If $alpha$ is a limit ordinal, then $operatorname{cf}(alpha)$ is a limit ordinal [duplicate]
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This question already has an answer here:
Why cofinality is a cardinal.
1 answer
In the textbook Introduction to Set Theory by Hrbacek and Jech, Section 9.2, the authors first introduce the definition of increasing sequence of ordinals:
Then they introduce cofinality:
My question: how do we prove that $operatorname{cf}(alpha)$ is a limit ordinal?
If I take a sequence $langle alpha_nu mid nu<1 rangle$ where $alpha_0=omega$. It is clear that the limit of this sequence is $omega$ and thus $operatorname{cf}(omega)=1$, which is a successor ordinal.
I don't know what's wrong with my reasoning.
elementary-set-theory cardinals ordinals
edited Dec 26 '18 at 16:13
Martin Sleziak
45k10122277
asked Dec 26 '18 at 15:44
Le Anh DungLe Anh Dung
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marked as duplicate by Holo, Namaste elementary-set-theory
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Dec 26 '18 at 18:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Why cofinality is a cardinal.
1 answer
In the textbook Introduction to Set Theory by Hrbacek and Jech, Section 9.2, the authors first introduce the definition of increasing sequence of ordinals:
Then they introduce cofinality:
My question: how do we prove that $operatorname{cf}(alpha)$ is a limit ordinal?
If I take a sequence $langle alpha_nu mid nu<1 rangle$ where $alpha_0=omega$. It is clear that the limit of this sequence is $omega$ and thus $operatorname{cf}(omega)=1$, which is a successor ordinal.
I don't know what's wrong with my reasoning.
elementary-set-theory cardinals ordinals
edited Dec 26 '18 at 16:13
Martin Sleziak
45k10122277
asked Dec 26 '18 at 15:44
Le Anh DungLe Anh Dung
1,4011621
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marked as duplicate by Holo, Namaste elementary-set-theory
Users with the elementary-set-theory badge can single-handedly close elementary-set-theory questions as duplicates and reopen them as needed.
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Dec 26 '18 at 18:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Hi @MartinSleziak, is it possible to deduce If $α$ is a limit ordinal, then $cf(α)$ is a limit ordinal from the authors' definition of increasing sequence and cofinality?
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:00
2
$begingroup$
From the quote above, it seems that the authors definite limit of a sequence of ordinal numbers only if $vartheta$ is a limit ordinal. (This perhaps does not add to clarify of the definition, but since they define cofinality only for limit ordinals, it is probably enough for their purposes.)
$endgroup$
– Martin Sleziak
Dec 26 '18 at 16:16
$begingroup$
Thank you so much @MartinSleziak! I got it.
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:18
$begingroup$
What's wrong with my question that I received a downvote?
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:30
add a comment |
$begingroup$
This question already has an answer here:
Why cofinality is a cardinal.
1 answer
In the textbook Introduction to Set Theory by Hrbacek and Jech, Section 9.2, the authors first introduce the definition of increasing sequence of ordinals:
Then they introduce cofinality:
My question: how do we prove that $operatorname{cf}(alpha)$ is a limit ordinal?
If I take a sequence $langle alpha_nu mid nu<1 rangle$ where $alpha_0=omega$. It is clear that the limit of this sequence is $omega$ and thus $operatorname{cf}(omega)=1$, which is a successor ordinal.
I don't know what's wrong with my reasoning.
elementary-set-theory cardinals ordinals
edited Dec 26 '18 at 16:13
Martin Sleziak
45k10122277
asked Dec 26 '18 at 15:44
Le Anh DungLe Anh Dung
1,4011621
$endgroup$
This question already has an answer here:
Why cofinality is a cardinal.
1 answer
In the textbook Introduction to Set Theory by Hrbacek and Jech, Section 9.2, the authors first introduce the definition of increasing sequence of ordinals:
Then they introduce cofinality:
My question: how do we prove that $operatorname{cf}(alpha)$ is a limit ordinal?
If I take a sequence $langle alpha_nu mid nu<1 rangle$ where $alpha_0=omega$. It is clear that the limit of this sequence is $omega$ and thus $operatorname{cf}(omega)=1$, which is a successor ordinal.
I don't know what's wrong with my reasoning.
This question already has an answer here:
Why cofinality is a cardinal.
