Each well-orderable set $X$ is equipotent to a unique initial ordinal
$begingroup$
Each well-orderable set $X$ is equipotent to a unique initial ordinal.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Lemma: Every well-ordered set is isomorphic to a unique ordinal.
Existence
By Lemma and Axiom of Choice, $X$ is equipotent to some ordinal $alpha$. Let $alpha_0$ be the least ordinal equipotent to $X$. Then $alpha_0$ is an initial ordinal. If not, $|alpha_0|=|beta|=|X|$ for some $beta<alpha_0$. This contradicts the minimality of $alpha_0$.
Uniqueness
Assume the contrary that $X$ is equipotent to initial ordinals $alpha_0,alpha_1$ such that $alpha_0neqalpha_1$. WLOG, we assume $alpha_0<alpha_1$. Furthermore, $|alpha_0|=|alpha_1|$. This contradicts the fact that $alpha_1$ is an initial ordinal.
Then the cardinal number of $X$, denoted by $|X|$, is defined as the unique initial ordinal equipotent to $X$.
proof-verification elementary-set-theory ordinals
asked Dec 2 '18 at 15:14
Le Anh DungLe Anh Dung
1,1921621
$endgroup$
add a comment |
$begingroup$
Each well-orderable set $X$ is equipotent to a unique initial ordinal.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Lemma: Every well-ordered set is isomorphic to a unique ordinal.
Existence
By Lemma and Axiom of Choice, $X$ is equipotent to some ordinal $alpha$. Let $alpha_0$ be the least ordinal equipotent to $X$. Then $alpha_0$ is an initial ordinal. If not, $|alpha_0|=|beta|=|X|$ for some $beta<alpha_0$. This contradicts the minimality of $alpha_0$.
Uniqueness
Assume the contrary that $X$ is equipotent to initial ordinals $alpha_0,alpha_1$ such that $alpha_0neqalpha_1$. WLOG, we assume $alpha_0<alpha_1$. Furthermore, $|alpha_0|=|alpha_1|$. This contradicts the fact that $alpha_1$ is an initial ordinal.
Then the cardinal number of $X$, denoted by $|X|$, is defined as the unique initial ordinal equipotent to $X$.
proof-verification elementary-set-theory ordinals
asked Dec 2 '18 at 15:14
Le Anh DungLe Anh Dung
1,1921621
$endgroup$
$begingroup$
Whether "It's clear that $X$ is equipotent to ..." is really clear depends a lot on your assumptions. For instance, if you have assumed the axiom of choice, then it's clear. Otherwise it looks good to me.
$endgroup$
– Arthur
Dec 2 '18 at 15:26
$begingroup$
It is not clear that there is such an ordinal. Refer back to the result that allows you to conclude that. Every time you post, the same problem is pointed out to you, and you keep ignoring it.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 15:54
$begingroup$
Hi @AndrésE.Caicedo! I have added a lemma as you suggested. Could you please more specific on my same problem? Is it that I appeal to a result without clearly mentioning it?
$endgroup$
– Le Anh Dung
Dec 2 '18 at 16:20
$begingroup$
Yes, that's precisely what I meant.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 16:36
add a comment |
$begingroup$
Each well-orderable set $X$ is equipotent to a unique initial ordinal.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Lemma: Every well-ordered set is isomorphic to a unique ordinal.
Existence
By Lemma and Axiom of Choice, $X$ is equipotent to some ordinal $alpha$. Let $alpha_0$ be the least ordinal equipotent to $X$. Then $alpha_0$ is an initial ordinal. If not, $|alpha_0|=|beta|=|X|$ for some $beta<alpha_0$. This contradicts the minimality of $alpha_0$.
Uniqueness
Assume the contrary that $X$ is equipotent to initial ordinals $alpha_0,alpha_1$ such that $alpha_0neqalpha_1$. WLOG, we assume $alpha_0<alpha_1$. Furthermore, $|alpha_0|=|alpha_1|$. This contradicts the fact that $alpha_1$ is an initial ordinal.
Then the cardinal number of $X$, denoted by $|X|$, is defined as the unique initial ordinal equipotent to $X$.
proof-verification elementary-set-theory ordinals
asked Dec 2 '18 at 15:14
Le Anh DungLe Anh Dung
1,1921621
$endgroup$
Each well-orderable set $X$ is equipotent to a unique initial ordinal.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Lemma: Every well-ordered set is isomorphic to a unique ordinal.
