Is my understanding of a proof from textbook Introduction to Set Theory by Hrbacek and Jech correct?
$begingroup$
3.8 Therorem Let us assume the Generalized Continuum Hypothesis. If $aleph_alpha$ is a regular cardinal, then $$aleph_alpha^{aleph_beta}=begin{cases}
aleph_alpha&text{if }beta<alpha\aleph_{beta+1}&text{if }betagealphaend{cases}$$
My textbook presents the theorem and its proof as follows:
I would like to ask if my understanding of the below statement is correct. $$B=bigcup_{delta<omega_alpha}mathcal{P}(delta)text{ be the collection of all bounded subsets of } omega_alpha$$
Thank you for your help!
$delta<omega_alpha implies delta$ is bounded
If not, there exists $delta<omega_alpha$ such that $sup delta=omega_alpha$. Let $(beta_ximidxi<lambda)$ be an increasing enumeration of $delta$. Then $|lambda|=|delta|ledelta<omega_alpha$ and $lim_{xitolambda}beta_xi=sup delta=omega_alpha$. It follows that $aleph_alpha$ is singular, which contradicts the fact that $aleph_alpha$ is regular.
$X$ is a bounded subset of $omega_alpha$ $implies Xsubseteqdelta$ for some $delta<omega_alpha$
Since $X$ is a bounded subset of $omega_alpha$, $sup X<omega_alpha$. We have $Xsubseteq {gamma intext{ Ord} mid gamma le sup X}$. It is clear that ${gamma intext{ Ord} mid gamma le sup X}$ is a proper initial segment of $omega_alpha$ and thus an ordinal. Then ${gamma intext{ Ord} mid gamma le sup X} =delta$ for some $delta<omega_alpha$.
$B$ is the collection of all bounded subsets of $omega_alpha$ $implies B=bigcup_{delta<omega_alpha}mathcal{P}(delta)$
$Xin B implies$ $X$ is a bounded subsets of $omega_alpha$ $implies$ $Xsubseteqdelta$ for some $delta<omega_alpha$ $implies$ $Xinmathcal{P}(delta)$ for some $delta<omega_alpha$ $implies$ $Xinbigcup_{delta<omega_alpha}mathcal{P}(delta)$.
$Xinbigcup_{delta<omega_alpha}mathcal{P}(delta)$ $implies$ $Xinmathcal{P}(delta)$ for some $delta<omega_alpha$ $implies$ $Xsubseteqdelta$ for some $delta<omega_alpha$ $implies$ $sup X le sup delta < omega_alpha$ [Since $delta$ is bounded] $implies$ $X$ is bounded.
elementary-set-theory cardinals ordinals
asked Dec 29 '18 at 5:02
Le Anh DungLe Anh Dung
1,4511621
$endgroup$
add a comment |
$begingroup$
3.8 Therorem Let us assume the Generalized Continuum Hypothesis. If $aleph_alpha$ is a regular cardinal, then $$aleph_alpha^{aleph_beta}=begin{cases}
aleph_alpha&text{if }beta<alpha\aleph_{beta+1}&text{if }betagealphaend{cases}$$
My textbook presents the theorem and its proof as follows:
I would like to ask if my understanding of the below statement is correct. $$B=bigcup_{delta<omega_alpha}mathcal{P}(delta)text{ be the collection of all bounded subsets of } omega_alpha$$
Thank you for your help!
$delta<omega_alpha implies delta$ is bounded
If not, there exists $delta<omega_alpha$ such that $sup delta=omega_alpha$. Let $(beta_ximidxi<lambda)$ be an increasing enumeration of $delta$. Then $|lambda|=|delta|ledelta<omega_alpha$ and $lim_{xitolambda}beta_xi=sup delta=omega_alpha$. It follows that $aleph_alpha$ is singular, which contradicts the fact that $aleph_alpha$ is regular.
$X$ is a bounded subset of $omega_alpha$ $implies Xsubseteqdelta$ for some $delta<omega_alpha$
Since $X$ is a bounded subset of $omega_alpha$, $sup X<omega_alpha$. We have $Xsubseteq {gamma intext{ Ord} mid gamma le sup X}$. It is clear that ${gamma intext{ Ord} mid gamma le sup X}$ is a proper initial segment of $omega_alpha$ and thus an ordinal. Then ${gamma intext{ Ord} mid gamma le sup X} =delta$ for some $delta<omega_alpha$.
