Mixing
Is every set relation $R$ that is extensional and well-founded, necessarily linear?
$begingroup$
Let $R$ be the relation. Due to extensionality and well-foundedness, exists a unique minimal element and a unique successor, And the one after that, and a unique limit point for them. Does this necessarily mean that the relation is a well-ordering? I cannot find any counterexample.
elementary-set-theory
edited Dec 22 '18 at 14:08
Holo
5,75521031
asked Dec 22 '18 at 14:06
Uri George PeterzilUri George Peterzil
929
$endgroup$
add a comment |
$begingroup$
Let $R$ be the relation. Due to extensionality and well-foundedness, exists a unique minimal element and a unique successor, And the one after that, and a unique limit point for them. Does this necessarily mean that the relation is a well-ordering? I cannot find any counterexample.
elementary-set-theory
edited Dec 22 '18 at 14:08
Holo
5,75521031
asked Dec 22 '18 at 14:06
Uri George PeterzilUri George Peterzil
929
$endgroup$
$begingroup$
What do you mean by $R$ is extensional?
$endgroup$
– Holo
Dec 22 '18 at 14:09
$begingroup$
If ${x|x<x_1 }={x|x<x_2 }$ then $x_1 =x_2$.
$endgroup$
– Uri George Peterzil
Dec 22 '18 at 14:11
2
$begingroup$
What's the definition of well-founded you're using? Because I'm familiar with this one, which does not give you that minimal elements are unique, or that you have a unique successor operation.
$endgroup$
– Malice Vidrine
Dec 22 '18 at 17:43
add a comment |
$begingroup$
Let $R$ be the relation. Due to extensionality and well-foundedness, exists a unique minimal element and a unique successor, And the one after that, and a unique limit point for them. Does this necessarily mean that the relation is a well-ordering? I cannot find any counterexample.
elementary-set-theory
edited Dec 22 '18 at 14:08
Holo
5,75521031
asked Dec 22 '18 at 14:06
Uri George PeterzilUri George Peterzil
929
$endgroup$
Let $R$ be the relation. Due to extensionality and well-foundedness, exists a unique minimal element and a unique successor, And the one after that, and a unique limit point for them. Does this necessarily mean that the relation is a well-ordering? I cannot find any counterexample.
elementary-set-theory
elementary-set-theory
edited Dec 22 '18 at 14:08
Holo
5,75521031
asked Dec 22 '18 at 14:06
Uri George PeterzilUri George Peterzil
929
edited Dec 22 '18 at 14:08
Holo
5,75521031
asked Dec 22 '18 at 14:06
Uri George PeterzilUri George Peterzil
929
edited Dec 22 '18 at 14:08
Holo
5,75521031
edited Dec 22 '18 at 14:08
Holo
5,75521031
edited Dec 22 '18 at 14:08
Holo
5,75521031
5,75521031
asked Dec 22 '18 at 14:06
Uri George PeterzilUri George Peterzil
929
asked Dec 22 '18 at 14:06
Uri George PeterzilUri George Peterzil
929
asked Dec 22 '18 at 14:06
Uri George PeterzilUri George Peterzil
929
929
$begingroup$
What do you mean by $R$ is extensional?
$endgroup$
– Holo
Dec 22 '18 at 14:09
$begingroup$
If ${x|x<x_1 }={x|x<x_2 }$ then $x_1 =x_2$.
$endgroup$
– Uri George Peterzil
Dec 22 '18 at 14:11
2
$begingroup$
What's the definition of well-founded you're using? Because I'm familiar with this one, which does not give you that minimal elements are unique, or that you have a unique successor operation.
$endgroup$
– Malice Vidrine
Dec 22 '18 at 17:43
add a comment |
$begingroup$
What do you mean by $R$ is extensional?
$endgroup$
– Holo
Dec 22 '18 at 14:09
$begingroup$
If ${x|x<x_1 }={x|x<x_2 }$ then $x_1 =x_2$.
$endgroup$
– Uri George Peterzil
Dec 22 '18 at 14:11
2
$begingroup$
What's the definition of well-founded you're using? Because I'm familiar with this one, which does not give you that minimal elements are unique, or that you have a unique successor operation.
$endgroup$
– Malice Vidrine
Dec 22 '18 at 17:43
$begingroup$
What do you mean by $R$ is extensional?
