A finite poset $P$ is inductive iff it has a least element The Next CEO of Stack OverflowProve that for any infinite poset there is an infinite subset which is either linearly ordered or antichain.Building an antichain in a finite posetIf $X$ is inductive, then, $U = x in X mid x $ is transitive and every nonempty $ z subset x$ has an $in$-minimal element $ $ is inductive.Prove $X$ is well ordered.Prob. 5, Sec. 4.1 in Kreyszig's functional analysis book: A finite partially ordered set has a maximal elementShow that the set of prime ideals of A has minimal element with respect to inclusionI'm trying to prove that any finite partially ordered set has a minimal element.Maximal chain with upper bound has at least one elementEvery nonempty subset of $mathbbN$ has a smallest element.If $A$ is an ordered set has the Least upper bound property iff it has the greatest lower bound property.
Traveling with my 5 year old daughter (as the father) without the mother from Germany to Mexico
Finitely generated matrix groups whose eigenvalues are all algebraic
Airship steam engine room - problems and conflict
Are British MPs missing the point, with these 'Indicative Votes'?
How should I connect my cat5 cable to connectors having an orange-green line?
Avoiding the "not like other girls" trope?
logical reads on global temp table, but not on session-level temp table
How can I replace x-axis labels with pre-determined symbols?
Prodigo = pro + ago?
Is the offspring between a demon and a celestial possible? If so what is it called and is it in a book somewhere?
Read/write a pipe-delimited file line by line with some simple text manipulation
Ising model simulation
What steps are necessary to read a Modern SSD in Medieval Europe?
Find a path from s to t using as few red nodes as possible
How do I secure a TV wall mount?
Do I need to write [sic] when including a quotation with a number less than 10 that isn't written out?
Upgrading From a 9 Speed Sora Derailleur?
Raspberry pi 3 B with Ubuntu 18.04 server arm64: what pi version
How can I prove that a state of equilibrium is unstable?
Does int main() need a declaration on C++?
How to coordinate airplane tickets?
Which acid/base does a strong base/acid react when added to a buffer solution?
What day is it again?
Calculate the Mean mean of two numbers
A finite poset $P$ is inductive iff it has a least element
The Next CEO of Stack OverflowProve that for any infinite poset there is an infinite subset which is either linearly ordered or antichain.Building an antichain in a finite posetIf $X$ is inductive, then, $U = x in X mid x $ is transitive and every nonempty $ z subset x$ has an $in$-minimal element $ $ is inductive.Prove $X$ is well ordered.Prob. 5, Sec. 4.1 in Kreyszig's functional analysis book: A finite partially ordered set has a maximal elementShow that the set of prime ideals of A has minimal element with respect to inclusionI'm trying to prove that any finite partially ordered set has a minimal element.Maximal chain with upper bound has at least one elementEvery nonempty subset of $mathbbN$ has a smallest element.If $A$ is an ordered set has the Least upper bound property iff it has the greatest lower bound property.
$begingroup$
The question (in the title) is from Notes on Set Theory by Moschovakis, 2nd edition chapter 6.
First, I am not exactly sure what they mean by "it has a least element." I guessed that the author means that "iff it has a subset $S$ such that $S$ has a least element and that no element of $S$ is comparable to any other element outside of it." But then, the set containing $a,b,c$ and with its nontrivial relations as only $a leq b$ and $c leq b$ is inductive, but it doesn't have a least element according to this interpretation... So I am forced to assume that the author means that "iff it has a subset $S$ such that $S$ has a least element." (this is my interpretation for the rest of this post.) But then, any nonempty $P$ has a least element if we take this interpretation, since given $a in P$, $a$ is a least element of $ a $ so that we are required to prove that every finite poset is inductive...
Assuming the second interpretation:
I tried induction on the number of elements $n$ of $P = x_1,...,x_n $. For $n=1$ the statement is true trivially. Assume the result for $k <n$.
Let $S$ be a chain in $P$. If $S neq P$ then by induction $S$ has a supremum. If $S=P$ then $S':=S - { x_n {$ has a supremum, say $x_1$. Since $S=P$ is a chain, either $x_1 leq x_n$ or $x_n x_1$, and the greater of the two is a shpremum.