$B$ is the collection of all bounded subsets of $omega_alpha$ $implies B=bigcup_{delta<omega_alpha}mathcal{P}(delta)$
$Xin B implies$ $X$ is a bounded subsets of $omega_alpha$ $implies$ $Xsubseteqdelta$ for some $delta<omega_alpha$ $implies$ $Xinmathcal{P}(delta)$ for some $delta<omega_alpha$ $implies$ $Xinbigcup_{delta<omega_alpha}mathcal{P}(delta)$.
$Xinbigcup_{delta<omega_alpha}mathcal{P}(delta)$ $implies$ $Xinmathcal{P}(delta)$ for some $delta<omega_alpha$ $implies$ $Xsubseteqdelta$ for some $delta<omega_alpha$ $implies$ $sup X le sup delta < omega_alpha$ [Since $delta$ is bounded] $implies$ $X$ is bounded.
elementary-set-theory cardinals ordinals
asked Dec 29 '18 at 5:02
Le Anh DungLe Anh Dung
1,4511621
$endgroup$
add a comment |
$begingroup$
3.8 Therorem Let us assume the Generalized Continuum Hypothesis. If $aleph_alpha$ is a regular cardinal, then $$aleph_alpha^{aleph_beta}=begin{cases}
aleph_alpha&text{if }beta<alpha\aleph_{beta+1}&text{if }betagealphaend{cases}$$
My textbook presents the theorem and its proof as follows:
I would like to ask if my understanding of the below statement is correct. $$B=bigcup_{delta<omega_alpha}mathcal{P}(delta)text{ be the collection of all bounded subsets of } omega_alpha$$
Thank you for your help!
$delta<omega_alpha implies delta$ is bounded
If not, there exists $delta<omega_alpha$ such that $sup delta=omega_alpha$. Let $(beta_ximidxi<lambda)$ be an increasing enumeration of $delta$. Then $|lambda|=|delta|ledelta<omega_alpha$ and $lim_{xitolambda}beta_xi=sup delta=omega_alpha$. It follows that $aleph_alpha$ is singular, which contradicts the fact that $aleph_alpha$ is regular.
$X$ is a bounded subset of $omega_alpha$ $implies Xsubseteqdelta$ for some $delta<omega_alpha$
Since $X$ is a bounded subset of $omega_alpha$, $sup X<omega_alpha$. We have $Xsubseteq {gamma intext{ Ord} mid gamma le sup X}$. It is clear that ${gamma intext{ Ord} mid gamma le sup X}$ is a proper initial segment of $omega_alpha$ and thus an ordinal. Then ${gamma intext{ Ord} mid gamma le sup X} =delta$ for some $delta<omega_alpha$.
$B$ is the collection of all bounded subsets of $omega_alpha$ $implies B=bigcup_{delta<omega_alpha}mathcal{P}(delta)$
$Xin B implies$ $X$ is a bounded subsets of $omega_alpha$ $implies$ $Xsubseteqdelta$ for some $delta<omega_alpha$ $implies$ $Xinmathcal{P}(delta)$ for some $delta<omega_alpha$ $implies$ $Xinbigcup_{delta<omega_alpha}mathcal{P}(delta)$.
$Xinbigcup_{delta<omega_alpha}mathcal{P}(delta)$ $implies$ $Xinmathcal{P}(delta)$ for some $delta<omega_alpha$ $implies$ $Xsubseteqdelta$ for some $delta<omega_alpha$ $implies$ $sup X le sup delta < omega_alpha$ [Since $delta$ is bounded] $implies$ $X$ is bounded.
elementary-set-theory cardinals ordinals
asked Dec 29 '18 at 5:02
Le Anh DungLe Anh Dung
1,4511621
$endgroup$
3.8 Therorem Let us assume the Generalized Continuum Hypothesis. If $aleph_alpha$ is a regular cardinal, then $$aleph_alpha^{aleph_beta}=begin{cases}
aleph_alpha&text{if }beta<alpha\aleph_{beta+1}&text{if }betagealphaend{cases}$$
My textbook presents the theorem and its proof as follows:
I would like to ask if my understanding of the below statement is correct. $$B=bigcup_{delta<omega_alpha}mathcal{P}(delta)text{ be the collection of all bounded subsets of } omega_alpha$$
Thank you for your help!