1 answer
elementary-set-theory cardinals ordinals
elementary-set-theory cardinals ordinals
edited Dec 26 '18 at 16:13
Martin Sleziak
45k10122277
asked Dec 26 '18 at 15:44
Le Anh DungLe Anh Dung
1,4011621
edited Dec 26 '18 at 16:13
Martin Sleziak
45k10122277
asked Dec 26 '18 at 15:44
Le Anh DungLe Anh Dung
1,4011621
edited Dec 26 '18 at 16:13
Martin Sleziak
45k10122277
edited Dec 26 '18 at 16:13
Martin Sleziak
45k10122277
edited Dec 26 '18 at 16:13
Martin Sleziak
45k10122277
45k10122277
asked Dec 26 '18 at 15:44
Le Anh DungLe Anh Dung
1,4011621
asked Dec 26 '18 at 15:44
Le Anh DungLe Anh Dung
1,4011621
asked Dec 26 '18 at 15:44
Le Anh DungLe Anh Dung
1,4011621
1,4011621
marked as duplicate by Holo, Namaste elementary-set-theory
Users with the elementary-set-theory badge can single-handedly close elementary-set-theory questions as duplicates and reopen them as needed.
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Dec 26 '18 at 18:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Holo, Namaste elementary-set-theory
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Dec 26 '18 at 18:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Hi @MartinSleziak, is it possible to deduce If $α$ is a limit ordinal, then $cf(α)$ is a limit ordinal from the authors' definition of increasing sequence and cofinality?
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:00
2
$begingroup$
From the quote above, it seems that the authors definite limit of a sequence of ordinal numbers only if $vartheta$ is a limit ordinal. (This perhaps does not add to clarify of the definition, but since they define cofinality only for limit ordinals, it is probably enough for their purposes.)
$endgroup$
– Martin Sleziak
Dec 26 '18 at 16:16
$begingroup$
Thank you so much @MartinSleziak! I got it.
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:18
$begingroup$
What's wrong with my question that I received a downvote?
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:30
add a comment |
$begingroup$
Hi @MartinSleziak, is it possible to deduce If $α$ is a limit ordinal, then $cf(α)$ is a limit ordinal from the authors' definition of increasing sequence and cofinality?
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:00
2
$begingroup$
From the quote above, it seems that the authors definite limit of a sequence of ordinal numbers only if $vartheta$ is a limit ordinal. (This perhaps does not add to clarify of the definition, but since they define cofinality only for limit ordinals, it is probably enough for their purposes.)
$endgroup$
– Martin Sleziak
Dec 26 '18 at 16:16
$begingroup$
Thank you so much @MartinSleziak! I got it.
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:18
$begingroup$
What's wrong with my question that I received a downvote?
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:30
$begingroup$
Hi @MartinSleziak, is it possible to deduce If $α$ is a limit ordinal, then $cf(α)$ is a limit ordinal from the authors' definition of increasing sequence and cofinality?
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:00
$begingroup$
Hi @MartinSleziak, is it possible to deduce If $α$ is a limit ordinal, then $cf(α)$ is a limit ordinal from the authors' definition of increasing sequence and cofinality?
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:00
2
2
$begingroup$
From the quote above, it seems that the authors definite limit of a sequence of ordinal numbers only if $vartheta$ is a limit ordinal. (This perhaps does not add to clarify of the definition, but since they define cofinality only for limit ordinals, it is probably enough for their purposes.)
$endgroup$
– Martin Sleziak
Dec 26 '18 at 16:16
$begingroup$
From the quote above, it seems that the authors definite limit of a sequence of ordinal numbers only if $vartheta$ is a limit ordinal. (This perhaps does not add to clarify of the definition, but since they define cofinality only for limit ordinals, it is probably enough for their purposes.)
$endgroup$
– Martin Sleziak
Dec 26 '18 at 16:16
$begingroup$
Thank you so much @MartinSleziak! I got it.
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:18
$begingroup$
Thank you so much @MartinSleziak! I got it.
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:18
$begingroup$
What's wrong with my question that I received a downvote?
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:30
$begingroup$
What's wrong with my question that I received a downvote?
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As Martin Sleziak observed, the definition of limit the text uses applies only to sequences of limit-ordinal length - so $langleomegarangle$ does not count as an increasing sequence of ordinals whose limit is $omega$.
Personally, I think this isn't quite optimal; another approach is to define $cf(alpha)$ as the smallest $lambda$ (limit or not) such that there is a sequence of ordinals $<alpha$ of length $lambda$ with every ordinal $<alpha$ being $le$ some term in the sequence. Then the cofinality of a successor ordinal is $1$ (consider the sequence $langle betarangle$ in $alpha=beta+1$), and the cofinality of a limit ordinal is a limit ordinal.
Note that if we replace "$le$" with "$<$" in the definition above, the cofinality of a successor ordinal becomes undefined since the limit of a sequence of ordinals $<beta+1$ is at most $beta$. This isn't really a problem, since nobody talks about the cofinality of successor ordinals, but it is in my opinion a bit annoying.