Is this solution correct?
Edit: For clarification, the author uses the definition that a poset is inductive iff every chain has a least upper bound; in a chain, every element in a chain is comparable.
proof-verification elementary-set-theory
asked Mar 28 at 10:23
Cute BrownieCute Brownie
1,048417
$endgroup$
|
show 4 more comments
$begingroup$
The question (in the title) is from Notes on Set Theory by Moschovakis, 2nd edition chapter 6.
First, I am not exactly sure what they mean by "it has a least element." I guessed that the author means that "iff it has a subset $S$ such that $S$ has a least element and that no element of $S$ is comparable to any other element outside of it." But then, the set containing $a,b,c$ and with its nontrivial relations as only $a leq b$ and $c leq b$ is inductive, but it doesn't have a least element according to this interpretation... So I am forced to assume that the author means that "iff it has a subset $S$ such that $S$ has a least element." (this is my interpretation for the rest of this post.) But then, any nonempty $P$ has a least element if we take this interpretation, since given $a in P$, $a$ is a least element of $ a $ so that we are required to prove that every finite poset is inductive...
Assuming the second interpretation:
I tried induction on the number of elements $n$ of $P = x_1,...,x_n $. For $n=1$ the statement is true trivially. Assume the result for $k <n$.
Let $S$ be a chain in $P$. If $S neq P$ then by induction $S$ has a supremum. If $S=P$ then $S':=S - { x_n {$ has a supremum, say $x_1$. Since $S=P$ is a chain, either $x_1 leq x_n$ or $x_n x_1$, and the greater of the two is a shpremum.
Is this solution correct?
Edit: For clarification, the author uses the definition that a poset is inductive iff every chain has a least upper bound; in a chain, every element in a chain is comparable.
proof-verification elementary-set-theory
asked Mar 28 at 10:23
Cute BrownieCute Brownie
1,048417
$endgroup$
$begingroup$
(P,<=) has a least element when exists a in P with for all x in P, a <= x.
$endgroup$
– William Elliot
Mar 28 at 10:40
$begingroup$
I don't understand, every nonempty finite poset is inductive : take a chain, either it's empty in which case any element of $P$ will do, or it's not, in which case it's a finite totally ordered set, hence has a maximum, hence an upper bound
$endgroup$
– Max
Mar 28 at 11:46
$begingroup$
@Max. What definition of inductive set are you using?
$endgroup$
– William Elliot
Mar 28 at 11:49
$begingroup$
@WilliamElliot the one from Zorn's lemma : a poset is inductive iff any chain has an upper bound
$endgroup$
– Max
Mar 28 at 12:00
1
$begingroup$
I'm going to guess that the definition of "inductive" that is intended in the question is "every chain has a least upper bound". Then finiteness ensures that this holds for every nonempty chain (since such a chain has a top element), and the additional assumption that the poset has a least element is exactly what's needed to take care of the empty chain.
$endgroup$
– Andreas Blass
Mar 29 at 2:04
|
show 4 more comments
$begingroup$
The question (in the title) is from Notes on Set Theory by Moschovakis, 2nd edition chapter 6.
First, I am not exactly sure what they mean by "it has a least element." I guessed that the author means that "iff it has a subset $S$ such that $S$ has a least element and that no element of $S$ is comparable to any other element outside of it." But then, the set containing $a,b,c$ and with its nontrivial relations as only $a leq b$ and $c leq b$ is inductive, but it doesn't have a least element according to this interpretation... So I am forced to assume that the author means that "iff it has a subset $S$ such that $S$ has a least element." (this is my interpretation for the rest of this post.) But then, any nonempty $P$ has a least element if we take this interpretation, since given $a in P$, $a$ is a least element of $ a $ so that we are required to prove that every finite poset is inductive...
Assuming the second interpretation:
I tried induction on the number of elements $n$ of $P = x_1,...,x_n $. For $n=1$ the statement is true trivially. Assume the result for $k <n$.
