A question regarding Brian M. Scott's proof that $text{cf}(aleph_{omega_1})=omega_1$
$begingroup$
$text{cf}(aleph_{omega_1})=omega_1$
From here, I quote Brian M. Scott's proof:
Suppose that $langlealpha_n:ninomegarangle$ is an increasing sequence cofinal in $omega_{omega_1}$. For each $xi<omega_1$ there is a least $n(xi)inomega$ such that $omega_xilealpha_{n(xi)}$. Clearly there is then an $minomega$ such $X={xi<omega_1:n(xi)=m}$ is uncountable, and it follows easily that $color{blue}{omega_{omega_1}=sup_{xiin X}omega_xi}lealpha_m<omega_{omega_1}$, which is absurd.
I would like to ask what is the reasoning behind $color{blue}{omega_{omega_1}=sup_{xiin X}omega_xi}$.
I think we can prove $X$ is cofinal in $omega_1$ from the fact that $Xsubseteq omega_1$ and $|X|=omega_1$. I have tried but unsuccessfully.
Thank you for your help!
elementary-set-theory proof-explanation cardinals ordinals
asked Dec 27 '18 at 16:31
Le Anh DungLe Anh Dung
1,4141621
$endgroup$
add a comment |
$begingroup$
$text{cf}(aleph_{omega_1})=omega_1$
From here, I quote Brian M. Scott's proof:
Suppose that $langlealpha_n:ninomegarangle$ is an increasing sequence cofinal in $omega_{omega_1}$. For each $xi<omega_1$ there is a least $n(xi)inomega$ such that $omega_xilealpha_{n(xi)}$. Clearly there is then an $minomega$ such $X={xi<omega_1:n(xi)=m}$ is uncountable, and it follows easily that $color{blue}{omega_{omega_1}=sup_{xiin X}omega_xi}lealpha_m<omega_{omega_1}$, which is absurd.
I would like to ask what is the reasoning behind $color{blue}{omega_{omega_1}=sup_{xiin X}omega_xi}$.
I think we can prove $X$ is cofinal in $omega_1$ from the fact that $Xsubseteq omega_1$ and $|X|=omega_1$. I have tried but unsuccessfully.
Thank you for your help!
elementary-set-theory proof-explanation cardinals ordinals
asked Dec 27 '18 at 16:31
Le Anh DungLe Anh Dung
1,4141621
$endgroup$
2
$begingroup$
Since a countable union of countable sets is countable, a countable partition of an uncountable set has one part (at least) uncountable. Yes choice is used, and is necessary.
$endgroup$
– Asaf Karagila♦
Dec 27 '18 at 16:41
$begingroup$
Hi @AsafKaragila, did you mean that to prove the existence of such set $X$, we must use the theorem a countable union of countable sets is countable, which in turn appeals to Axiom of Countable Choice?
$endgroup$
– Le Anh Dung
Dec 28 '18 at 0:30
1
$begingroup$
Yes, exactly. Countable choice is needed.
$endgroup$
– Asaf Karagila♦
Dec 28 '18 at 0:33
$begingroup$
Thank you so much @AsafKaragila, I got it.
$endgroup$
– Le Anh Dung
Dec 28 '18 at 0:35
add a comment |
$begingroup$
$text{cf}(aleph_{omega_1})=omega_1$
From here, I quote Brian M. Scott's proof:
Suppose that $langlealpha_n:ninomegarangle$ is an increasing sequence cofinal in $omega_{omega_1}$. For each $xi<omega_1$ there is a least $n(xi)inomega$ such that $omega_xilealpha_{n(xi)}$. Clearly there is then an $minomega$ such $X={xi<omega_1:n(xi)=m}$ is uncountable, and it follows easily that $color{blue}{omega_{omega_1}=sup_{xiin X}omega_xi}lealpha_m<omega_{omega_1}$, which is absurd.
I would like to ask what is the reasoning behind $color{blue}{omega_{omega_1}=sup_{xiin X}omega_xi}$.
I think we can prove $X$ is cofinal in $omega_1$ from the fact that $Xsubseteq omega_1$ and $|X|=omega_1$. I have tried but unsuccessfully.
Thank you for your help!
elementary-set-theory proof-explanation cardinals ordinals
asked Dec 27 '18 at 16:31
Le Anh DungLe Anh Dung
1,4141621
$endgroup$
$text{cf}(aleph_{omega_1})=omega_1$
From here, I quote Brian M. Scott's proof:
Suppose that $langlealpha_n:ninomegarangle$ is an increasing sequence cofinal in $omega_{omega_1}$. For each $xi<omega_1$ there is a least $n(xi)inomega$ such that $omega_xilealpha_{n(xi)}$. Clearly there is then an $minomega$ such $X={xi<omega_1:n(xi)=m}$ is uncountable, and it follows easily that $color{blue}{omega_{omega_1}=sup_{xiin X}omega_xi}lealpha_m<omega_{omega_1}$, which is absurd.
