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$begingroup$

I have the following problem:

Let $X=2, 3, 4, 5, ldots$ be ordered by division, that is:

$$xleq yLeftrightarrow xmid y,$$
and let $mathcalS=(A, leq_A): Asubset X$ where $leq_A$ is the order induced on $A$ by $leq$ via restriction. Then we can order $mathcalS$ via inclusion, that is:

$$Aleq_mathcalS BLeftrightarrow Asubset B.$$

Find the minimal and maximal elements of $mathcalS$ via $leq_mathcalS$.

What I don't understand about this problem is: why should one consider $mathcalS$ as subsets of $X$ ordered by the induced order? I mean, why can't it just be: Let $mathcalS$ be the subsets of $X$ ordered by inclusion?

The minimal elements should be the unitary sets and the maximal element should be $X$, right?

abstract-algebra order-theory

share|cite|improve this question

edited yesterday

user26857

39.3k124183

asked yesterday

PtFPtF

4,04821734

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can you give us a reference for this question?
  $endgroup$
  – user26857
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But $mathcal S$ is indeed ordered by inclusion. The elements of $mathcal S$, which are subsets of $X$ are the ones which are ordered by the order induced from $X$ by inclusion. About the other question, if $varnothing$ is an element of $mathcal S$ (and so it seems), then it is the only minimal element; the only maximal one is $X$, as you suspected.
  $endgroup$
  – amrsa
  yesterday
  

  
  
  
  

add a comment | 

0

$begingroup$

I have the following problem:

Let $X=2, 3, 4, 5, ldots$ be ordered by division, that is:

$$xleq yLeftrightarrow xmid y,$$
and let $mathcalS=(A, leq_A): Asubset X$ where $leq_A$ is the order induced on $A$ by $leq$ via restriction. Then we can order $mathcalS$ via inclusion, that is:

$$Aleq_mathcalS BLeftrightarrow Asubset B.$$

Find the minimal and maximal elements of $mathcalS$ via $leq_mathcalS$.

What I don't understand about this problem is: why should one consider $mathcalS$ as subsets of $X$ ordered by the induced order? I mean, why can't it just be: Let $mathcalS$ be the subsets of $X$ ordered by inclusion?

The minimal elements should be the unitary sets and the maximal element should be $X$, right?

abstract-algebra order-theory

share|cite|improve this question

edited yesterday

user26857

39.3k124183

asked yesterday

PtFPtF

4,04821734

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can you give us a reference for this question?
  $endgroup$
  – user26857
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But $mathcal S$ is indeed ordered by inclusion. The elements of $mathcal S$, which are subsets of $X$ are the ones which are ordered by the order induced from $X$ by inclusion. About the other question, if $varnothing$ is an element of $mathcal S$ (and so it seems), then it is the only minimal element; the only maximal one is $X$, as you suspected.
  $endgroup$
  – amrsa
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

I have the following problem:

Let $X=2, 3, 4, 5, ldots$ be ordered by division, that is:

$$xleq yLeftrightarrow xmid y,$$
and let $mathcalS=(A, leq_A): Asubset X$ where $leq_A$ is the order induced on $A$ by $leq$ via restriction. Then we can order $mathcalS$ via inclusion, that is:

$$Aleq_mathcalS BLeftrightarrow Asubset B.$$

Find the minimal and maximal elements of $mathcalS$ via $leq_mathcalS$.

What I don't understand about this problem is: why should one consider $mathcalS$ as subsets of $X$ ordered by the induced order? I mean, why can't it just be: Let $mathcalS$ be the subsets of $X$ ordered by inclusion?

The minimal elements should be the unitary sets and the maximal element should be $X$, right?

abstract-algebra order-theory

share|cite|improve this question

edited yesterday

user26857

39.3k124183

asked yesterday

PtFPtF

4,04821734

$endgroup$

share|cite|improve this question

edited yesterday

user26857

39.3k124183

asked yesterday

PtFPtF

4,04821734

share|cite|improve this question

edited yesterday

user26857

39.3k124183

asked yesterday

PtFPtF

4,04821734

edited yesterday

user26857

39.3k124183

edited yesterday

user26857

39.3k124183

asked yesterday

PtFPtF

4,04821734

asked yesterday

PtFPtF

4,04821734