Note also that the definition above does apply to $0$, and gives the "correct" answer $cf(0)=0$, via the empty sequence.
answered Dec 26 '18 at 18:48
Noah SchweberNoah Schweber
128k10152294
$endgroup$
$begingroup$
Hi @Noah, I don't understand the sentence ... with every ordinal $<α$ being $≤$ some term in the sequence. If every term in the sequence is less than $alpha$, how can ... being $≤$ some term in the sequence?
$endgroup$
– Le Anh Dung
Dec 27 '18 at 1:12
$begingroup$
@LeAnhDung The point is that if $gamma<alpha$, then there needs to be some term $eta$ in the sequence such that $gammaleeta$.
$endgroup$
– Noah Schweber
Dec 27 '18 at 1:23
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As Martin Sleziak observed, the definition of limit the text uses applies only to sequences of limit-ordinal length - so $langleomegarangle$ does not count as an increasing sequence of ordinals whose limit is $omega$.
Personally, I think this isn't quite optimal; another approach is to define $cf(alpha)$ as the smallest $lambda$ (limit or not) such that there is a sequence of ordinals $<alpha$ of length $lambda$ with every ordinal $<alpha$ being $le$ some term in the sequence. Then the cofinality of a successor ordinal is $1$ (consider the sequence $langle betarangle$ in $alpha=beta+1$), and the cofinality of a limit ordinal is a limit ordinal.
Note that if we replace "$le$" with "$<$" in the definition above, the cofinality of a successor ordinal becomes undefined since the limit of a sequence of ordinals $<beta+1$ is at most $beta$. This isn't really a problem, since nobody talks about the cofinality of successor ordinals, but it is in my opinion a bit annoying.
Note also that the definition above does apply to $0$, and gives the "correct" answer $cf(0)=0$, via the empty sequence.
answered Dec 26 '18 at 18:48
Noah SchweberNoah Schweber
128k10152294
$endgroup$
$begingroup$
Hi @Noah, I don't understand the sentence ... with every ordinal $<α$ being $≤$ some term in the sequence. If every term in the sequence is less than $alpha$, how can ... being $≤$ some term in the sequence?
$endgroup$
– Le Anh Dung
Dec 27 '18 at 1:12
$begingroup$
@LeAnhDung The point is that if $gamma<alpha$, then there needs to be some term $eta$ in the sequence such that $gammaleeta$.
$endgroup$
– Noah Schweber
Dec 27 '18 at 1:23
add a comment |
$begingroup$
As Martin Sleziak observed, the definition of limit the text uses applies only to sequences of limit-ordinal length - so $langleomegarangle$ does not count as an increasing sequence of ordinals whose limit is $omega$.
Personally, I think this isn't quite optimal; another approach is to define $cf(alpha)$ as the smallest $lambda$ (limit or not) such that there is a sequence of ordinals $<alpha$ of length $lambda$ with every ordinal $<alpha$ being $le$ some term in the sequence. Then the cofinality of a successor ordinal is $1$ (consider the sequence $langle betarangle$ in $alpha=beta+1$), and the cofinality of a limit ordinal is a limit ordinal.
Note that if we replace "$le$" with "$<$" in the definition above, the cofinality of a successor ordinal becomes undefined since the limit of a sequence of ordinals $<beta+1$ is at most $beta$. This isn't really a problem, since nobody talks about the cofinality of successor ordinals, but it is in my opinion a bit annoying.
Note also that the definition above does apply to $0$, and gives the "correct" answer $cf(0)=0$, via the empty sequence.
answered Dec 26 '18 at 18:48
Noah SchweberNoah Schweber
128k10152294
$endgroup$
$begingroup$
Hi @Noah, I don't understand the sentence ... with every ordinal $<α$ being $≤$ some term in the sequence. If every term in the sequence is less than $alpha$, how can ... being $≤$ some term in the sequence?
$endgroup$
– Le Anh Dung
Dec 27 '18 at 1:12
$begingroup$
@LeAnhDung The point is that if $gamma<alpha$, then there needs to be some term $eta$ in the sequence such that $gammaleeta$.
$endgroup$
– Noah Schweber
Dec 27 '18 at 1:23
add a comment |
$begingroup$
As Martin Sleziak observed, the definition of limit the text uses applies only to sequences of limit-ordinal length - so $langleomegarangle$ does not count as an increasing sequence of ordinals whose limit is $omega$.
Personally, I think this isn't quite optimal; another approach is to define $cf(alpha)$ as the smallest $lambda$ (limit or not) such that there is a sequence of ordinals $<alpha$ of length $lambda$ with every ordinal $<alpha$ being $le$ some term in the sequence. Then the cofinality of a successor ordinal is $1$ (consider the sequence $langle betarangle$ in $alpha=beta+1$), and the cofinality of a limit ordinal is a limit ordinal.
