Well ordered sets: the example of real numbers [on hold] The Next CEO of Stack Overflowquestions about well-orderWith Choice, is any linearly ordered set well-ordered if no subset has order type $omega^*$?Existence of a well-ordered set with a special elementDoes there exist an amazing poset that is neither totally-ordered nor well-founded?Injection from a Well-Ordered Set to $mathbbQ$Can a Proper Partial Order have a Totally-Ordered 'Spine'?Does every nonempty well-ordered set have a least element?Well-founded ordering on natural numbersWell ordering on the quotient of well ordered setsParticular union of well ordered sets that is well ordered
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Well ordered sets: the example of real numbers [on hold]
The Next CEO of Stack Overflowquestions about well-orderWith Choice, is any linearly ordered set well-ordered if no subset has order type $omega^*$?Existence of a well-ordered set with a special elementDoes there exist an amazing poset that is neither totally-ordered nor well-founded?Injection from a Well-Ordered Set to $mathbbQ$Can a Proper Partial Order have a Totally-Ordered 'Spine'?Does every nonempty well-ordered set have a least element?Well-founded ordering on natural numbersWell ordering on the quotient of well ordered setsParticular union of well ordered sets that is well ordered
$begingroup$
I don't understand why $mathbbR$ equipped with $leq$ is considered not well ordered?
I know that the usual answer is "it does not have a least element"
But that conflicts with the definition that states that a set A with the operation $leq$ "is said to be well ordered (or have a well-founded order) iff every nonempty subset of A has a least element "
If I take any subset of the real numbers it has by definition a least element: its lower bound ... can someone explain where is my mistake?
This question is interesting firstly to gain a better understanding of the notion of well ordering and total ordering, for the purpose of mathematical education. In my opinion, just wanting to delve deeper into a question is totally relevant and I do not understand the "put on hold"
Source: http://mathworld.wolfram.com/WellOrderedSet.html
order-theory
asked Mar 27 at 19:37
MachupicchuMachupicchu
239
$endgroup$
put on hold as off-topic by Andrés E. Caicedo, zz20s, Leucippus, dantopa, Abcd 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – zz20s, Leucippus, dantopa, Abcd
|
show 12 more comments
$begingroup$
I don't understand why $mathbbR$ equipped with $leq$ is considered not well ordered?
I know that the usual answer is "it does not have a least element"
But that conflicts with the definition that states that a set A with the operation $leq$ "is said to be well ordered (or have a well-founded order) iff every nonempty subset of A has a least element "
If I take any subset of the real numbers it has by definition a least element: its lower bound ... can someone explain where is my mistake?
This question is interesting firstly to gain a better understanding of the notion of well ordering and total ordering, for the purpose of mathematical education. In my opinion, just wanting to delve deeper into a question is totally relevant and I do not understand the "put on hold"
Source: http://mathworld.wolfram.com/WellOrderedSet.html
order-theory
asked Mar 27 at 19:37
MachupicchuMachupicchu
239
$endgroup$
put on hold as off-topic by Andrés E. Caicedo, zz20s, Leucippus, dantopa, Abcd 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – zz20s, Leucippus, dantopa, Abcd
$begingroup$
For me, 0 would be the "least" element of the set (0,1) what is my mistake?
$endgroup$
– Machupicchu
Mar 27 at 19:43
$begingroup$
My first comment was wrong. $0$ is not in $(0,1)$, so it is not a least element of $(0,1).$ It is the greatest lower bound. Although the subset of negative real numbers has no infimum.
$endgroup$
– Thomas Andrews
Mar 27 at 19:44
$begingroup$
A least element of a set has to be contained in this set, which is not always true of its lower bounds. For instance, $(0,1)$ has lower bounds, but none of them is contained in the set. Besides, some sets don't even have lower bounds, like $(-infty,0)$.
$endgroup$
– Wojowu
Mar 27 at 19:46
$begingroup$
I like to think of well-ordered sets as a set of race results. Given a set $R$ of racers in a race, we might want to find the first (lowest) amongst all the racers, or the first woman, or the first in age 65-70, or the first with a wheelchair. Given anynon-empty subset of $R,$ we might want to know which racer "won" amongst that subset.
$endgroup$
– Thomas Andrews
Mar 27 at 19:48
2
$begingroup$
What is the "real number just after" another real number? If $r<r'$ are real numbers, we can show that there is a real number between them, namely the average $r<fracr+r'2<r'.$ So there is no "next" real number after $r.$
$endgroup$
– Thomas Andrews
Mar 27 at 19:57
|
show 12 more comments
$begingroup$
I don't understand why $mathbbR$ equipped with $leq$ is considered not well ordered?