$delta<omega_alpha implies delta$ is bounded
If not, there exists $delta<omega_alpha$ such that $sup delta=omega_alpha$. Let $(beta_ximidxi<lambda)$ be an increasing enumeration of $delta$. Then $|lambda|=|delta|ledelta<omega_alpha$ and $lim_{xitolambda}beta_xi=sup delta=omega_alpha$. It follows that $aleph_alpha$ is singular, which contradicts the fact that $aleph_alpha$ is regular.
$X$ is a bounded subset of $omega_alpha$ $implies Xsubseteqdelta$ for some $delta<omega_alpha$
Since $X$ is a bounded subset of $omega_alpha$, $sup X<omega_alpha$. We have $Xsubseteq {gamma intext{ Ord} mid gamma le sup X}$. It is clear that ${gamma intext{ Ord} mid gamma le sup X}$ is a proper initial segment of $omega_alpha$ and thus an ordinal. Then ${gamma intext{ Ord} mid gamma le sup X} =delta$ for some $delta<omega_alpha$.
$B$ is the collection of all bounded subsets of $omega_alpha$ $implies B=bigcup_{delta<omega_alpha}mathcal{P}(delta)$
$Xin B implies$ $X$ is a bounded subsets of $omega_alpha$ $implies$ $Xsubseteqdelta$ for some $delta<omega_alpha$ $implies$ $Xinmathcal{P}(delta)$ for some $delta<omega_alpha$ $implies$ $Xinbigcup_{delta<omega_alpha}mathcal{P}(delta)$.
$Xinbigcup_{delta<omega_alpha}mathcal{P}(delta)$ $implies$ $Xinmathcal{P}(delta)$ for some $delta<omega_alpha$ $implies$ $Xsubseteqdelta$ for some $delta<omega_alpha$ $implies$ $sup X le sup delta < omega_alpha$ [Since $delta$ is bounded] $implies$ $X$ is bounded.
elementary-set-theory cardinals ordinals
elementary-set-theory cardinals ordinals
asked Dec 29 '18 at 5:02
Le Anh DungLe Anh Dung
1,4511621
asked Dec 29 '18 at 5:02
Le Anh DungLe Anh Dung
1,4511621
edited Dec 29 '18 at 5:15
Le Anh Dung
asked Dec 29 '18 at 5:02
Le Anh DungLe Anh Dung
1,4511621
asked Dec 29 '18 at 5:02
Le Anh DungLe Anh Dung
1,4511621
asked Dec 29 '18 at 5:02
Le Anh DungLe Anh Dung
1,4511621
1,4511621
add a comment |
add a comment |
1 Answer
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$begingroup$
- This doesn't really make any sense. Of course an ordinal is bounded in any ordinal that it is less than: $supdelta le delta < omega_alpha.$ And this has nothing to do with regularity.
- Yes, this is correct.
- Yes, this is correct.
You could have made part 3 a little slicker by showing in part 2 that the reverse implication holds as well.
answered Dec 29 '18 at 6:11
spaceisdarkgreenspaceisdarkgreen
33.5k21753
$endgroup$
$begingroup$
If you don't know already, a good thing to think about next would be where in the proof regularity of $aleph_alpha$ is used.
$endgroup$
– spaceisdarkgreen
Dec 29 '18 at 6:17
$begingroup$
I have fixed 1. as follows: $forallgammaindelta:gamma<delta implies sup delta le delta < omega_alpha implies delta$ is bounded. I guess you have some implicit assumption when writing $color{blue}{supdelta = delta < omega_alpha}$. I have a counter-example: $5={0,1,2,3,4}$ and thus $sup 5= 4 neq 5$. I think the regularity of $aleph_alpha$ is used in the statement every $Xin S$ is a bounded subset of $omega_alpha$. Please have a check on my above reasoning. Thank you for your help!
$endgroup$
– Le Anh Dung
Dec 29 '18 at 7:39
$begingroup$
@LeAnhDung Yes, that's all correct and I edited to fix the error. It's easy to forget that successor ordinals exist.
$endgroup$
– spaceisdarkgreen
Dec 29 '18 at 7:45
$begingroup$
Thank you for all of your dedicated help!
$endgroup$
– Le Anh Dung
Dec 29 '18 at 7:57
add a comment |
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$begingroup$
- This doesn't really make any sense. Of course an ordinal is bounded in any ordinal that it is less than: $supdelta le delta < omega_alpha.$ And this has nothing to do with regularity.
- Yes, this is correct.
- Yes, this is correct.
