For any set $A$, the Hartogs number of $A$ is an initial ordinal
$begingroup$
The Hartogs number of $A$ is the least ordinal which is not equipotent to any subset of $A$. We denote the Hartogs number of $A$ by $h(A)$.
Theorem: For any set $A$, $h(A)$ is an initial ordinal.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Assume the contrary that $h(A)$ is not an initial ordinal. Then $|h(A)|=|alpha|$ for some ordinal $alpha<h(A)$.
Since $alpha<h(A)$, $alpha$ is equipotent to some subset of $A$. Thus $h(A)$ is equipotent to some subset of $A$. This contradicts the definition of $h(A)$.
proof-verification elementary-set-theory ordinals
asked Dec 2 '18 at 15:50
Le Anh DungLe Anh Dung
1,0391521
$endgroup$
|
show 2 more comments
$begingroup$
The Hartogs number of $A$ is the least ordinal which is not equipotent to any subset of $A$. We denote the Hartogs number of $A$ by $h(A)$.
Theorem: For any set $A$, $h(A)$ is an initial ordinal.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Assume the contrary that $h(A)$ is not an initial ordinal. Then $|h(A)|=|alpha|$ for some ordinal $alpha<h(A)$.
Since $alpha<h(A)$, $alpha$ is equipotent to some subset of $A$. Thus $h(A)$ is equipotent to some subset of $A$. This contradicts the definition of $h(A)$.
proof-verification elementary-set-theory ordinals
asked Dec 2 '18 at 15:50
Le Anh DungLe Anh Dung
1,0391521
$endgroup$
1
$begingroup$
If you cannot verify a proof like this, you need to reexamine your mathematical confidence.
$endgroup$
– Asaf Karagila♦
Dec 2 '18 at 16:40
1
$begingroup$
Hmm, isn't a good way to gain mathematical confidence, and stock up on that celebrated staple mathematical maturity, posting questions like this?
$endgroup$
– Michael Weiss
Dec 2 '18 at 17:55
1
$begingroup$
@Michael I don't think so.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 17:58
1
$begingroup$
Hey you guys, be nice and back off. He did give a well writen proof, a proof better than than a lot of stuff that's presented.
$endgroup$
– William Elliot
Dec 3 '18 at 0:11
$begingroup$
Okay, let's ask this in a different way. Where do you think that your proof might be wrong?
$endgroup$
– Asaf Karagila♦
Dec 3 '18 at 8:44
|
show 2 more comments
$begingroup$
The Hartogs number of $A$ is the least ordinal which is not equipotent to any subset of $A$. We denote the Hartogs number of $A$ by $h(A)$.
Theorem: For any set $A$, $h(A)$ is an initial ordinal.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Assume the contrary that $h(A)$ is not an initial ordinal. Then $|h(A)|=|alpha|$ for some ordinal $alpha<h(A)$.
Since $alpha<h(A)$, $alpha$ is equipotent to some subset of $A$. Thus $h(A)$ is equipotent to some subset of $A$. This contradicts the definition of $h(A)$.
proof-verification elementary-set-theory ordinals
asked Dec 2 '18 at 15:50
Le Anh DungLe Anh Dung
1,0391521
$endgroup$
The Hartogs number of $A$ is the least ordinal which is not equipotent to any subset of $A$. We denote the Hartogs number of $A$ by $h(A)$.
Theorem: For any set $A$, $h(A)$ is an initial ordinal.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Assume the contrary that $h(A)$ is not an initial ordinal. Then $|h(A)|=|alpha|$ for some ordinal $alpha<h(A)$.
Since $alpha<h(A)$, $alpha$ is equipotent to some subset of $A$. Thus $h(A)$ is equipotent to some subset of $A$. This contradicts the definition of $h(A)$.
proof-verification elementary-set-theory ordinals
proof-verification elementary-set-theory ordinals
asked Dec 2 '18 at 15:50
Le Anh DungLe Anh Dung
1,0391521
asked Dec 2 '18 at 15:50
Le Anh DungLe Anh Dung
1,0391521
edited Dec 2 '18 at 16:36
Le Anh Dung
asked Dec 2 '18 at 15:50
Le Anh DungLe Anh Dung
1,0391521
asked Dec 2 '18 at 15:50
Le Anh DungLe Anh Dung
1,0391521
asked Dec 2 '18 at 15:50
Le Anh DungLe Anh Dung
1,0391521
1,0391521
1
$begingroup$
If you cannot verify a proof like this, you need to reexamine your mathematical confidence.
