Property of meets and joins (infima and suprema) in a set

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

From a book:

Suppose that $X$ has binary meets and a top element. Then by induction it is easy to see that $X$ has meets of all finite subsets.

So, every 2-element subset of $X$ has a meet (infimum). Why does this imply that every finite subset has a meet too?

A "top element" is just defined to be the infimum of the empty subset, which the author considers finite.

order-theory supremum-and-infimum lattice-orders

share|cite|improve this question

edited Jul 26 at 21:59

Taroccoesbrocco

3,36941331

asked Jul 26 at 21:31

Mathemert

189313

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It is not just the author who considers the empty set finite.
  – Andrés E. Caicedo
  Jul 26 at 21:52
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

From a book:

Suppose that $X$ has binary meets and a top element. Then by induction it is easy to see that $X$ has meets of all finite subsets.

So, every 2-element subset of $X$ has a meet (infimum). Why does this imply that every finite subset has a meet too?

A "top element" is just defined to be the infimum of the empty subset, which the author considers finite.

order-theory supremum-and-infimum lattice-orders

share|cite|improve this question

edited Jul 26 at 21:59

Taroccoesbrocco

3,36941331

asked Jul 26 at 21:31

Mathemert

189313

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It is not just the author who considers the empty set finite.
  – Andrés E. Caicedo
  Jul 26 at 21:52
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

From a book:

Suppose that $X$ has binary meets and a top element. Then by induction it is easy to see that $X$ has meets of all finite subsets.

So, every 2-element subset of $X$ has a meet (infimum). Why does this imply that every finite subset has a meet too?

A "top element" is just defined to be the infimum of the empty subset, which the author considers finite.

order-theory supremum-and-infimum lattice-orders

share|cite|improve this question

edited Jul 26 at 21:59

Taroccoesbrocco

3,36941331

asked Jul 26 at 21:31

Mathemert

189313

From a book:

Suppose that $X$ has binary meets and a top element. Then by induction it is easy to see that $X$ has meets of all finite subsets.

So, every 2-element subset of $X$ has a meet (infimum). Why does this imply that every finite subset has a meet too?

A "top element" is just defined to be the infimum of the empty subset, which the author considers finite.

order-theory supremum-and-infimum lattice-orders

share|cite|improve this question

edited Jul 26 at 21:59

Taroccoesbrocco

3,36941331

asked Jul 26 at 21:31

Mathemert

189313

share|cite|improve this question

share|cite|improve this question

edited Jul 26 at 21:59

Taroccoesbrocco

3,36941331

edited Jul 26 at 21:59

Taroccoesbrocco

3,36941331

edited Jul 26 at 21:59

Taroccoesbrocco

3,36941331

3,36941331

asked Jul 26 at 21:31

Mathemert

189313

asked Jul 26 at 21:31

Mathemert

189313

asked Jul 26 at 21:31

Mathemert

189313

189313

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It is not just the author who considers the empty set finite.
  – Andrés E. Caicedo
  Jul 26 at 21:52
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It is not just the author who considers the empty set finite.
  – Andrés E. Caicedo
  Jul 26 at 21:52
  

  
  
  
  

2

2

It is not just the author who considers the empty set finite.
– Andrés E. Caicedo
Jul 26 at 21:52

It is not just the author who considers the empty set finite.
– Andrés E. Caicedo
Jul 26 at 21:52

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote

The book is right: roughly speaking, the idea is that you have to compare pairwise the elements of the finite subset of $X$. Let us see that more formally.

Given a finite $Y subseteq X$, I prove by induction on the cardinality $n$ of $Y$ that $X$ has the meet of $Y$.

If $n = 0$, then $Y = emptyset$ and the meet of $Y$ is the top element (which is an element of $X$ by hypothesis).

Let $n > 0$: then, $Y = y_1, dots, y_n$ where the $y_i$'s are pairwise distinct.
By induction hypothesis, $X$ has the meet, say $x$, of $y_1, dots, y_n-1$. By hypothesis, $X$ has the meet, say $x'$, of $x$ and $y_n$. It is easy to prove that $x'$ is the meet of $Y$.
Indeed:

• $x'$ is a lower bound of $Y$, since it is a lower bound of $y_n$ and $x$, and hence of $y_1, dots, y_n-1$ by transitivity (since $x$ is a lower bound of $y_1, dots, y_n-1$);




• any lower bound $z$ of $Y$ is a lower bound of $y_1, dots, y_n-1$ and then $z leq x$ by maximality of $x$, thus $z leq x'$ by transitivity; therefore, $x'$ is greater than or equal to any other lower bound of $Y$.

share|cite|improve this answer

edited Jul 26 at 22:11

answered Jul 26 at 21:47

Taroccoesbrocco

3,36941331

add a comment | 

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863840%2fproperty-of-meets-and-joins-infima-and-suprema-in-a-set%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

The book is right: roughly speaking, the idea is that you have to compare pairwise the elements of the finite subset of $X$. Let us see that more formally.

Given a finite $Y subseteq X$, I prove by induction on the cardinality $n$ of $Y$ that $X$ has the meet of $Y$.

If $n = 0$, then $Y = emptyset$ and the meet of $Y$ is the top element (which is an element of $X$ by hypothesis).