I would like to ask what is the reasoning behind $color{blue}{omega_{omega_1}=sup_{xiin X}omega_xi}$.
I think we can prove $X$ is cofinal in $omega_1$ from the fact that $Xsubseteq omega_1$ and $|X|=omega_1$. I have tried but unsuccessfully.
Thank you for your help!
elementary-set-theory proof-explanation cardinals ordinals
elementary-set-theory proof-explanation cardinals ordinals
asked Dec 27 '18 at 16:31
Le Anh DungLe Anh Dung
1,4141621
asked Dec 27 '18 at 16:31
Le Anh DungLe Anh Dung
1,4141621
asked Dec 27 '18 at 16:31
Le Anh DungLe Anh Dung
1,4141621
asked Dec 27 '18 at 16:31
Le Anh DungLe Anh Dung
1,4141621
asked Dec 27 '18 at 16:31
Le Anh DungLe Anh Dung
1,4141621
1,4141621
2
$begingroup$
Since a countable union of countable sets is countable, a countable partition of an uncountable set has one part (at least) uncountable. Yes choice is used, and is necessary.
$endgroup$
– Asaf Karagila♦
Dec 27 '18 at 16:41
$begingroup$
Hi @AsafKaragila, did you mean that to prove the existence of such set $X$, we must use the theorem a countable union of countable sets is countable, which in turn appeals to Axiom of Countable Choice?
$endgroup$
– Le Anh Dung
Dec 28 '18 at 0:30
1
$begingroup$
Yes, exactly. Countable choice is needed.
$endgroup$
– Asaf Karagila♦
Dec 28 '18 at 0:33
$begingroup$
Thank you so much @AsafKaragila, I got it.
$endgroup$
– Le Anh Dung
Dec 28 '18 at 0:35
add a comment |
2
$begingroup$
Since a countable union of countable sets is countable, a countable partition of an uncountable set has one part (at least) uncountable. Yes choice is used, and is necessary.
$endgroup$
– Asaf Karagila♦
Dec 27 '18 at 16:41
$begingroup$
Hi @AsafKaragila, did you mean that to prove the existence of such set $X$, we must use the theorem a countable union of countable sets is countable, which in turn appeals to Axiom of Countable Choice?
$endgroup$
– Le Anh Dung
Dec 28 '18 at 0:30
1
$begingroup$
Yes, exactly. Countable choice is needed.
$endgroup$
– Asaf Karagila♦
Dec 28 '18 at 0:33
$begingroup$
Thank you so much @AsafKaragila, I got it.
$endgroup$
– Le Anh Dung
Dec 28 '18 at 0:35
2
2
$begingroup$
Since a countable union of countable sets is countable, a countable partition of an uncountable set has one part (at least) uncountable. Yes choice is used, and is necessary.
$endgroup$
– Asaf Karagila♦
Dec 27 '18 at 16:41
$begingroup$
Since a countable union of countable sets is countable, a countable partition of an uncountable set has one part (at least) uncountable. Yes choice is used, and is necessary.
$endgroup$
– Asaf Karagila♦
Dec 27 '18 at 16:41
$begingroup$
Hi @AsafKaragila, did you mean that to prove the existence of such set $X$, we must use the theorem a countable union of countable sets is countable, which in turn appeals to Axiom of Countable Choice?
$endgroup$
– Le Anh Dung
Dec 28 '18 at 0:30
$begingroup$
Hi @AsafKaragila, did you mean that to prove the existence of such set $X$, we must use the theorem a countable union of countable sets is countable, which in turn appeals to Axiom of Countable Choice?
$endgroup$
– Le Anh Dung
Dec 28 '18 at 0:30
1
1
$begingroup$
Yes, exactly. Countable choice is needed.
$endgroup$
– Asaf Karagila♦
Dec 28 '18 at 0:33
$begingroup$
Yes, exactly. Countable choice is needed.
$endgroup$
– Asaf Karagila♦
Dec 28 '18 at 0:33
$begingroup$
Thank you so much @AsafKaragila, I got it.
$endgroup$
– Le Anh Dung
Dec 28 '18 at 0:35
$begingroup$
Thank you so much @AsafKaragila, I got it.
$endgroup$
– Le Anh Dung
Dec 28 '18 at 0:35
add a comment |
1 Answer
1
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$begingroup$
Every initial segment of $omega_1$ is countable, so $X$ is unbounded in $omega_1$
Like Asaf commented, the fact that there exists such $X$ require choice, without some kind of choice it is not necessarily correct
answered Dec 27 '18 at 16:43
HoloHolo
6,08421131
$endgroup$
$begingroup$
Thank you so much! Everything is now clear.
