set of all bijections of the form from $B^omega$ to $B$ where $Bsubset A$ ($F$ is not empty, since $(2^omega)^omega=2^omega$);
Order $F$ using extension of function, apply Zorn's Lemma for a maximal function $g:B^omegato B$;
To show that $|B|=|A|$, notice that otherwise $Asetminus B$ contains a set $C$ with the same cardinality as $B$, and so $g$ can be extended to $(Bcup C)^omegato (Bcup C)$, which violates the maximality of $B$;
Since $|B|=|A|=a$ and $|B^omega|=|B|$, we have $a^omega=a$.
This looks very fishy, as I failed to discover a theorem that directly implies this. Is the proof above correct?
cardinals
asked Jul 28 at 13:08
Ginger88895
898
add a comment |Â
up vote
0
down vote
favorite
Here is my proof, given that $|A|=a$ and $omega=|mathbbN|$(also commonly denoted as $aleph_0$):
Let $F$ be the set of all bijections of the form from $B^omega$ to $B$ where $Bsubset A$ ($F$ is not empty, since $(2^omega)^omega=2^omega$);
Order $F$ using extension of function, apply Zorn's Lemma for a maximal function $g:B^omegato B$;
To show that $|B|=|A|$, notice that otherwise $Asetminus B$ contains a set $C$ with the same cardinality as $B$, and so $g$ can be extended to $(Bcup C)^omegato (Bcup C)$, which violates the maximality of $B$;
Since $|B|=|A|=a$ and $|B^omega|=|B|$, we have $a^omega=a$.
This looks very fishy, as I failed to discover a theorem that directly implies this. Is the proof above correct?
cardinals
asked Jul 28 at 13:08
Ginger88895
898
3
How do you know that Zorn's lemma applies? The union of a chain of bijections $f$ for which there is a $Bsubseteq A$ such that $f!:B^aleph_0to B$ does not need to have this form.
– Andrés E. Caicedo
Jul 28 at 13:55
Ah I see... taking $bigcup B_i$ does not work since $B^aleph_0$ is an infinite sequence. Thanks for the help! :D
– Ginger88895
Jul 28 at 14:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Here is my proof, given that $|A|=a$ and $omega=|mathbbN|$(also commonly denoted as $aleph_0$):
Let $F$ be the set of all bijections of the form from $B^omega$ to $B$ where $Bsubset A$ ($F$ is not empty, since $(2^omega)^omega=2^omega$);
Order $F$ using extension of function, apply Zorn's Lemma for a maximal function $g:B^omegato B$;
To show that $|B|=|A|$, notice that otherwise $Asetminus B$ contains a set $C$ with the same cardinality as $B$, and so $g$ can be extended to $(Bcup C)^omegato (Bcup C)$, which violates the maximality of $B$;
Since $|B|=|A|=a$ and $|B^omega|=|B|$, we have $a^omega=a$.
This looks very fishy, as I failed to discover a theorem that directly implies this. Is the proof above correct?
cardinals
asked Jul 28 at 13:08
Ginger88895
898
Here is my proof, given that $|A|=a$ and $omega=|mathbbN|$(also commonly denoted as $aleph_0$):
Let $F$ be the set of all bijections of the form from $B^omega$ to $B$ where $Bsubset A$ ($F$ is not empty, since $(2^omega)^omega=2^omega$);
Order $F$ using extension of function, apply Zorn's Lemma for a maximal function $g:B^omegato B$;
To show that $|B|=|A|$, notice that otherwise $Asetminus B$ contains a set $C$ with the same cardinality as $B$, and so $g$ can be extended to $(Bcup C)^omegato (Bcup C)$, which violates the maximality of $B$;
Since $|B|=|A|=a$ and $|B^omega|=|B|$, we have $a^omega=a$.
This looks very fishy, as I failed to discover a theorem that directly implies this. Is the proof above correct?
cardinals
asked Jul 28 at 13:08
Ginger88895
898
edited Jul 28 at 13:18
asked Jul 28 at 13:08
Ginger88895
898
asked Jul 28 at 13:08
Ginger88895
898
asked Jul 28 at 13:08
Ginger88895
898
898
3
How do you know that Zorn's lemma applies? The union of a chain of bijections $f$ for which there is a $Bsubseteq A$ such that $f!:B^aleph_0to B$ does not need to have this form.
– Andrés E. Caicedo
Jul 28 at 13:55
Ah I see... taking $bigcup B_i$ does not work since $B^aleph_0$ is an infinite sequence. Thanks for the help! :D
– Ginger88895
Jul 28 at 14:12
add a comment |Â
3
How do you know that Zorn's lemma applies? The union of a chain of bijections $f$ for which there is a $Bsubseteq A$ such that $f!:B^aleph_0to B$ does not need to have this form.