Let $S$ be a chain in $P$. If $S neq P$ then by induction $S$ has a supremum. If $S=P$ then $S':=S - { x_n {$ has a supremum, say $x_1$. Since $S=P$ is a chain, either $x_1 leq x_n$ or $x_n x_1$, and the greater of the two is a shpremum.
Is this solution correct?
Edit: For clarification, the author uses the definition that a poset is inductive iff every chain has a least upper bound; in a chain, every element in a chain is comparable.
proof-verification elementary-set-theory
asked Mar 28 at 10:23
Cute BrownieCute Brownie
1,048417
$endgroup$
The question (in the title) is from Notes on Set Theory by Moschovakis, 2nd edition chapter 6.
First, I am not exactly sure what they mean by "it has a least element." I guessed that the author means that "iff it has a subset $S$ such that $S$ has a least element and that no element of $S$ is comparable to any other element outside of it." But then, the set containing $a,b,c$ and with its nontrivial relations as only $a leq b$ and $c leq b$ is inductive, but it doesn't have a least element according to this interpretation... So I am forced to assume that the author means that "iff it has a subset $S$ such that $S$ has a least element." (this is my interpretation for the rest of this post.) But then, any nonempty $P$ has a least element if we take this interpretation, since given $a in P$, $a$ is a least element of $ a $ so that we are required to prove that every finite poset is inductive...
Assuming the second interpretation:
I tried induction on the number of elements $n$ of $P = x_1,...,x_n $. For $n=1$ the statement is true trivially. Assume the result for $k <n$.
Let $S$ be a chain in $P$. If $S neq P$ then by induction $S$ has a supremum. If $S=P$ then $S':=S - { x_n {$ has a supremum, say $x_1$. Since $S=P$ is a chain, either $x_1 leq x_n$ or $x_n x_1$, and the greater of the two is a shpremum.
Is this solution correct?
Edit: For clarification, the author uses the definition that a poset is inductive iff every chain has a least upper bound; in a chain, every element in a chain is comparable.
proof-verification elementary-set-theory
proof-verification elementary-set-theory
asked Mar 28 at 10:23
Cute BrownieCute Brownie
1,048417
asked Mar 28 at 10:23
Cute BrownieCute Brownie
1,048417
edited Mar 29 at 3:38
Cute Brownie
asked Mar 28 at 10:23
Cute BrownieCute Brownie
1,048417
asked Mar 28 at 10:23
Cute BrownieCute Brownie
1,048417
asked Mar 28 at 10:23
Cute BrownieCute Brownie
1,048417
1,048417
$begingroup$
(P,<=) has a least element when exists a in P with for all x in P, a <= x.
$endgroup$
– William Elliot
Mar 28 at 10:40
$begingroup$
I don't understand, every nonempty finite poset is inductive : take a chain, either it's empty in which case any element of $P$ will do, or it's not, in which case it's a finite totally ordered set, hence has a maximum, hence an upper bound
$endgroup$
– Max
Mar 28 at 11:46
$begingroup$
@Max. What definition of inductive set are you using?
$endgroup$
– William Elliot
Mar 28 at 11:49
$begingroup$
@WilliamElliot the one from Zorn's lemma : a poset is inductive iff any chain has an upper bound
$endgroup$
– Max
Mar 28 at 12:00
1
$begingroup$
I'm going to guess that the definition of "inductive" that is intended in the question is "every chain has a least upper bound". Then finiteness ensures that this holds for every nonempty chain (since such a chain has a top element), and the additional assumption that the poset has a least element is exactly what's needed to take care of the empty chain.
$endgroup$
– Andreas Blass
Mar 29 at 2:04
|
show 4 more comments
$begingroup$
(P,<=) has a least element when exists a in P with for all x in P, a <= x.
$endgroup$
– William Elliot
Mar 28 at 10:40
$begingroup$
I don't understand, every nonempty finite poset is inductive : take a chain, either it's empty in which case any element of $P$ will do, or it's not, in which case it's a finite totally ordered set, hence has a maximum, hence an upper bound
$endgroup$
– Max
Mar 28 at 11:46
$begingroup$
@Max. What definition of inductive set are you using?