Existence
By Lemma and Axiom of Choice, $X$ is equipotent to some ordinal $alpha$. Let $alpha_0$ be the least ordinal equipotent to $X$. Then $alpha_0$ is an initial ordinal. If not, $|alpha_0|=|beta|=|X|$ for some $beta<alpha_0$. This contradicts the minimality of $alpha_0$.
Uniqueness
Assume the contrary that $X$ is equipotent to initial ordinals $alpha_0,alpha_1$ such that $alpha_0neqalpha_1$. WLOG, we assume $alpha_0<alpha_1$. Furthermore, $|alpha_0|=|alpha_1|$. This contradicts the fact that $alpha_1$ is an initial ordinal.
Then the cardinal number of $X$, denoted by $|X|$, is defined as the unique initial ordinal equipotent to $X$.
proof-verification elementary-set-theory ordinals
proof-verification elementary-set-theory ordinals
asked Dec 2 '18 at 15:14
Le Anh DungLe Anh Dung
1,1921621
asked Dec 2 '18 at 15:14
Le Anh DungLe Anh Dung
1,1921621
edited Dec 2 '18 at 16:17
Le Anh Dung
asked Dec 2 '18 at 15:14
Le Anh DungLe Anh Dung
1,1921621
asked Dec 2 '18 at 15:14
Le Anh DungLe Anh Dung
1,1921621
asked Dec 2 '18 at 15:14
Le Anh DungLe Anh Dung
1,1921621
1,1921621
$begingroup$
Whether "It's clear that $X$ is equipotent to ..." is really clear depends a lot on your assumptions. For instance, if you have assumed the axiom of choice, then it's clear. Otherwise it looks good to me.
$endgroup$
– Arthur
Dec 2 '18 at 15:26
$begingroup$
It is not clear that there is such an ordinal. Refer back to the result that allows you to conclude that. Every time you post, the same problem is pointed out to you, and you keep ignoring it.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 15:54
$begingroup$
Hi @AndrésE.Caicedo! I have added a lemma as you suggested. Could you please more specific on my same problem? Is it that I appeal to a result without clearly mentioning it?
$endgroup$
– Le Anh Dung
Dec 2 '18 at 16:20
$begingroup$
Yes, that's precisely what I meant.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 16:36
add a comment |
$begingroup$
Whether "It's clear that $X$ is equipotent to ..." is really clear depends a lot on your assumptions. For instance, if you have assumed the axiom of choice, then it's clear. Otherwise it looks good to me.
$endgroup$
– Arthur
Dec 2 '18 at 15:26
$begingroup$
It is not clear that there is such an ordinal. Refer back to the result that allows you to conclude that. Every time you post, the same problem is pointed out to you, and you keep ignoring it.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 15:54
$begingroup$
Hi @AndrésE.Caicedo! I have added a lemma as you suggested. Could you please more specific on my same problem? Is it that I appeal to a result without clearly mentioning it?
$endgroup$
– Le Anh Dung
Dec 2 '18 at 16:20
$begingroup$
Yes, that's precisely what I meant.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 16:36
$begingroup$
Whether "It's clear that $X$ is equipotent to ..." is really clear depends a lot on your assumptions. For instance, if you have assumed the axiom of choice, then it's clear. Otherwise it looks good to me.
$endgroup$
– Arthur
Dec 2 '18 at 15:26
$begingroup$
Whether "It's clear that $X$ is equipotent to ..." is really clear depends a lot on your assumptions. For instance, if you have assumed the axiom of choice, then it's clear. Otherwise it looks good to me.
$endgroup$
– Arthur
Dec 2 '18 at 15:26
$begingroup$
It is not clear that there is such an ordinal. Refer back to the result that allows you to conclude that. Every time you post, the same problem is pointed out to you, and you keep ignoring it.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 15:54
$begingroup$
It is not clear that there is such an ordinal. Refer back to the result that allows you to conclude that. Every time you post, the same problem is pointed out to you, and you keep ignoring it.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 15:54
$begingroup$
Hi @AndrésE.Caicedo! I have added a lemma as you suggested. Could you please more specific on my same problem? Is it that I appeal to a result without clearly mentioning it?