– Andrés E. Caicedo
Jul 28 at 13:55
Ah I see... taking $bigcup B_i$ does not work since $B^aleph_0$ is an infinite sequence. Thanks for the help! :D
– Ginger88895
Jul 28 at 14:12
3
3
How do you know that Zorn's lemma applies? The union of a chain of bijections $f$ for which there is a $Bsubseteq A$ such that $f!:B^aleph_0to B$ does not need to have this form.
– Andrés E. Caicedo
Jul 28 at 13:55
How do you know that Zorn's lemma applies? The union of a chain of bijections $f$ for which there is a $Bsubseteq A$ such that $f!:B^aleph_0to B$ does not need to have this form.
– Andrés E. Caicedo
Jul 28 at 13:55
Ah I see... taking $bigcup B_i$ does not work since $B^aleph_0$ is an infinite sequence. Thanks for the help! :D
– Ginger88895
Jul 28 at 14:12
Ah I see... taking $bigcup B_i$ does not work since $B^aleph_0$ is an infinite sequence. Thanks for the help! :D
– Ginger88895
Jul 28 at 14:12
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The claim is false in the first place. A consequence of Konig's theorem is that for all infinite cardinals $kappa$, it holds $$kappa<kappa^operatornamecofkappa$$ and there is plenty of cardinals $kappa$ such that $kappa>2^aleph_0$ and $operatornamecofkappa=aleph_0$. Notably, $beth_omega$ (see here for the definition of beth-sequence).
answered Jul 28 at 13:28
Saucy O'Path
2,394217
Thanks... after some further research I am aware that Konig's theorem disproves this. However, I am very curious about where my proof fails :)
– Ginger88895
Jul 28 at 13:29
@Ginger88895 It is quite unclear how you can extend $g$ in point (3). You seem to be under the false impression that $(Bcup C)^omega=B^omegacup C^omega$, while actually the LHS is much larger: there are all the functions that send some natural numbers to elements of $B$ and the other ones to elements of $C$.
– Saucy O'Path
Jul 28 at 13:37
Actually I picked $C$ so that $|C|=|B|$... then since $B$ is infinite, $|Bcup C|=|B|$, and also $|(Bcup C)^omegasetminus B^omega|=|C|$, which indicates the existence of a bijection from $(Bcup C)^omegasetminus B^omega$ to $C$...
– Ginger88895
Jul 28 at 13:40
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The claim is false in the first place. A consequence of Konig's theorem is that for all infinite cardinals $kappa$, it holds $$kappa<kappa^operatornamecofkappa$$ and there is plenty of cardinals $kappa$ such that $kappa>2^aleph_0$ and $operatornamecofkappa=aleph_0$. Notably, $beth_omega$ (see here for the definition of beth-sequence).
answered Jul 28 at 13:28
Saucy O'Path
2,394217
Thanks... after some further research I am aware that Konig's theorem disproves this. However, I am very curious about where my proof fails :)
– Ginger88895
Jul 28 at 13:29
@Ginger88895 It is quite unclear how you can extend $g$ in point (3). You seem to be under the false impression that $(Bcup C)^omega=B^omegacup C^omega$, while actually the LHS is much larger: there are all the functions that send some natural numbers to elements of $B$ and the other ones to elements of $C$.
– Saucy O'Path
Jul 28 at 13:37
Actually I picked $C$ so that $|C|=|B|$... then since $B$ is infinite, $|Bcup C|=|B|$, and also $|(Bcup C)^omegasetminus B^omega|=|C|$, which indicates the existence of a bijection from $(Bcup C)^omegasetminus B^omega$ to $C$...
– Ginger88895
Jul 28 at 13:40
add a comment |Â
up vote
1
down vote
accepted
The claim is false in the first place. A consequence of Konig's theorem is that for all infinite cardinals $kappa$, it holds $$kappa<kappa^operatornamecofkappa$$ and there is plenty of cardinals $kappa$ such that $kappa>2^aleph_0$ and $operatornamecofkappa=aleph_0$. Notably, $beth_omega$ (see here for the definition of beth-sequence).
answered Jul 28 at 13:28
Saucy O'Path
2,394217
Thanks... after some further research I am aware that Konig's theorem disproves this. However, I am very curious about where my proof fails :)
– Ginger88895
Jul 28 at 13:29
@Ginger88895 It is quite unclear how you can extend $g$ in point (3). You seem to be under the false impression that $(Bcup C)^omega=B^omegacup C^omega$, while actually the LHS is much larger: there are all the functions that send some natural numbers to elements of $B$ and the other ones to elements of $C$.
– Saucy O'Path
Jul 28 at 13:37
Actually I picked $C$ so that $|C|=|B|$... then since $B$ is infinite, $|Bcup C|=|B|$, and also $|(Bcup C)^omegasetminus B^omega|=|C|$, which indicates the existence of a bijection from $(Bcup C)^omegasetminus B^omega$ to $C$...