$endgroup$
– William Elliot
Mar 28 at 11:49
$begingroup$
@WilliamElliot the one from Zorn's lemma : a poset is inductive iff any chain has an upper bound
$endgroup$
– Max
Mar 28 at 12:00
1
$begingroup$
I'm going to guess that the definition of "inductive" that is intended in the question is "every chain has a least upper bound". Then finiteness ensures that this holds for every nonempty chain (since such a chain has a top element), and the additional assumption that the poset has a least element is exactly what's needed to take care of the empty chain.
$endgroup$
– Andreas Blass
Mar 29 at 2:04
$begingroup$
(P,<=) has a least element when exists a in P with for all x in P, a <= x.
$endgroup$
– William Elliot
Mar 28 at 10:40
$begingroup$
(P,<=) has a least element when exists a in P with for all x in P, a <= x.
$endgroup$
– William Elliot
Mar 28 at 10:40
$begingroup$
I don't understand, every nonempty finite poset is inductive : take a chain, either it's empty in which case any element of $P$ will do, or it's not, in which case it's a finite totally ordered set, hence has a maximum, hence an upper bound
$endgroup$
– Max
Mar 28 at 11:46
$begingroup$
I don't understand, every nonempty finite poset is inductive : take a chain, either it's empty in which case any element of $P$ will do, or it's not, in which case it's a finite totally ordered set, hence has a maximum, hence an upper bound
$endgroup$
– Max
Mar 28 at 11:46
$begingroup$
@Max. What definition of inductive set are you using?
$endgroup$
– William Elliot
Mar 28 at 11:49
$begingroup$
@Max. What definition of inductive set are you using?
$endgroup$
– William Elliot
Mar 28 at 11:49
$begingroup$
@WilliamElliot the one from Zorn's lemma : a poset is inductive iff any chain has an upper bound
$endgroup$
– Max
Mar 28 at 12:00
$begingroup$
@WilliamElliot the one from Zorn's lemma : a poset is inductive iff any chain has an upper bound
$endgroup$
– Max
Mar 28 at 12:00
1
1
$begingroup$
I'm going to guess that the definition of "inductive" that is intended in the question is "every chain has a least upper bound". Then finiteness ensures that this holds for every nonempty chain (since such a chain has a top element), and the additional assumption that the poset has a least element is exactly what's needed to take care of the empty chain.
$endgroup$
– Andreas Blass
Mar 29 at 2:04
$begingroup$
I'm going to guess that the definition of "inductive" that is intended in the question is "every chain has a least upper bound". Then finiteness ensures that this holds for every nonempty chain (since such a chain has a top element), and the additional assumption that the poset has a least element is exactly what's needed to take care of the empty chain.
$endgroup$
– Andreas Blass
Mar 29 at 2:04
|
show 4 more comments
0
active
oldest
votes
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%2f3165715%2fa-finite-poset-p-is-inductive-iff-it-has-a-least-element%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%2f3165715%2fa-finite-poset-p-is-inductive-iff-it-has-a-least-element%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
$begingroup$
(P,<=) has a least element when exists a in P with for all x in P, a <= x.
$endgroup$
– William Elliot
Mar 28 at 10:40
$begingroup$
I don't understand, every nonempty finite poset is inductive : take a chain, either it's empty in which case any element of $P$ will do, or it's not, in which case it's a finite totally ordered set, hence has a maximum, hence an upper bound
$endgroup$
– Max
Mar 28 at 11:46
$begingroup$
@Max. What definition of inductive set are you using?
$endgroup$
– William Elliot
Mar 28 at 11:49
$begingroup$
@WilliamElliot the one from Zorn's lemma : a poset is inductive iff any chain has an upper bound
$endgroup$
– Max
Mar 28 at 12:00
1
$begingroup$
I'm going to guess that the definition of "inductive" that is intended in the question is "every chain has a least upper bound". Then finiteness ensures that this holds for every nonempty chain (since such a chain has a top element), and the additional assumption that the poset has a least element is exactly what's needed to take care of the empty chain.
$endgroup$
– Andreas Blass
Mar 29 at 2:04