You could have made part 3 a little slicker by showing in part 2 that the reverse implication holds as well.
answered Dec 29 '18 at 6:11
spaceisdarkgreenspaceisdarkgreen
33.5k21753
$endgroup$
$begingroup$
If you don't know already, a good thing to think about next would be where in the proof regularity of $aleph_alpha$ is used.
$endgroup$
– spaceisdarkgreen
Dec 29 '18 at 6:17
$begingroup$
I have fixed 1. as follows: $forallgammaindelta:gamma<delta implies sup delta le delta < omega_alpha implies delta$ is bounded. I guess you have some implicit assumption when writing $color{blue}{supdelta = delta < omega_alpha}$. I have a counter-example: $5={0,1,2,3,4}$ and thus $sup 5= 4 neq 5$. I think the regularity of $aleph_alpha$ is used in the statement every $Xin S$ is a bounded subset of $omega_alpha$. Please have a check on my above reasoning. Thank you for your help!
$endgroup$
– Le Anh Dung
Dec 29 '18 at 7:39
$begingroup$
@LeAnhDung Yes, that's all correct and I edited to fix the error. It's easy to forget that successor ordinals exist.
$endgroup$
– spaceisdarkgreen
Dec 29 '18 at 7:45
$begingroup$
Thank you for all of your dedicated help!
$endgroup$
– Le Anh Dung
Dec 29 '18 at 7:57
add a comment |
$begingroup$
- This doesn't really make any sense. Of course an ordinal is bounded in any ordinal that it is less than: $supdelta le delta < omega_alpha.$ And this has nothing to do with regularity.
- Yes, this is correct.
- Yes, this is correct.
You could have made part 3 a little slicker by showing in part 2 that the reverse implication holds as well.
answered Dec 29 '18 at 6:11
spaceisdarkgreenspaceisdarkgreen
33.5k21753
$endgroup$
$begingroup$
If you don't know already, a good thing to think about next would be where in the proof regularity of $aleph_alpha$ is used.
$endgroup$
– spaceisdarkgreen
Dec 29 '18 at 6:17
$begingroup$
I have fixed 1. as follows: $forallgammaindelta:gamma<delta implies sup delta le delta < omega_alpha implies delta$ is bounded. I guess you have some implicit assumption when writing $color{blue}{supdelta = delta < omega_alpha}$. I have a counter-example: $5={0,1,2,3,4}$ and thus $sup 5= 4 neq 5$. I think the regularity of $aleph_alpha$ is used in the statement every $Xin S$ is a bounded subset of $omega_alpha$. Please have a check on my above reasoning. Thank you for your help!
$endgroup$
– Le Anh Dung
Dec 29 '18 at 7:39
$begingroup$
@LeAnhDung Yes, that's all correct and I edited to fix the error. It's easy to forget that successor ordinals exist.
$endgroup$
– spaceisdarkgreen
Dec 29 '18 at 7:45
$begingroup$
Thank you for all of your dedicated help!
$endgroup$
– Le Anh Dung
Dec 29 '18 at 7:57
add a comment |
$begingroup$
- This doesn't really make any sense. Of course an ordinal is bounded in any ordinal that it is less than: $supdelta le delta < omega_alpha.$ And this has nothing to do with regularity.
- Yes, this is correct.
- Yes, this is correct.
You could have made part 3 a little slicker by showing in part 2 that the reverse implication holds as well.
answered Dec 29 '18 at 6:11
spaceisdarkgreenspaceisdarkgreen
33.5k21753
$endgroup$
- This doesn't really make any sense. Of course an ordinal is bounded in any ordinal that it is less than: $supdelta le delta < omega_alpha.$ And this has nothing to do with regularity.
- Yes, this is correct.
- Yes, this is correct.
You could have made part 3 a little slicker by showing in part 2 that the reverse implication holds as well.
answered Dec 29 '18 at 6:11
spaceisdarkgreenspaceisdarkgreen
33.5k21753
edited Dec 29 '18 at 7:46
answered Dec 29 '18 at 6:11
spaceisdarkgreenspaceisdarkgreen
33.5k21753
answered Dec 29 '18 at 6:11
spaceisdarkgreenspaceisdarkgreen
33.5k21753
answered Dec 29 '18 at 6:11
spaceisdarkgreenspaceisdarkgreen
33.5k21753
33.5k21753
$begingroup$
If you don't know already, a good thing to think about next would be where in the proof regularity of $aleph_alpha$ is used.
