Question in Proof That Every Well-Ordering is a Total-Ordering












0












$begingroup$


I am trying to make sense of the following:



Theorem: Let $R$ be a relation on a set $A$.
If $R$ is a well-ordered relation on $A$, then $R$ is a total-order relation on $A$.



Proof: Suppose $x,yin A.$ Let $R={x,y}not=emptyset$. By assumption, $R$ has a least element. If it is $x$, then $xleq y$. If it is $y$, then $yleq x$.



$$text{__________________________________}$$



For a relation to be total ordered, then it must be that for each $x,yin A$, $xRy$ and $yRx$. In this case we are comparing $x , y$ where one or the other is a least element. What about if neither is a least element though? Do we then simply say that $x,y$ must relate because if we consider whatever is the least element, say $z$, then both $x,y$ can relate to $z$ and therefore must relate to each other?






elementary-set-theory relations well-orders







share|cite|improve this question









$endgroup$












  • $begingroup$
    One of them must be the least element of ${x,y}$ because there are no other elements in the set, and it has a least element.
    $endgroup$
    – dbx
    Dec 5 '18 at 15:13










  • $begingroup$
    Where does this proof come from? In the definitions as I've learned them, well-ordered relations are defined to be both well-founded and totally ordered, and well-founded relations need not be totally ordered. See also math.stackexchange.com/questions/116642/…
    $endgroup$
    – J.G.
    Dec 5 '18 at 15:19










  • $begingroup$
    The definition my book uses denotes well-ordered relation as a partially-ordered relation on $A$ such that each non-empty subset of $A$ has a least element with respect to $R$.
    $endgroup$
    – M. Damon
    Dec 5 '18 at 15:26










  • $begingroup$
    @J.G.: Well-founded relations have a minimal element in each non-empty set; well-orders have a minimum in each non-empty set.
    $endgroup$
    – Asaf Karagila♦
    Dec 5 '18 at 15:27










  • $begingroup$
    @AsafKaragila In other words, the minimal element of the non-empty set is only required to be unique in the latter case? Fair enough.
    $endgroup$
    – J.G.
    Dec 5 '18 at 15:31
















0












$begingroup$


I am trying to make sense of the following:



Theorem: Let $R$ be a relation on a set $A$.
If $R$ is a well-ordered relation on $A$, then $R$ is a total-order relation on $A$.



Proof: Suppose $x,yin A.$ Let $R={x,y}not=emptyset$. By assumption, $R$ has a least element. If it is $x$, then $xleq y$. If it is $y$, then $yleq x$.



$$text{__________________________________}$$



For a relation to be total ordered, then it must be that for each $x,yin A$, $xRy$ and $yRx$. In this case we are comparing $x , y$ where one or the other is a least element. What about if neither is a least element though? Do we then simply say that $x,y$ must relate because if we consider whatever is the least element, say $z$, then both $x,y$ can relate to $z$ and therefore must relate to each other?






elementary-set-theory relations well-orders







share|cite|improve this question









$endgroup$












  • $begingroup$
    One of them must be the least element of ${x,y}$ because there are no other elements in the set, and it has a least element.
    $endgroup$
    – dbx
    Dec 5 '18 at 15:13










  • $begingroup$
    Where does this proof come from? In the definitions as I've learned them, well-ordered relations are defined to be both well-founded and totally ordered, and well-founded relations need not be totally ordered. See also math.stackexchange.com/questions/116642/…
    $endgroup$
    – J.G.
    Dec 5 '18 at 15:19










  • $begingroup$
    The definition my book uses denotes well-ordered relation as a partially-ordered relation on $A$ such that each non-empty subset of $A$ has a least element with respect to $R$.
    $endgroup$
    – M. Damon
    Dec 5 '18 at 15:26










  • $begingroup$
    @J.G.: Well-founded relations have a minimal element in each non-empty set; well-orders have a minimum in each non-empty set.
    $endgroup$
    – Asaf Karagila♦
    Dec 5 '18 at 15:27










  • $begingroup$
    @AsafKaragila In other words, the minimal element of the non-empty set is only required to be unique in the latter case? Fair enough.
    $endgroup$
    – J.G.
    Dec 5 '18 at 15:31














0












0








0





$begingroup$


I am trying to make sense of the following:



Theorem: Let $R$ be a relation on a set $A$.
If $R$ is a well-ordered relation on $A$, then $R$ is a total-order relation on $A$.