I know that the usual answer is "it does not have a least element"
But that conflicts with the definition that states that a set A with the operation $leq$ "is said to be well ordered (or have a well-founded order) iff every nonempty subset of A has a least element "
If I take any subset of the real numbers it has by definition a least element: its lower bound ... can someone explain where is my mistake?
This question is interesting firstly to gain a better understanding of the notion of well ordering and total ordering, for the purpose of mathematical education. In my opinion, just wanting to delve deeper into a question is totally relevant and I do not understand the "put on hold"
Source: http://mathworld.wolfram.com/WellOrderedSet.html
order-theory
asked Mar 27 at 19:37
MachupicchuMachupicchu
239
$endgroup$
I don't understand why $mathbbR$ equipped with $leq$ is considered not well ordered?
I know that the usual answer is "it does not have a least element"
But that conflicts with the definition that states that a set A with the operation $leq$ "is said to be well ordered (or have a well-founded order) iff every nonempty subset of A has a least element "
If I take any subset of the real numbers it has by definition a least element: its lower bound ... can someone explain where is my mistake?
This question is interesting firstly to gain a better understanding of the notion of well ordering and total ordering, for the purpose of mathematical education. In my opinion, just wanting to delve deeper into a question is totally relevant and I do not understand the "put on hold"
Source: http://mathworld.wolfram.com/WellOrderedSet.html
order-theory
order-theory
asked Mar 27 at 19:37
MachupicchuMachupicchu
239
asked Mar 27 at 19:37
MachupicchuMachupicchu
239
edited 2 days ago
Machupicchu
asked Mar 27 at 19:37
MachupicchuMachupicchu
239
asked Mar 27 at 19:37
MachupicchuMachupicchu
239
asked Mar 27 at 19:37
MachupicchuMachupicchu
239
239
put on hold as off-topic by Andrés E. Caicedo, zz20s, Leucippus, dantopa, Abcd 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – zz20s, Leucippus, dantopa, Abcd
put on hold as off-topic by Andrés E. Caicedo, zz20s, Leucippus, dantopa, Abcd 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – zz20s, Leucippus, dantopa, Abcd
$begingroup$
For me, 0 would be the "least" element of the set (0,1) what is my mistake?
$endgroup$
– Machupicchu
Mar 27 at 19:43
$begingroup$
My first comment was wrong. $0$ is not in $(0,1)$, so it is not a least element of $(0,1).$ It is the greatest lower bound. Although the subset of negative real numbers has no infimum.
$endgroup$
– Thomas Andrews
Mar 27 at 19:44
$begingroup$
A least element of a set has to be contained in this set, which is not always true of its lower bounds. For instance, $(0,1)$ has lower bounds, but none of them is contained in the set. Besides, some sets don't even have lower bounds, like $(-infty,0)$.
$endgroup$
– Wojowu
Mar 27 at 19:46
$begingroup$
I like to think of well-ordered sets as a set of race results. Given a set $R$ of racers in a race, we might want to find the first (lowest) amongst all the racers, or the first woman, or the first in age 65-70, or the first with a wheelchair. Given anynon-empty subset of $R,$ we might want to know which racer "won" amongst that subset.
$endgroup$
– Thomas Andrews
Mar 27 at 19:48
2
$begingroup$
What is the "real number just after" another real number? If $r<r'$ are real numbers, we can show that there is a real number between them, namely the average $r<fracr+r'2<r'.$ So there is no "next" real number after $r.$
$endgroup$
– Thomas Andrews
Mar 27 at 19:57
|
show 12 more comments
$begingroup$
For me, 0 would be the "least" element of the set (0,1) what is my mistake?
$endgroup$
– Machupicchu
Mar 27 at 19:43
$begingroup$
My first comment was wrong. $0$ is not in $(0,1)$, so it is not a least element of $(0,1).$ It is the greatest lower bound. Although the subset of negative real numbers has no infimum.
$endgroup$
– Thomas Andrews
Mar 27 at 19:44
$begingroup$
A least element of a set has to be contained in this set, which is not always true of its lower bounds. For instance, $(0,1)$ has lower bounds, but none of them is contained in the set. Besides, some sets don't even have lower bounds, like $(-infty,0)$.
$endgroup$
– Wojowu
Mar 27 at 19:46
$begingroup$
I like to think of well-ordered sets as a set of race results. Given a set $R$ of racers in a race, we might want to find the first (lowest) amongst all the racers, or the first woman, or the first in age 65-70, or the first with a wheelchair. Given anynon-empty subset of $R,$ we might want to know which racer "won" amongst that subset.
