A question about a proof of Hausdorff's Formula
$begingroup$
My textbook Introduction to Set Theory by Hrbacek and Jech presents Hausdorff's Formula:
and its corresponding proof:
I am unable to deduce 3. from 1. and 2. as stated in the proof.
Each function $f:omega_beta to omega_{alpha+1}$ is bounded
The definition of ordinal exponentiation: $omega_{alpha+1}^{omega_beta}=sup {omega_{alpha+1}^{lambda} mid lambda< omega_beta}=bigcup_{lambda< omega_beta}omega_{alpha+1}^{lambda}$
$omega_{alpha+1}^{omega_beta}=bigcup_{gamma < omega_{alpha+1}}gamma^{omega_beta}$
Could you please elaborate on this point? Thank you for your help!
elementary-set-theory proof-explanation cardinals ordinals
asked Dec 30 '18 at 3:42
Le Anh DungLe Anh Dung
1,0701521
$endgroup$
add a comment |
$begingroup$
My textbook Introduction to Set Theory by Hrbacek and Jech presents Hausdorff's Formula:
and its corresponding proof:
I am unable to deduce 3. from 1. and 2. as stated in the proof.
Each function $f:omega_beta to omega_{alpha+1}$ is bounded
The definition of ordinal exponentiation: $omega_{alpha+1}^{omega_beta}=sup {omega_{alpha+1}^{lambda} mid lambda< omega_beta}=bigcup_{lambda< omega_beta}omega_{alpha+1}^{lambda}$
$omega_{alpha+1}^{omega_beta}=bigcup_{gamma < omega_{alpha+1}}gamma^{omega_beta}$
Could you please elaborate on this point? Thank you for your help!
elementary-set-theory proof-explanation cardinals ordinals
asked Dec 30 '18 at 3:42
Le Anh DungLe Anh Dung
1,0701521
$endgroup$
add a comment |
$begingroup$
My textbook Introduction to Set Theory by Hrbacek and Jech presents Hausdorff's Formula:
and its corresponding proof:
I am unable to deduce 3. from 1. and 2. as stated in the proof.
Each function $f:omega_beta to omega_{alpha+1}$ is bounded
The definition of ordinal exponentiation: $omega_{alpha+1}^{omega_beta}=sup {omega_{alpha+1}^{lambda} mid lambda< omega_beta}=bigcup_{lambda< omega_beta}omega_{alpha+1}^{lambda}$
$omega_{alpha+1}^{omega_beta}=bigcup_{gamma < omega_{alpha+1}}gamma^{omega_beta}$
Could you please elaborate on this point? Thank you for your help!
elementary-set-theory proof-explanation cardinals ordinals
asked Dec 30 '18 at 3:42
Le Anh DungLe Anh Dung
1,0701521
$endgroup$
My textbook Introduction to Set Theory by Hrbacek and Jech presents Hausdorff's Formula:
and its corresponding proof:
I am unable to deduce 3. from 1. and 2. as stated in the proof.
Each function $f:omega_beta to omega_{alpha+1}$ is bounded
The definition of ordinal exponentiation: $omega_{alpha+1}^{omega_beta}=sup {omega_{alpha+1}^{lambda} mid lambda< omega_beta}=bigcup_{lambda< omega_beta}omega_{alpha+1}^{lambda}$
$omega_{alpha+1}^{omega_beta}=bigcup_{gamma < omega_{alpha+1}}gamma^{omega_beta}$
Could you please elaborate on this point? Thank you for your help!
elementary-set-theory proof-explanation cardinals ordinals
elementary-set-theory proof-explanation cardinals ordinals
asked Dec 30 '18 at 3:42
Le Anh DungLe Anh Dung
1,0701521
asked Dec 30 '18 at 3:42
Le Anh DungLe Anh Dung
1,0701521
edited Dec 30 '18 at 3:56
Le Anh Dung
asked Dec 30 '18 at 3:42
Le Anh DungLe Anh Dung
1,0701521
asked Dec 30 '18 at 3:42
Le Anh DungLe Anh Dung
1,0701521
asked Dec 30 '18 at 3:42
Le Anh DungLe Anh Dung
1,0701521
1,0701521
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is not ordinal exponentiation, it is cardinal exponentiation, i.e. $omega_{alpha+1}^{omega_beta}$ is the set of all functions $omega_betato omega_{alpha+1}.$ If every such function is bounded, then every such function is a function $omega_betato gamma$ for some $gamma < omega_{alpha}$ and hence $ omega_{alpha+1}^{omega_beta} = bigcup_{gamma < omega_{alpha+1}}gamma^{omega_beta}.$
answered Dec 30 '18 at 4:11
spaceisdarkgreenspaceisdarkgreen
32.5k21753
$endgroup$
$begingroup$
Thank you so much for your answer! I am now clear. To avoid this confusion, it is better to denote the set of all functions $f:X to Y$ by $^X Y$ rather than $Y^X$. Although this denotation is very uncommon, I have seen some users doing so.