Proof: Suppose $x,yin A.$ Let $R={x,y}not=emptyset$. By assumption, $R$ has a least element. If it is $x$, then $xleq y$. If it is $y$, then $yleq x$.



$$text{__________________________________}$$



For a relation to be total ordered, then it must be that for each $x,yin A$, $xRy$ and $yRx$. In this case we are comparing $x , y$ where one or the other is a least element. What about if neither is a least element though? Do we then simply say that $x,y$ must relate because if we consider whatever is the least element, say $z$, then both $x,y$ can relate to $z$ and therefore must relate to each other?






elementary-set-theory relations well-orders







share|cite|improve this question









$endgroup$




I am trying to make sense of the following:



Theorem: Let $R$ be a relation on a set $A$.
If $R$ is a well-ordered relation on $A$, then $R$ is a total-order relation on $A$.



Proof: Suppose $x,yin A.$ Let $R={x,y}not=emptyset$. By assumption, $R$ has a least element. If it is $x$, then $xleq y$. If it is $y$, then $yleq x$.



$$text{__________________________________}$$



For a relation to be total ordered, then it must be that for each $x,yin A$, $xRy$ and $yRx$. In this case we are comparing $x , y$ where one or the other is a least element. What about if neither is a least element though? Do we then simply say that $x,y$ must relate because if we consider whatever is the least element, say $z$, then both $x,y$ can relate to $z$ and therefore must relate to each other?






elementary-set-theory relations well-orders




elementary-set-theory relations well-orders






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 15:12









M. Damon M. Damon

235




235












  • $begingroup$
    One of them must be the least element of ${x,y}$ because there are no other elements in the set, and it has a least element.
    $endgroup$
    – dbx
    Dec 5 '18 at 15:13










  • $begingroup$
    Where does this proof come from? In the definitions as I've learned them, well-ordered relations are defined to be both well-founded and totally ordered, and well-founded relations need not be totally ordered. See also math.stackexchange.com/questions/116642/…
    $endgroup$
    – J.G.
    Dec 5 '18 at 15:19










  • $begingroup$
    The definition my book uses denotes well-ordered relation as a partially-ordered relation on $A$ such that each non-empty subset of $A$ has a least element with respect to $R$.
    $endgroup$
    – M. Damon
    Dec 5 '18 at 15:26










  • $begingroup$
    @J.G.: Well-founded relations have a minimal element in each non-empty set; well-orders have a minimum in each non-empty set.
    $endgroup$
    – Asaf Karagila♦
    Dec 5 '18 at 15:27










  • $begingroup$
    @AsafKaragila In other words, the minimal element of the non-empty set is only required to be unique in the latter case? Fair enough.
    $endgroup$
    – J.G.
    Dec 5 '18 at 15:31


















  • $begingroup$
    One of them must be the least element of ${x,y}$ because there are no other elements in the set, and it has a least element.
    $endgroup$
    – dbx
    Dec 5 '18 at 15:13










  • $begingroup$
    Where does this proof come from? In the definitions as I've learned them, well-ordered relations are defined to be both well-founded and totally ordered, and well-founded relations need not be totally ordered. See also math.stackexchange.com/questions/116642/…
    $endgroup$
    – J.G.
    Dec 5 '18 at 15:19










  • $begingroup$
    The definition my book uses denotes well-ordered relation as a partially-ordered relation on $A$ such that each non-empty subset of $A$ has a least element with respect to $R$.
    $endgroup$
    – M. Damon
    Dec 5 '18 at 15:26










  • $begingroup$
    @J.G.: Well-founded relations have a minimal element in each non-empty set; well-orders have a minimum in each non-empty set.
    $endgroup$
    – Asaf Karagila♦
    Dec 5 '18 at 15:27










  • $begingroup$
    @AsafKaragila In other words, the minimal element of the non-empty set is only required to be unique in the latter case? Fair enough.
    $endgroup$
    – J.G.
    Dec 5 '18 at 15:31
















$begingroup$
One of them must be the least element of ${x,y}$ because there are no other elements in the set, and it has a least element.
$endgroup$
– dbx
Dec 5 '18 at 15:13




$begingroup$
One of them must be the least element of ${x,y}$ because there are no other elements in the set, and it has a least element.
$endgroup$
– dbx
Dec 5 '18 at 15:13












$begingroup$
Where does this proof come from? In the definitions as I've learned them, well-ordered relations are defined to be both well-founded and totally ordered, and well-founded relations need not be totally ordered. See also math.stackexchange.com/questions/116642/…
$endgroup$
– J.G.
Dec 5 '18 at 15:19