$endgroup$
– Le Anh Dung
Dec 27 '18 at 16:48
$begingroup$
It is unfortunate that I forget an useful property that an initial segment of an ordinal is an ordinal.
$endgroup$
– Le Anh Dung
Dec 27 '18 at 16:50
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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$begingroup$
Every initial segment of $omega_1$ is countable, so $X$ is unbounded in $omega_1$
Like Asaf commented, the fact that there exists such $X$ require choice, without some kind of choice it is not necessarily correct
answered Dec 27 '18 at 16:43
HoloHolo
6,08421131
$endgroup$
$begingroup$
Thank you so much! Everything is now clear.
$endgroup$
– Le Anh Dung
Dec 27 '18 at 16:48
$begingroup$
It is unfortunate that I forget an useful property that an initial segment of an ordinal is an ordinal.
$endgroup$
– Le Anh Dung
Dec 27 '18 at 16:50
add a comment |
$begingroup$
Every initial segment of $omega_1$ is countable, so $X$ is unbounded in $omega_1$
Like Asaf commented, the fact that there exists such $X$ require choice, without some kind of choice it is not necessarily correct
answered Dec 27 '18 at 16:43
HoloHolo
6,08421131
$endgroup$
$begingroup$
Thank you so much! Everything is now clear.
$endgroup$
– Le Anh Dung
Dec 27 '18 at 16:48
$begingroup$
It is unfortunate that I forget an useful property that an initial segment of an ordinal is an ordinal.
$endgroup$
– Le Anh Dung
Dec 27 '18 at 16:50
add a comment |
$begingroup$
Every initial segment of $omega_1$ is countable, so $X$ is unbounded in $omega_1$
Like Asaf commented, the fact that there exists such $X$ require choice, without some kind of choice it is not necessarily correct
answered Dec 27 '18 at 16:43
HoloHolo
6,08421131
$endgroup$
Every initial segment of $omega_1$ is countable, so $X$ is unbounded in $omega_1$
Like Asaf commented, the fact that there exists such $X$ require choice, without some kind of choice it is not necessarily correct
answered Dec 27 '18 at 16:43
HoloHolo
6,08421131
answered Dec 27 '18 at 16:43
HoloHolo
6,08421131
answered Dec 27 '18 at 16:43
HoloHolo
6,08421131
answered Dec 27 '18 at 16:43
HoloHolo
6,08421131
6,08421131
$begingroup$
Thank you so much! Everything is now clear.
$endgroup$
– Le Anh Dung
Dec 27 '18 at 16:48
$begingroup$
It is unfortunate that I forget an useful property that an initial segment of an ordinal is an ordinal.
$endgroup$
– Le Anh Dung
Dec 27 '18 at 16:50
add a comment |
$begingroup$
Thank you so much! Everything is now clear.
$endgroup$
– Le Anh Dung
Dec 27 '18 at 16:48
$begingroup$
It is unfortunate that I forget an useful property that an initial segment of an ordinal is an ordinal.
$endgroup$
– Le Anh Dung
Dec 27 '18 at 16:50
$begingroup$
Thank you so much! Everything is now clear.
$endgroup$
– Le Anh Dung
Dec 27 '18 at 16:48
$begingroup$
Thank you so much! Everything is now clear.
$endgroup$
– Le Anh Dung
Dec 27 '18 at 16:48
$begingroup$
It is unfortunate that I forget an useful property that an initial segment of an ordinal is an ordinal.
$endgroup$
– Le Anh Dung
Dec 27 '18 at 16:50
$begingroup$
It is unfortunate that I forget an useful property that an initial segment of an ordinal is an ordinal.
$endgroup$
– Le Anh Dung
Dec 27 '18 at 16:50
add a comment |
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$begingroup$
Since a countable union of countable sets is countable, a countable partition of an uncountable set has one part (at least) uncountable. Yes choice is used, and is necessary.
$endgroup$
– Asaf Karagila♦
Dec 27 '18 at 16:41
$begingroup$
Hi @AsafKaragila, did you mean that to prove the existence of such set $X$, we must use the theorem a countable union of countable sets is countable, which in turn appeals to Axiom of Countable Choice?
$endgroup$
– Le Anh Dung
Dec 28 '18 at 0:30
1
$begingroup$
Yes, exactly. Countable choice is needed.
$endgroup$
– Asaf Karagila♦
Dec 28 '18 at 0:33
$begingroup$
Thank you so much @AsafKaragila, I got it.
$endgroup$
– Le Anh Dung
Dec 28 '18 at 0:35