$endgroup$
– Le Anh Dung
Dec 30 '18 at 4:18
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It is not ordinal exponentiation, it is cardinal exponentiation, i.e. $omega_{alpha+1}^{omega_beta}$ is the set of all functions $omega_betato omega_{alpha+1}.$ If every such function is bounded, then every such function is a function $omega_betato gamma$ for some $gamma < omega_{alpha}$ and hence $ omega_{alpha+1}^{omega_beta} = bigcup_{gamma < omega_{alpha+1}}gamma^{omega_beta}.$
answered Dec 30 '18 at 4:11
spaceisdarkgreenspaceisdarkgreen
32.5k21753
$endgroup$
$begingroup$
Thank you so much for your answer! I am now clear. To avoid this confusion, it is better to denote the set of all functions $f:X to Y$ by $^X Y$ rather than $Y^X$. Although this denotation is very uncommon, I have seen some users doing so.
$endgroup$
– Le Anh Dung
Dec 30 '18 at 4:18
add a comment |
$begingroup$
It is not ordinal exponentiation, it is cardinal exponentiation, i.e. $omega_{alpha+1}^{omega_beta}$ is the set of all functions $omega_betato omega_{alpha+1}.$ If every such function is bounded, then every such function is a function $omega_betato gamma$ for some $gamma < omega_{alpha}$ and hence $ omega_{alpha+1}^{omega_beta} = bigcup_{gamma < omega_{alpha+1}}gamma^{omega_beta}.$
answered Dec 30 '18 at 4:11
spaceisdarkgreenspaceisdarkgreen
32.5k21753
$endgroup$
$begingroup$
Thank you so much for your answer! I am now clear. To avoid this confusion, it is better to denote the set of all functions $f:X to Y$ by $^X Y$ rather than $Y^X$. Although this denotation is very uncommon, I have seen some users doing so.
$endgroup$
– Le Anh Dung
Dec 30 '18 at 4:18
add a comment |
$begingroup$
It is not ordinal exponentiation, it is cardinal exponentiation, i.e. $omega_{alpha+1}^{omega_beta}$ is the set of all functions $omega_betato omega_{alpha+1}.$ If every such function is bounded, then every such function is a function $omega_betato gamma$ for some $gamma < omega_{alpha}$ and hence $ omega_{alpha+1}^{omega_beta} = bigcup_{gamma < omega_{alpha+1}}gamma^{omega_beta}.$
answered Dec 30 '18 at 4:11
spaceisdarkgreenspaceisdarkgreen
32.5k21753
$endgroup$
It is not ordinal exponentiation, it is cardinal exponentiation, i.e. $omega_{alpha+1}^{omega_beta}$ is the set of all functions $omega_betato omega_{alpha+1}.$ If every such function is bounded, then every such function is a function $omega_betato gamma$ for some $gamma < omega_{alpha}$ and hence $ omega_{alpha+1}^{omega_beta} = bigcup_{gamma < omega_{alpha+1}}gamma^{omega_beta}.$
answered Dec 30 '18 at 4:11
spaceisdarkgreenspaceisdarkgreen
32.5k21753
answered Dec 30 '18 at 4:11
spaceisdarkgreenspaceisdarkgreen
32.5k21753
answered Dec 30 '18 at 4:11
spaceisdarkgreenspaceisdarkgreen
32.5k21753
answered Dec 30 '18 at 4:11
spaceisdarkgreenspaceisdarkgreen
32.5k21753
32.5k21753
$begingroup$
Thank you so much for your answer! I am now clear. To avoid this confusion, it is better to denote the set of all functions $f:X to Y$ by $^X Y$ rather than $Y^X$. Although this denotation is very uncommon, I have seen some users doing so.
$endgroup$
– Le Anh Dung
Dec 30 '18 at 4:18
add a comment |
$begingroup$
Thank you so much for your answer! I am now clear. To avoid this confusion, it is better to denote the set of all functions $f:X to Y$ by $^X Y$ rather than $Y^X$. Although this denotation is very uncommon, I have seen some users doing so.
$endgroup$
– Le Anh Dung
Dec 30 '18 at 4:18
$begingroup$
Thank you so much for your answer! I am now clear. To avoid this confusion, it is better to denote the set of all functions $f:X to Y$ by $^X Y$ rather than $Y^X$. Although this denotation is very uncommon, I have seen some users doing so.
$endgroup$
– Le Anh Dung
Dec 30 '18 at 4:18
$begingroup$
Thank you so much for your answer! I am now clear. To avoid this confusion, it is better to denote the set of all functions $f:X to Y$ by $^X Y$ rather than $Y^X$. Although this denotation is very uncommon, I have seen some users doing so.
$endgroup$
– Le Anh Dung
Dec 30 '18 at 4:18
add a comment |
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