$begingroup$
Where does this proof come from? In the definitions as I've learned them, well-ordered relations are defined to be both well-founded and totally ordered, and well-founded relations need not be totally ordered. See also math.stackexchange.com/questions/116642/…
$endgroup$
– J.G.
Dec 5 '18 at 15:19












$begingroup$
The definition my book uses denotes well-ordered relation as a partially-ordered relation on $A$ such that each non-empty subset of $A$ has a least element with respect to $R$.
$endgroup$
– M. Damon
Dec 5 '18 at 15:26




$begingroup$
The definition my book uses denotes well-ordered relation as a partially-ordered relation on $A$ such that each non-empty subset of $A$ has a least element with respect to $R$.
$endgroup$
– M. Damon
Dec 5 '18 at 15:26












$begingroup$
@J.G.: Well-founded relations have a minimal element in each non-empty set; well-orders have a minimum in each non-empty set.
$endgroup$
– Asaf Karagila♦
Dec 5 '18 at 15:27




$begingroup$
@J.G.: Well-founded relations have a minimal element in each non-empty set; well-orders have a minimum in each non-empty set.
$endgroup$
– Asaf Karagila♦
Dec 5 '18 at 15:27












$begingroup$
@AsafKaragila In other words, the minimal element of the non-empty set is only required to be unique in the latter case? Fair enough.
$endgroup$
– J.G.
Dec 5 '18 at 15:31




$begingroup$
@AsafKaragila In other words, the minimal element of the non-empty set is only required to be unique in the latter case? Fair enough.
$endgroup$
– J.G.
Dec 5 '18 at 15:31










1 Answer
1






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oldest

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1












$begingroup$

How could none of them be the last element? There is no other element in ${x,y}$ besides $x$ and $y$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    +1 To expand on this answer, recall that ${x,y}$ is nonempty since $x$ and $y$ are elements (it is still possible $x$ and $y$ are equal). Because $R$ is well-ordered, there must exist a $R$-least-element in ${x,y}$.
    $endgroup$
    – Alberto Takase
    Dec 6 '18 at 10:39











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

How could none of them be the last element? There is no other element in ${x,y}$ besides $x$ and $y$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    +1 To expand on this answer, recall that ${x,y}$ is nonempty since $x$ and $y$ are elements (it is still possible $x$ and $y$ are equal). Because $R$ is well-ordered, there must exist a $R$-least-element in ${x,y}$.
    $endgroup$
    – Alberto Takase
    Dec 6 '18 at 10:39
















1












$begingroup$

How could none of them be the last element? There is no other element in ${x,y}$ besides $x$ and $y$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    +1 To expand on this answer, recall that ${x,y}$ is nonempty since $x$ and $y$ are elements (it is still possible $x$ and $y$ are equal). Because $R$ is well-ordered, there must exist a $R$-least-element in ${x,y}$.
    $endgroup$
    – Alberto Takase
    Dec 6 '18 at 10:39














1












1








1





$begingroup$

How could none of them be the last element? There is no other element in ${x,y}$ besides $x$ and $y$.






share|cite|improve this answer









$endgroup$



How could none of them be the last element? There is no other element in ${x,y}$ besides $x$ and $y$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 15:16









José Carlos SantosJosé Carlos Santos

156k22126227




156k22126227












  • $begingroup$
    +1 To expand on this answer, recall that ${x,y}$ is nonempty since $x$ and $y$ are elements (it is still possible $x$ and $y$ are equal). Because $R$ is well-ordered, there must exist a $R$-least-element in ${x,y}$.
    $endgroup$
    – Alberto Takase
    Dec 6 '18 at 10:39


















  • $begingroup$
    +1 To expand on this answer, recall that ${x,y}$ is nonempty since $x$ and $y$ are elements (it is still possible $x$ and $y$ are equal). Because $R$ is well-ordered, there must exist a $R$-least-element in ${x,y}$.
    $endgroup$
    – Alberto Takase
    Dec 6 '18 at 10:39
















$begingroup$
+1 To expand on this answer, recall that ${x,y}$ is nonempty since $x$ and $y$ are elements (it is still possible $x$ and $y$ are equal). Because $R$ is well-ordered, there must exist a $R$-least-element in ${x,y}$.
$endgroup$
– Alberto Takase
Dec 6 '18 at 10:39




$begingroup$
+1 To expand on this answer, recall that ${x,y}$ is nonempty since $x$ and $y$ are elements (it is still possible $x$ and $y$ are equal). Because $R$ is well-ordered, there must exist a $R$-least-element in ${x,y}$.
$endgroup$
– Alberto Takase
Dec 6 '18 at 10:39


















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