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
1,0871521
$endgroup$
|
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
1,0871521
$endgroup$
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
1,0871521
$endgroup$
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
1,0871521
asked Dec 2 '18 at 15:50
Le Anh DungLe Anh Dung
1,0871521
edited Dec 2 '18 at 16:36
Le Anh Dung
asked Dec 2 '18 at 15:50
Le Anh DungLe Anh Dung
1,0871521
asked Dec 2 '18 at 15:50
Le Anh DungLe Anh Dung
1,0871521
asked Dec 2 '18 at 15:50
Le Anh DungLe Anh Dung
1,0871521
1,0871521
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
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022789%2ffor-any-set-a-the-hartogs-number-of-a-is-an-initial-ordinal%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022789%2ffor-any-set-a-the-hartogs-number-of-a-is-an-initial-ordinal%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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