Note that if we replace "$le$" with "$<$" in the definition above, the cofinality of a successor ordinal becomes undefined since the limit of a sequence of ordinals $<beta+1$ is at most $beta$. This isn't really a problem, since nobody talks about the cofinality of successor ordinals, but it is in my opinion a bit annoying.
Note also that the definition above does apply to $0$, and gives the "correct" answer $cf(0)=0$, via the empty sequence.
answered Dec 26 '18 at 18:48
Noah SchweberNoah Schweber
128k10152294
$endgroup$
As Martin Sleziak observed, the definition of limit the text uses applies only to sequences of limit-ordinal length - so $langleomegarangle$ does not count as an increasing sequence of ordinals whose limit is $omega$.
Personally, I think this isn't quite optimal; another approach is to define $cf(alpha)$ as the smallest $lambda$ (limit or not) such that there is a sequence of ordinals $<alpha$ of length $lambda$ with every ordinal $<alpha$ being $le$ some term in the sequence. Then the cofinality of a successor ordinal is $1$ (consider the sequence $langle betarangle$ in $alpha=beta+1$), and the cofinality of a limit ordinal is a limit ordinal.
Note that if we replace "$le$" with "$<$" in the definition above, the cofinality of a successor ordinal becomes undefined since the limit of a sequence of ordinals $<beta+1$ is at most $beta$. This isn't really a problem, since nobody talks about the cofinality of successor ordinals, but it is in my opinion a bit annoying.
Note also that the definition above does apply to $0$, and gives the "correct" answer $cf(0)=0$, via the empty sequence.
answered Dec 26 '18 at 18:48
Noah SchweberNoah Schweber
128k10152294
answered Dec 26 '18 at 18:48
Noah SchweberNoah Schweber
128k10152294
answered Dec 26 '18 at 18:48
Noah SchweberNoah Schweber
128k10152294
answered Dec 26 '18 at 18:48
Noah SchweberNoah Schweber
128k10152294
128k10152294
$begingroup$
Hi @Noah, I don't understand the sentence ... with every ordinal $<α$ being $≤$ some term in the sequence. If every term in the sequence is less than $alpha$, how can ... being $≤$ some term in the sequence?
$endgroup$
– Le Anh Dung
Dec 27 '18 at 1:12
$begingroup$
@LeAnhDung The point is that if $gamma<alpha$, then there needs to be some term $eta$ in the sequence such that $gammaleeta$.
$endgroup$
– Noah Schweber
Dec 27 '18 at 1:23
add a comment |
$begingroup$
Hi @Noah, I don't understand the sentence ... with every ordinal $<α$ being $≤$ some term in the sequence. If every term in the sequence is less than $alpha$, how can ... being $≤$ some term in the sequence?
$endgroup$
– Le Anh Dung
Dec 27 '18 at 1:12
$begingroup$
@LeAnhDung The point is that if $gamma<alpha$, then there needs to be some term $eta$ in the sequence such that $gammaleeta$.
$endgroup$
– Noah Schweber
Dec 27 '18 at 1:23
$begingroup$
Hi @Noah, I don't understand the sentence ... with every ordinal $<α$ being $≤$ some term in the sequence. If every term in the sequence is less than $alpha$, how can ... being $≤$ some term in the sequence?
$endgroup$
– Le Anh Dung
Dec 27 '18 at 1:12
$begingroup$
Hi @Noah, I don't understand the sentence ... with every ordinal $<α$ being $≤$ some term in the sequence. If every term in the sequence is less than $alpha$, how can ... being $≤$ some term in the sequence?
$endgroup$
– Le Anh Dung
Dec 27 '18 at 1:12
$begingroup$
@LeAnhDung The point is that if $gamma<alpha$, then there needs to be some term $eta$ in the sequence such that $gammaleeta$.
$endgroup$
– Noah Schweber
Dec 27 '18 at 1:23
$begingroup$
@LeAnhDung The point is that if $gamma<alpha$, then there needs to be some term $eta$ in the sequence such that $gammaleeta$.
$endgroup$
– Noah Schweber
Dec 27 '18 at 1:23
add a comment |
$begingroup$
Hi @MartinSleziak, is it possible to deduce If $α$ is a limit ordinal, then $cf(α)$ is a limit ordinal from the authors' definition of increasing sequence and cofinality?
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:00
2
$begingroup$
From the quote above, it seems that the authors definite limit of a sequence of ordinal numbers only if $vartheta$ is a limit ordinal. (This perhaps does not add to clarify of the definition, but since they define cofinality only for limit ordinals, it is probably enough for their purposes.)
$endgroup$
– Martin Sleziak
Dec 26 '18 at 16:16
$begingroup$
Thank you so much @MartinSleziak! I got it.
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:18
$begingroup$
What's wrong with my question that I received a downvote?
$endgroup$
– Le Anh Dung
Dec 26 '18 at 16:30