Mixing
Let $E$ be an inductive ordered set, then $forall ain E$ ($exists$ a maximal element $min E$ $ni$ $aleq m$)
$begingroup$
I read the following in a book:
Theorem: Let $E$ be an inductive ordered set and let $a$ be and element of $E$. Then there exists a maximal element $m$ of $E$ such that $mgeq a$.
If I understand correctly, an inductive ordered set does not necessarily have to be totally ordered. So $a$ can be an element that's not comparable to any other element of $E$. Then there exists no $m$ such that $mgeq a$. Shouldn't we add a condition: $a$ is comparable to some element in $E$?
elementary-set-theory
edited Feb 3 at 1:21
Andrés E. Caicedo
66k8160252
asked Feb 2 at 21:34
AliAli
1057
$endgroup$
add a comment |
$begingroup$
I read the following in a book:
Theorem: Let $E$ be an inductive ordered set and let $a$ be and element of $E$. Then there exists a maximal element $m$ of $E$ such that $mgeq a$.
If I understand correctly, an inductive ordered set does not necessarily have to be totally ordered. So $a$ can be an element that's not comparable to any other element of $E$. Then there exists no $m$ such that $mgeq a$. Shouldn't we add a condition: $a$ is comparable to some element in $E$?
elementary-set-theory
edited Feb 3 at 1:21
Andrés E. Caicedo
66k8160252
asked Feb 2 at 21:34
AliAli
1057
$endgroup$
3
$begingroup$
$a$ is always comparable to $a$
$endgroup$
– Max
Feb 2 at 21:47
$begingroup$
@Max That's it! Thanks.
$endgroup$
– Ali
Feb 2 at 22:13
$begingroup$
The positive integers N, are inductively ordered and though 1 in N, there is no maximal number > 1.
$endgroup$
– William Elliot
Feb 3 at 0:22
$begingroup$
@WilliamElliot Interesting! Apparently this theorem only works for finite sets.
$endgroup$
– Ali
Feb 3 at 0:29
add a comment |
$begingroup$
I read the following in a book:
Theorem: Let $E$ be an inductive ordered set and let $a$ be and element of $E$. Then there exists a maximal element $m$ of $E$ such that $mgeq a$.
If I understand correctly, an inductive ordered set does not necessarily have to be totally ordered. So $a$ can be an element that's not comparable to any other element of $E$. Then there exists no $m$ such that $mgeq a$. Shouldn't we add a condition: $a$ is comparable to some element in $E$?
elementary-set-theory
edited Feb 3 at 1:21
Andrés E. Caicedo
66k8160252
asked Feb 2 at 21:34
AliAli
1057
$endgroup$
I read the following in a book:
Theorem: Let $E$ be an inductive ordered set and let $a$ be and element of $E$. Then there exists a maximal element $m$ of $E$ such that $mgeq a$.
If I understand correctly, an inductive ordered set does not necessarily have to be totally ordered. So $a$ can be an element that's not comparable to any other element of $E$. Then there exists no $m$ such that $mgeq a$. Shouldn't we add a condition: $a$ is comparable to some element in $E$?
elementary-set-theory
elementary-set-theory
edited Feb 3 at 1:21
Andrés E. Caicedo
66k8160252
asked Feb 2 at 21:34
AliAli
1057
edited Feb 3 at 1:21
Andrés E. Caicedo
66k8160252
asked Feb 2 at 21:34
AliAli
1057
edited Feb 3 at 1:21
Andrés E. Caicedo
66k8160252
edited Feb 3 at 1:21
Andrés E. Caicedo
66k8160252
edited Feb 3 at 1:21
Andrés E. Caicedo
66k8160252
66k8160252
asked Feb 2 at 21:34
AliAli
1057
asked Feb 2 at 21:34
AliAli
1057
asked Feb 2 at 21:34
AliAli
1057
1057
3
$begingroup$
$a$ is always comparable to $a$
$endgroup$
– Max
Feb 2 at 21:47
$begingroup$
@Max That's it! Thanks.
$endgroup$
– Ali
Feb 2 at 22:13
$begingroup$
The positive integers N, are inductively ordered and though 1 in N, there is no maximal number > 1.
$endgroup$
– William Elliot
Feb 3 at 0:22
$begingroup$
@WilliamElliot Interesting! Apparently this theorem only works for finite sets.
$endgroup$
– Ali
Feb 3 at 0:29
add a comment |
3
$begingroup$
$a$ is always comparable to $a$
$endgroup$
– Max
Feb 2 at 21:47
$begingroup$
@Max That's it! Thanks.
$endgroup$
– Ali
Feb 2 at 22:13
$begingroup$
The positive integers N, are inductively ordered and though 1 in N, there is no maximal number > 1.
$endgroup$
– William Elliot
Feb 3 at 0:22
$begingroup$
@WilliamElliot Interesting! Apparently this theorem only works for finite sets.
$endgroup$
– Ali
Feb 3 at 0:29
3
3
$begingroup$
$a$ is always comparable to $a$
$endgroup$
– Max
Feb 2 at 21:47
$begingroup$
$a$ is always comparable to $a$
$endgroup$
– Max
Feb 2 at 21:47
$begingroup$
@Max That's it! Thanks.
$endgroup$
– Ali
Feb 2 at 22:13
$begingroup$
@Max That's it! Thanks.
$endgroup$
– Ali
Feb 2 at 22:13
$begingroup$
The positive integers N, are inductively ordered and though 1 in N, there is no maximal number > 1.
$endgroup$
– William Elliot
Feb 3 at 0:22
$begingroup$
The positive integers N, are inductively ordered and though 1 in N, there is no maximal number > 1.
$endgroup$
– William Elliot
Feb 3 at 0:22
$begingroup$
@WilliamElliot Interesting! Apparently this theorem only works for finite sets.
$endgroup$
– Ali
Feb 3 at 0:29
$begingroup$
@WilliamElliot Interesting! Apparently this theorem only works for finite sets.
$endgroup$
– Ali
Feb 3 at 0:29
add a comment |
0
active
oldest
votes
Your Answer
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%2f3097827%2flet-e-be-an-inductive-ordered-set-then-forall-a-in-e-exists-a-maximal%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%2f3097827%2flet-e-be-an-inductive-ordered-set-then-forall-a-in-e-exists-a-maximal%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
3
$begingroup$
$a$ is always comparable to $a$
$endgroup$
– Max
Feb 2 at 21:47
$begingroup$
@Max That's it! Thanks.
$endgroup$
– Ali
Feb 2 at 22:13
$begingroup$
The positive integers N, are inductively ordered and though 1 in N, there is no maximal number > 1.
$endgroup$
– William Elliot
Feb 3 at 0:22
$begingroup$
@WilliamElliot Interesting! Apparently this theorem only works for finite sets.
$endgroup$
– Ali
Feb 3 at 0:29