$endgroup$
– Le Anh Dung
Dec 2 '18 at 16:20
$begingroup$
Hi @AndrésE.Caicedo! I have added a lemma as you suggested. Could you please more specific on my same problem? Is it that I appeal to a result without clearly mentioning it?
$endgroup$
– Le Anh Dung
Dec 2 '18 at 16:20
$begingroup$
Yes, that's precisely what I meant.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 16:36
$begingroup$
Yes, that's precisely what I meant.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 16:36
add a comment |
1 Answer
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$begingroup$
Assuming you indeed know that the class of ordinals is linearly well-ordered, this proof looks fine to me. I would say: $alpha_0$ is initial, because if $beta < alpha_0$ it cannot be equipotent to $X$ by minimality of $alpha_0$ and as $X simeq alpha_0$, $beta notsimeq alpha_0$ as otherwise $beta simeq X$; and so $alpha_0$ is initial. It comes down to the same thing, namely using that $alpha$ is initial iff $forall beta < alpha: beta notsimeq alpha$.
answered Dec 2 '18 at 15:26
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Assuming you indeed know that the class of ordinals is linearly well-ordered, this proof looks fine to me. I would say: $alpha_0$ is initial, because if $beta < alpha_0$ it cannot be equipotent to $X$ by minimality of $alpha_0$ and as $X simeq alpha_0$, $beta notsimeq alpha_0$ as otherwise $beta simeq X$; and so $alpha_0$ is initial. It comes down to the same thing, namely using that $alpha$ is initial iff $forall beta < alpha: beta notsimeq alpha$.
answered Dec 2 '18 at 15:26
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
Assuming you indeed know that the class of ordinals is linearly well-ordered, this proof looks fine to me. I would say: $alpha_0$ is initial, because if $beta < alpha_0$ it cannot be equipotent to $X$ by minimality of $alpha_0$ and as $X simeq alpha_0$, $beta notsimeq alpha_0$ as otherwise $beta simeq X$; and so $alpha_0$ is initial. It comes down to the same thing, namely using that $alpha$ is initial iff $forall beta < alpha: beta notsimeq alpha$.
answered Dec 2 '18 at 15:26
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
Assuming you indeed know that the class of ordinals is linearly well-ordered, this proof looks fine to me. I would say: $alpha_0$ is initial, because if $beta < alpha_0$ it cannot be equipotent to $X$ by minimality of $alpha_0$ and as $X simeq alpha_0$, $beta notsimeq alpha_0$ as otherwise $beta simeq X$; and so $alpha_0$ is initial. It comes down to the same thing, namely using that $alpha$ is initial iff $forall beta < alpha: beta notsimeq alpha$.
answered Dec 2 '18 at 15:26
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
Assuming you indeed know that the class of ordinals is linearly well-ordered, this proof looks fine to me. I would say: $alpha_0$ is initial, because if $beta < alpha_0$ it cannot be equipotent to $X$ by minimality of $alpha_0$ and as $X simeq alpha_0$, $beta notsimeq alpha_0$ as otherwise $beta simeq X$; and so $alpha_0$ is initial. It comes down to the same thing, namely using that $alpha$ is initial iff $forall beta < alpha: beta notsimeq alpha$.
answered Dec 2 '18 at 15:26
Henno BrandsmaHenno Brandsma
110k347116
answered Dec 2 '18 at 15:26
Henno BrandsmaHenno Brandsma
110k347116
answered Dec 2 '18 at 15:26
Henno BrandsmaHenno Brandsma
110k347116
answered Dec 2 '18 at 15:26
Henno BrandsmaHenno Brandsma
110k347116
110k347116
add a comment |
add a comment |
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$begingroup$
Whether "It's clear that $X$ is equipotent to ..." is really clear depends a lot on your assumptions. For instance, if you have assumed the axiom of choice, then it's clear. Otherwise it looks good to me.
$endgroup$
– Arthur
Dec 2 '18 at 15:26
$begingroup$
It is not clear that there is such an ordinal. Refer back to the result that allows you to conclude that. Every time you post, the same problem is pointed out to you, and you keep ignoring it.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 15:54
$begingroup$
Hi @AndrésE.Caicedo! I have added a lemma as you suggested. Could you please more specific on my same problem? Is it that I appeal to a result without clearly mentioning it?
$endgroup$
– Le Anh Dung
Dec 2 '18 at 16:20
$begingroup$
Yes, that's precisely what I meant.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 16:36