$endgroup$
– spaceisdarkgreen
Dec 29 '18 at 6:17
$begingroup$
I have fixed 1. as follows: $forallgammaindelta:gamma<delta implies sup delta le delta < omega_alpha implies delta$ is bounded. I guess you have some implicit assumption when writing $color{blue}{supdelta = delta < omega_alpha}$. I have a counter-example: $5={0,1,2,3,4}$ and thus $sup 5= 4 neq 5$. I think the regularity of $aleph_alpha$ is used in the statement every $Xin S$ is a bounded subset of $omega_alpha$. Please have a check on my above reasoning. Thank you for your help!
$endgroup$
– Le Anh Dung
Dec 29 '18 at 7:39
$begingroup$
@LeAnhDung Yes, that's all correct and I edited to fix the error. It's easy to forget that successor ordinals exist.
$endgroup$
– spaceisdarkgreen
Dec 29 '18 at 7:45
$begingroup$
Thank you for all of your dedicated help!
$endgroup$
– Le Anh Dung
Dec 29 '18 at 7:57
add a comment |
$begingroup$
If you don't know already, a good thing to think about next would be where in the proof regularity of $aleph_alpha$ is used.
$endgroup$
– spaceisdarkgreen
Dec 29 '18 at 6:17
$begingroup$
I have fixed 1. as follows: $forallgammaindelta:gamma<delta implies sup delta le delta < omega_alpha implies delta$ is bounded. I guess you have some implicit assumption when writing $color{blue}{supdelta = delta < omega_alpha}$. I have a counter-example: $5={0,1,2,3,4}$ and thus $sup 5= 4 neq 5$. I think the regularity of $aleph_alpha$ is used in the statement every $Xin S$ is a bounded subset of $omega_alpha$. Please have a check on my above reasoning. Thank you for your help!
$endgroup$
– Le Anh Dung
Dec 29 '18 at 7:39
$begingroup$
@LeAnhDung Yes, that's all correct and I edited to fix the error. It's easy to forget that successor ordinals exist.
$endgroup$
– spaceisdarkgreen
Dec 29 '18 at 7:45
$begingroup$
Thank you for all of your dedicated help!
$endgroup$
– Le Anh Dung
Dec 29 '18 at 7:57
$begingroup$
If you don't know already, a good thing to think about next would be where in the proof regularity of $aleph_alpha$ is used.
$endgroup$
– spaceisdarkgreen
Dec 29 '18 at 6:17
$begingroup$
If you don't know already, a good thing to think about next would be where in the proof regularity of $aleph_alpha$ is used.
$endgroup$
– spaceisdarkgreen
Dec 29 '18 at 6:17
$begingroup$
I have fixed 1. as follows: $forallgammaindelta:gamma<delta implies sup delta le delta < omega_alpha implies delta$ is bounded. I guess you have some implicit assumption when writing $color{blue}{supdelta = delta < omega_alpha}$. I have a counter-example: $5={0,1,2,3,4}$ and thus $sup 5= 4 neq 5$. I think the regularity of $aleph_alpha$ is used in the statement every $Xin S$ is a bounded subset of $omega_alpha$. Please have a check on my above reasoning. Thank you for your help!
$endgroup$
– Le Anh Dung
Dec 29 '18 at 7:39
$begingroup$
I have fixed 1. as follows: $forallgammaindelta:gamma<delta implies sup delta le delta < omega_alpha implies delta$ is bounded. I guess you have some implicit assumption when writing $color{blue}{supdelta = delta < omega_alpha}$. I have a counter-example: $5={0,1,2,3,4}$ and thus $sup 5= 4 neq 5$. I think the regularity of $aleph_alpha$ is used in the statement every $Xin S$ is a bounded subset of $omega_alpha$. Please have a check on my above reasoning. Thank you for your help!
$endgroup$
– Le Anh Dung
Dec 29 '18 at 7:39
$begingroup$
@LeAnhDung Yes, that's all correct and I edited to fix the error. It's easy to forget that successor ordinals exist.
$endgroup$
– spaceisdarkgreen
Dec 29 '18 at 7:45
$begingroup$
@LeAnhDung Yes, that's all correct and I edited to fix the error. It's easy to forget that successor ordinals exist.
$endgroup$
– spaceisdarkgreen
Dec 29 '18 at 7:45
$begingroup$
Thank you for all of your dedicated help!
$endgroup$
– Le Anh Dung
Dec 29 '18 at 7:57
$begingroup$
Thank you for all of your dedicated help!
$endgroup$
– Le Anh Dung
Dec 29 '18 at 7:57
add a comment |
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