$endgroup$
– Thomas Andrews
Mar 27 at 19:48
2
$begingroup$
What is the "real number just after" another real number? If $r<r'$ are real numbers, we can show that there is a real number between them, namely the average $r<fracr+r'2<r'.$ So there is no "next" real number after $r.$
$endgroup$
– Thomas Andrews
Mar 27 at 19:57
$begingroup$
For me, 0 would be the "least" element of the set (0,1) what is my mistake?
$endgroup$
– Machupicchu
Mar 27 at 19:43
$begingroup$
For me, 0 would be the "least" element of the set (0,1) what is my mistake?
$endgroup$
– Machupicchu
Mar 27 at 19:43
$begingroup$
My first comment was wrong. $0$ is not in $(0,1)$, so it is not a least element of $(0,1).$ It is the greatest lower bound. Although the subset of negative real numbers has no infimum.
$endgroup$
– Thomas Andrews
Mar 27 at 19:44
$begingroup$
My first comment was wrong. $0$ is not in $(0,1)$, so it is not a least element of $(0,1).$ It is the greatest lower bound. Although the subset of negative real numbers has no infimum.
$endgroup$
– Thomas Andrews
Mar 27 at 19:44
$begingroup$
A least element of a set has to be contained in this set, which is not always true of its lower bounds. For instance, $(0,1)$ has lower bounds, but none of them is contained in the set. Besides, some sets don't even have lower bounds, like $(-infty,0)$.
$endgroup$
– Wojowu
Mar 27 at 19:46
$begingroup$
A least element of a set has to be contained in this set, which is not always true of its lower bounds. For instance, $(0,1)$ has lower bounds, but none of them is contained in the set. Besides, some sets don't even have lower bounds, like $(-infty,0)$.
$endgroup$
– Wojowu
Mar 27 at 19:46
$begingroup$
I like to think of well-ordered sets as a set of race results. Given a set $R$ of racers in a race, we might want to find the first (lowest) amongst all the racers, or the first woman, or the first in age 65-70, or the first with a wheelchair. Given anynon-empty subset of $R,$ we might want to know which racer "won" amongst that subset.
$endgroup$
– Thomas Andrews
Mar 27 at 19:48
$begingroup$
I like to think of well-ordered sets as a set of race results. Given a set $R$ of racers in a race, we might want to find the first (lowest) amongst all the racers, or the first woman, or the first in age 65-70, or the first with a wheelchair. Given anynon-empty subset of $R,$ we might want to know which racer "won" amongst that subset.
$endgroup$
– Thomas Andrews
Mar 27 at 19:48
2
2
$begingroup$
What is the "real number just after" another real number? If $r<r'$ are real numbers, we can show that there is a real number between them, namely the average $r<fracr+r'2<r'.$ So there is no "next" real number after $r.$
$endgroup$
– Thomas Andrews
Mar 27 at 19:57
$begingroup$
What is the "real number just after" another real number? If $r<r'$ are real numbers, we can show that there is a real number between them, namely the average $r<fracr+r'2<r'.$ So there is no "next" real number after $r.$
$endgroup$
– Thomas Andrews
Mar 27 at 19:57
|
show 12 more comments
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$begingroup$
For me, 0 would be the "least" element of the set (0,1) what is my mistake?
$endgroup$
– Machupicchu
Mar 27 at 19:43
$begingroup$
My first comment was wrong. $0$ is not in $(0,1)$, so it is not a least element of $(0,1).$ It is the greatest lower bound. Although the subset of negative real numbers has no infimum.
$endgroup$
– Thomas Andrews
Mar 27 at 19:44
$begingroup$
A least element of a set has to be contained in this set, which is not always true of its lower bounds. For instance, $(0,1)$ has lower bounds, but none of them is contained in the set. Besides, some sets don't even have lower bounds, like $(-infty,0)$.
$endgroup$
– Wojowu
Mar 27 at 19:46
$begingroup$
I like to think of well-ordered sets as a set of race results. Given a set $R$ of racers in a race, we might want to find the first (lowest) amongst all the racers, or the first woman, or the first in age 65-70, or the first with a wheelchair. Given anynon-empty subset of $R,$ we might want to know which racer "won" amongst that subset.
$endgroup$
– Thomas Andrews
Mar 27 at 19:48
2
$begingroup$
What is the "real number just after" another real number? If $r<r'$ are real numbers, we can show that there is a real number between them, namely the average $r<fracr+r'2<r'.$ So there is no "next" real number after $r.$
$endgroup$
– Thomas Andrews
Mar 27 at 19:57