– Ginger88895
Jul 28 at 13:40
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The claim is false in the first place. A consequence of Konig's theorem is that for all infinite cardinals $kappa$, it holds $$kappa<kappa^operatornamecofkappa$$ and there is plenty of cardinals $kappa$ such that $kappa>2^aleph_0$ and $operatornamecofkappa=aleph_0$. Notably, $beth_omega$ (see here for the definition of beth-sequence).
answered Jul 28 at 13:28
Saucy O'Path
2,394217
The claim is false in the first place. A consequence of Konig's theorem is that for all infinite cardinals $kappa$, it holds $$kappa<kappa^operatornamecofkappa$$ and there is plenty of cardinals $kappa$ such that $kappa>2^aleph_0$ and $operatornamecofkappa=aleph_0$. Notably, $beth_omega$ (see here for the definition of beth-sequence).
answered Jul 28 at 13:28
Saucy O'Path
2,394217
edited Jul 28 at 14:12
answered Jul 28 at 13:28
Saucy O'Path
2,394217
answered Jul 28 at 13:28
Saucy O'Path
2,394217
answered Jul 28 at 13:28
Saucy O'Path
2,394217
2,394217
Thanks... after some further research I am aware that Konig's theorem disproves this. However, I am very curious about where my proof fails :)
– Ginger88895
Jul 28 at 13:29
@Ginger88895 It is quite unclear how you can extend $g$ in point (3). You seem to be under the false impression that $(Bcup C)^omega=B^omegacup C^omega$, while actually the LHS is much larger: there are all the functions that send some natural numbers to elements of $B$ and the other ones to elements of $C$.
– Saucy O'Path
Jul 28 at 13:37
Actually I picked $C$ so that $|C|=|B|$... then since $B$ is infinite, $|Bcup C|=|B|$, and also $|(Bcup C)^omegasetminus B^omega|=|C|$, which indicates the existence of a bijection from $(Bcup C)^omegasetminus B^omega$ to $C$...
– Ginger88895
Jul 28 at 13:40
add a comment |Â
Thanks... after some further research I am aware that Konig's theorem disproves this. However, I am very curious about where my proof fails :)
– Ginger88895
Jul 28 at 13:29
@Ginger88895 It is quite unclear how you can extend $g$ in point (3). You seem to be under the false impression that $(Bcup C)^omega=B^omegacup C^omega$, while actually the LHS is much larger: there are all the functions that send some natural numbers to elements of $B$ and the other ones to elements of $C$.
– Saucy O'Path
Jul 28 at 13:37
Actually I picked $C$ so that $|C|=|B|$... then since $B$ is infinite, $|Bcup C|=|B|$, and also $|(Bcup C)^omegasetminus B^omega|=|C|$, which indicates the existence of a bijection from $(Bcup C)^omegasetminus B^omega$ to $C$...
– Ginger88895
Jul 28 at 13:40
Thanks... after some further research I am aware that Konig's theorem disproves this. However, I am very curious about where my proof fails :)
– Ginger88895
Jul 28 at 13:29
Thanks... after some further research I am aware that Konig's theorem disproves this. However, I am very curious about where my proof fails :)
– Ginger88895
Jul 28 at 13:29
@Ginger88895 It is quite unclear how you can extend $g$ in point (3). You seem to be under the false impression that $(Bcup C)^omega=B^omegacup C^omega$, while actually the LHS is much larger: there are all the functions that send some natural numbers to elements of $B$ and the other ones to elements of $C$.
– Saucy O'Path
Jul 28 at 13:37
@Ginger88895 It is quite unclear how you can extend $g$ in point (3). You seem to be under the false impression that $(Bcup C)^omega=B^omegacup C^omega$, while actually the LHS is much larger: there are all the functions that send some natural numbers to elements of $B$ and the other ones to elements of $C$.
– Saucy O'Path
Jul 28 at 13:37
Actually I picked $C$ so that $|C|=|B|$... then since $B$ is infinite, $|Bcup C|=|B|$, and also $|(Bcup C)^omegasetminus B^omega|=|C|$, which indicates the existence of a bijection from $(Bcup C)^omegasetminus B^omega$ to $C$...
– Ginger88895
Jul 28 at 13:40
Actually I picked $C$ so that $|C|=|B|$... then since $B$ is infinite, $|Bcup C|=|B|$, and also $|(Bcup C)^omegasetminus B^omega|=|C|$, which indicates the existence of a bijection from $(Bcup C)^omegasetminus B^omega$ to $C$...
– Ginger88895
Jul 28 at 13:40
add a comment |Â
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3
How do you know that Zorn's lemma applies? The union of a chain of bijections $f$ for which there is a $Bsubseteq A$ such that $f!:B^aleph_0to B$ does not need to have this form.
– Andrés E. Caicedo
Jul 28 at 13:55
Ah I see... taking $bigcup B_i$ does not work since $B^aleph_0$ is an infinite sequence. Thanks for the help! :D
– Ginger88895
Jul 28 at 14:12