Let $n > 0$: then, $Y = y_1, dots, y_n$ where the $y_i$'s are pairwise distinct.
By induction hypothesis, $X$ has the meet, say $x$, of $y_1, dots, y_n-1$. By hypothesis, $X$ has the meet, say $x'$, of $x$ and $y_n$. It is easy to prove that $x'$ is the meet of $Y$.
Indeed:

• $x'$ is a lower bound of $Y$, since it is a lower bound of $y_n$ and $x$, and hence of $y_1, dots, y_n-1$ by transitivity (since $x$ is a lower bound of $y_1, dots, y_n-1$);




• any lower bound $z$ of $Y$ is a lower bound of $y_1, dots, y_n-1$ and then $z leq x$ by maximality of $x$, thus $z leq x'$ by transitivity; therefore, $x'$ is greater than or equal to any other lower bound of $Y$.

share|cite|improve this answer

edited Jul 26 at 22:11

answered Jul 26 at 21:47

Taroccoesbrocco

3,36941331

add a comment | 

up vote
2
down vote

The book is right: roughly speaking, the idea is that you have to compare pairwise the elements of the finite subset of $X$. Let us see that more formally.

Given a finite $Y subseteq X$, I prove by induction on the cardinality $n$ of $Y$ that $X$ has the meet of $Y$.

If $n = 0$, then $Y = emptyset$ and the meet of $Y$ is the top element (which is an element of $X$ by hypothesis).

Let $n > 0$: then, $Y = y_1, dots, y_n$ where the $y_i$'s are pairwise distinct.
By induction hypothesis, $X$ has the meet, say $x$, of $y_1, dots, y_n-1$. By hypothesis, $X$ has the meet, say $x'$, of $x$ and $y_n$. It is easy to prove that $x'$ is the meet of $Y$.
Indeed:

• $x'$ is a lower bound of $Y$, since it is a lower bound of $y_n$ and $x$, and hence of $y_1, dots, y_n-1$ by transitivity (since $x$ is a lower bound of $y_1, dots, y_n-1$);




• any lower bound $z$ of $Y$ is a lower bound of $y_1, dots, y_n-1$ and then $z leq x$ by maximality of $x$, thus $z leq x'$ by transitivity; therefore, $x'$ is greater than or equal to any other lower bound of $Y$.

share|cite|improve this answer

edited Jul 26 at 22:11

answered Jul 26 at 21:47

Taroccoesbrocco

3,36941331

add a comment | 

up vote
2
down vote

up vote
2
down vote

The book is right: roughly speaking, the idea is that you have to compare pairwise the elements of the finite subset of $X$. Let us see that more formally.

Given a finite $Y subseteq X$, I prove by induction on the cardinality $n$ of $Y$ that $X$ has the meet of $Y$.

If $n = 0$, then $Y = emptyset$ and the meet of $Y$ is the top element (which is an element of $X$ by hypothesis).

Let $n > 0$: then, $Y = y_1, dots, y_n$ where the $y_i$'s are pairwise distinct.
By induction hypothesis, $X$ has the meet, say $x$, of $y_1, dots, y_n-1$. By hypothesis, $X$ has the meet, say $x'$, of $x$ and $y_n$. It is easy to prove that $x'$ is the meet of $Y$.
Indeed:

• $x'$ is a lower bound of $Y$, since it is a lower bound of $y_n$ and $x$, and hence of $y_1, dots, y_n-1$ by transitivity (since $x$ is a lower bound of $y_1, dots, y_n-1$);




• any lower bound $z$ of $Y$ is a lower bound of $y_1, dots, y_n-1$ and then $z leq x$ by maximality of $x$, thus $z leq x'$ by transitivity; therefore, $x'$ is greater than or equal to any other lower bound of $Y$.

share|cite|improve this answer

edited Jul 26 at 22:11

answered Jul 26 at 21:47

Taroccoesbrocco

3,36941331

The book is right: roughly speaking, the idea is that you have to compare pairwise the elements of the finite subset of $X$. Let us see that more formally.

Given a finite $Y subseteq X$, I prove by induction on the cardinality $n$ of $Y$ that $X$ has the meet of $Y$.

If $n = 0$, then $Y = emptyset$ and the meet of $Y$ is the top element (which is an element of $X$ by hypothesis).

Let $n > 0$: then, $Y = y_1, dots, y_n$ where the $y_i$'s are pairwise distinct.
By induction hypothesis, $X$ has the meet, say $x$, of $y_1, dots, y_n-1$. By hypothesis, $X$ has the meet, say $x'$, of $x$ and $y_n$. It is easy to prove that $x'$ is the meet of $Y$.
Indeed:

• $x'$ is a lower bound of $Y$, since it is a lower bound of $y_n$ and $x$, and hence of $y_1, dots, y_n-1$ by transitivity (since $x$ is a lower bound of $y_1, dots, y_n-1$);




• any lower bound $z$ of $Y$ is a lower bound of $y_1, dots, y_n-1$ and then $z leq x$ by maximality of $x$, thus $z leq x'$ by transitivity; therefore, $x'$ is greater than or equal to any other lower bound of $Y$.

share|cite|improve this answer

edited Jul 26 at 22:11

answered Jul 26 at 21:47

Taroccoesbrocco

3,36941331

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 22:11

edited Jul 26 at 22:11

edited Jul 26 at 22:11

answered Jul 26 at 21:47

Taroccoesbrocco

3,36941331

answered Jul 26 at 21:47

Taroccoesbrocco

3,36941331

answered Jul 26 at 21:47

Taroccoesbrocco

3,36941331

3,36941331

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863840%2fproperty-of-meets-and-joins-infima-and-suprema-in-a-set%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

Name Email

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

Name Email

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

Name Email

Name Email

Name

Email

Name Email

Name

Email