$endgroup$
– Asaf Karagila♦
Dec 2 '18 at 16:40
1
$begingroup$
Hmm, isn't a good way to gain mathematical confidence, and stock up on that celebrated staple mathematical maturity, posting questions like this?
$endgroup$
– Michael Weiss
Dec 2 '18 at 17:55
1
$begingroup$
@Michael I don't think so.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 17:58
1
$begingroup$
Hey you guys, be nice and back off. He did give a well writen proof, a proof better than than a lot of stuff that's presented.
$endgroup$
– William Elliot
Dec 3 '18 at 0:11
$begingroup$
Okay, let's ask this in a different way. Where do you think that your proof might be wrong?
$endgroup$
– Asaf Karagila♦
Dec 3 '18 at 8:44
|
show 2 more comments
1
$begingroup$
If you cannot verify a proof like this, you need to reexamine your mathematical confidence.
$endgroup$
– Asaf Karagila♦
Dec 2 '18 at 16:40
1
$begingroup$
Hmm, isn't a good way to gain mathematical confidence, and stock up on that celebrated staple mathematical maturity, posting questions like this?
$endgroup$
– Michael Weiss
Dec 2 '18 at 17:55
1
$begingroup$
@Michael I don't think so.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 17:58
1
$begingroup$
Hey you guys, be nice and back off. He did give a well writen proof, a proof better than than a lot of stuff that's presented.
$endgroup$
– William Elliot
Dec 3 '18 at 0:11
$begingroup$
Okay, let's ask this in a different way. Where do you think that your proof might be wrong?
$endgroup$
– Asaf Karagila♦
Dec 3 '18 at 8:44
1
1
$begingroup$
If you cannot verify a proof like this, you need to reexamine your mathematical confidence.
$endgroup$
– Asaf Karagila♦
Dec 2 '18 at 16:40
$begingroup$
If you cannot verify a proof like this, you need to reexamine your mathematical confidence.
$endgroup$
– Asaf Karagila♦
Dec 2 '18 at 16:40
1
1
$begingroup$
Hmm, isn't a good way to gain mathematical confidence, and stock up on that celebrated staple mathematical maturity, posting questions like this?
$endgroup$
– Michael Weiss
Dec 2 '18 at 17:55
$begingroup$
Hmm, isn't a good way to gain mathematical confidence, and stock up on that celebrated staple mathematical maturity, posting questions like this?
$endgroup$
– Michael Weiss
Dec 2 '18 at 17:55
1
1
$begingroup$
@Michael I don't think so.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 17:58
$begingroup$
@Michael I don't think so.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 17:58
1
1
$begingroup$
Hey you guys, be nice and back off. He did give a well writen proof, a proof better than than a lot of stuff that's presented.
$endgroup$
– William Elliot
Dec 3 '18 at 0:11
$begingroup$
Hey you guys, be nice and back off. He did give a well writen proof, a proof better than than a lot of stuff that's presented.
$endgroup$
– William Elliot
Dec 3 '18 at 0:11
$begingroup$
Okay, let's ask this in a different way. Where do you think that your proof might be wrong?
$endgroup$
– Asaf Karagila♦
Dec 3 '18 at 8:44
$begingroup$
Okay, let's ask this in a different way. Where do you think that your proof might be wrong?
$endgroup$
– Asaf Karagila♦
Dec 3 '18 at 8:44
|
show 2 more comments
0
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1
$begingroup$
If you cannot verify a proof like this, you need to reexamine your mathematical confidence.
$endgroup$
– Asaf Karagila♦
Dec 2 '18 at 16:40
1
$begingroup$
Hmm, isn't a good way to gain mathematical confidence, and stock up on that celebrated staple mathematical maturity, posting questions like this?
$endgroup$
– Michael Weiss
Dec 2 '18 at 17:55
1
$begingroup$
@Michael I don't think so.
$endgroup$
– Andrés E. Caicedo
Dec 2 '18 at 17:58
1
$begingroup$
Hey you guys, be nice and back off. He did give a well writen proof, a proof better than than a lot of stuff that's presented.
$endgroup$
– William Elliot
Dec 3 '18 at 0:11
$begingroup$
Okay, let's ask this in a different way. Where do you think that your proof might be wrong?
$endgroup$
– Asaf Karagila♦
Dec 3 '18 at 8:44