A question regarding Brian M. Scott's proof that $text{cf}(aleph_{omega_1})=omega_1$












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







share|cite|improve this question









$endgroup$








  • 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
















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







share|cite|improve this question









$endgroup$








  • 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














1












1








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







share|cite|improve this question









$endgroup$





$text{cf}(aleph_{omega_1})=omega_1$




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






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share|cite|improve this question











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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














  • 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










1 Answer
1






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oldest

votes


















2












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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






share|cite|improve this answer









$endgroup$



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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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