$endgroup$
– Holo
Dec 22 '18 at 14:09
$begingroup$
What do you mean by $R$ is extensional?
$endgroup$
– Holo
Dec 22 '18 at 14:09
$begingroup$
If ${x|x<x_1 }={x|x<x_2 }$ then $x_1 =x_2$.
$endgroup$
– Uri George Peterzil
Dec 22 '18 at 14:11
$begingroup$
If ${x|x<x_1 }={x|x<x_2 }$ then $x_1 =x_2$.
$endgroup$
– Uri George Peterzil
Dec 22 '18 at 14:11
2
2
$begingroup$
What's the definition of well-founded you're using? Because I'm familiar with this one, which does not give you that minimal elements are unique, or that you have a unique successor operation.
$endgroup$
– Malice Vidrine
Dec 22 '18 at 17:43
$begingroup$
What's the definition of well-founded you're using? Because I'm familiar with this one, which does not give you that minimal elements are unique, or that you have a unique successor operation.
$endgroup$
– Malice Vidrine
Dec 22 '18 at 17:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A relation that is extensional and well-founded is not necessarily a well-order.
For instance take relation $in_A$ on transitive set: $$A={0,1,2,{1,2}}$$ where $0:=varnothing$, $1:={0}$ and $2:={0,1}$.
The elements $0$ and ${1,2}$ are incomparable.
A relation that is extensional and well-founded will be a well-order iff the relation is also transitive.
Further you might be interested in the Mostowski collapse lemma.
answered Dec 22 '18 at 14:42
drhabdrhab
101k544130
$endgroup$
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A relation that is extensional and well-founded is not necessarily a well-order.
For instance take relation $in_A$ on transitive set: $$A={0,1,2,{1,2}}$$ where $0:=varnothing$, $1:={0}$ and $2:={0,1}$.
The elements $0$ and ${1,2}$ are incomparable.
A relation that is extensional and well-founded will be a well-order iff the relation is also transitive.
Further you might be interested in the Mostowski collapse lemma.
answered Dec 22 '18 at 14:42
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
A relation that is extensional and well-founded is not necessarily a well-order.
For instance take relation $in_A$ on transitive set: $$A={0,1,2,{1,2}}$$ where $0:=varnothing$, $1:={0}$ and $2:={0,1}$.
The elements $0$ and ${1,2}$ are incomparable.
A relation that is extensional and well-founded will be a well-order iff the relation is also transitive.
Further you might be interested in the Mostowski collapse lemma.
answered Dec 22 '18 at 14:42
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
A relation that is extensional and well-founded is not necessarily a well-order.
For instance take relation $in_A$ on transitive set: $$A={0,1,2,{1,2}}$$ where $0:=varnothing$, $1:={0}$ and $2:={0,1}$.
The elements $0$ and ${1,2}$ are incomparable.
A relation that is extensional and well-founded will be a well-order iff the relation is also transitive.
Further you might be interested in the Mostowski collapse lemma.
answered Dec 22 '18 at 14:42
drhabdrhab
101k544130
$endgroup$
A relation that is extensional and well-founded is not necessarily a well-order.
For instance take relation $in_A$ on transitive set: $$A={0,1,2,{1,2}}$$ where $0:=varnothing$, $1:={0}$ and $2:={0,1}$.
The elements $0$ and ${1,2}$ are incomparable.
A relation that is extensional and well-founded will be a well-order iff the relation is also transitive.
Further you might be interested in the Mostowski collapse lemma.
answered Dec 22 '18 at 14:42
drhabdrhab
101k544130
edited Jan 12 at 7:49
answered Dec 22 '18 at 14:42
drhabdrhab
101k544130
answered Dec 22 '18 at 14:42
drhabdrhab
101k544130
answered Dec 22 '18 at 14:42
drhabdrhab
101k544130
101k544130
add a comment |
add a comment |
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$begingroup$
What do you mean by $R$ is extensional?
$endgroup$
– Holo
Dec 22 '18 at 14:09
$begingroup$
If ${x|x<x_1 }={x|x<x_2 }$ then $x_1 =x_2$.
$endgroup$
– Uri George Peterzil
Dec 22 '18 at 14:11
2
$begingroup$
What's the definition of well-founded you're using? Because I'm familiar with this one, which does not give you that minimal elements are unique, or that you have a unique successor operation.
$endgroup$
– Malice Vidrine
Dec 22 '18 at 17:43