Poset is complete iff it is cocomplete
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In Awodey's Category Theory, page 130, he says:
A poset is (co)-complete if it is so as a category, thus if it has all set-indexed meets (resp. joins). For posets, completeness and cocompleteness are equivalent. A lattice, Heyting algebra, Boolean algebra, etc. is called complete if it is so as a poset.
As an example, I'll take the poset $omega$, which is well-ordered. A subset of a well-ordered set has a least element, so $omega$ has all meets.
However, $omega$ itself (as a subset of $omega$) has no join. Indeed, a join would be $a$ such that for all $c in omega$, have "all $a_i leq c$" iff $a leq c$. Since there is no $c in omega$ such that for all $a_i in omega$, $a_i leq c$, we must therefore have $a not leq c$ for all $c in omega$. But no $a$ has this property, so $omega$ doesn't have a join.
In summary, $omega$ has all meets, but doesn't have all joins. That is, it is complete but not cocomplete.
What have I done wrong? Awodey makes the distinction between a Boolean algebra and a poset here, so we don't necessarily have access to the nice properties of Boolean algebras.
category-theory boolean-algebra
asked Oct 1 '15 at 9:38
Patrick Stevens
26.9k52669
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up vote
2
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In Awodey's Category Theory, page 130, he says:
A poset is (co)-complete if it is so as a category, thus if it has all set-indexed meets (resp. joins). For posets, completeness and cocompleteness are equivalent. A lattice, Heyting algebra, Boolean algebra, etc. is called complete if it is so as a poset.
As an example, I'll take the poset $omega$, which is well-ordered. A subset of a well-ordered set has a least element, so $omega$ has all meets.
However, $omega$ itself (as a subset of $omega$) has no join. Indeed, a join would be $a$ such that for all $c in omega$, have "all $a_i leq c$" iff $a leq c$. Since there is no $c in omega$ such that for all $a_i in omega$, $a_i leq c$, we must therefore have $a not leq c$ for all $c in omega$. But no $a$ has this property, so $omega$ doesn't have a join.
In summary, $omega$ has all meets, but doesn't have all joins. That is, it is complete but not cocomplete.
What have I done wrong? Awodey makes the distinction between a Boolean algebra and a poset here, so we don't necessarily have access to the nice properties of Boolean algebras.
category-theory boolean-algebra
asked Oct 1 '15 at 9:38
Patrick Stevens
26.9k52669
2
You need to consider degenerate cases like nullary meets/joins. $omega$ does not even have finite meets: it does not have a maximum element.
– Zhen Lin
Oct 1 '15 at 10:07
In fact, a category has finite products iff it has binary products and a terminal object (AKA the empty product).
– Najib Idrissi
Oct 1 '15 at 14:17
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In Awodey's Category Theory, page 130, he says:
A poset is (co)-complete if it is so as a category, thus if it has all set-indexed meets (resp. joins). For posets, completeness and cocompleteness are equivalent. A lattice, Heyting algebra, Boolean algebra, etc. is called complete if it is so as a poset.
As an example, I'll take the poset $omega$, which is well-ordered. A subset of a well-ordered set has a least element, so $omega$ has all meets.
However, $omega$ itself (as a subset of $omega$) has no join. Indeed, a join would be $a$ such that for all $c in omega$, have "all $a_i leq c$" iff $a leq c$. Since there is no $c in omega$ such that for all $a_i in omega$, $a_i leq c$, we must therefore have $a not leq c$ for all $c in omega$. But no $a$ has this property, so $omega$ doesn't have a join.
In summary, $omega$ has all meets, but doesn't have all joins. That is, it is complete but not cocomplete.
What have I done wrong? Awodey makes the distinction between a Boolean algebra and a poset here, so we don't necessarily have access to the nice properties of Boolean algebras.
category-theory boolean-algebra
asked Oct 1 '15 at 9:38
Patrick Stevens
26.9k52669
In Awodey's Category Theory, page 130, he says:
A poset is (co)-complete if it is so as a category, thus if it has all set-indexed meets (resp. joins). For posets, completeness and cocompleteness are equivalent. A lattice, Heyting algebra, Boolean algebra, etc. is called complete if it is so as a poset.
As an example, I'll take the poset $omega$, which is well-ordered. A subset of a well-ordered set has a least element, so $omega$ has all meets.
However, $omega$ itself (as a subset of $omega$) has no join. Indeed, a join would be $a$ such that for all $c in omega$, have "all $a_i leq c$" iff $a leq c$. Since there is no $c in omega$ such that for all $a_i in omega$, $a_i leq c$, we must therefore have $a not leq c$ for all $c in omega$. But no $a$ has this property, so $omega$ doesn't have a join.
In summary, $omega$ has all meets, but doesn't have all joins. That is, it is complete but not cocomplete.
What have I done wrong? Awodey makes the distinction between a Boolean algebra and a poset here, so we don't necessarily have access to the nice properties of Boolean algebras.
category-theory boolean-algebra
asked Oct 1 '15 at 9:38
Patrick Stevens
26.9k52669
asked Oct 1 '15 at 9:38
Patrick Stevens
26.9k52669
asked Oct 1 '15 at 9:38
Patrick Stevens
26.9k52669
asked Oct 1 '15 at 9:38
Patrick Stevens
26.9k52669
26.9k52669
2
You need to consider degenerate cases like nullary meets/joins. $omega$ does not even have finite meets: it does not have a maximum element.
– Zhen Lin
Oct 1 '15 at 10:07
In fact, a category has finite products iff it has binary products and a terminal object (AKA the empty product).
– Najib Idrissi
Oct 1 '15 at 14:17
add a comment |Â
2
You need to consider degenerate cases like nullary meets/joins. $omega$ does not even have finite meets: it does not have a maximum element.
– Zhen Lin
Oct 1 '15 at 10:07
In fact, a category has finite products iff it has binary products and a terminal object (AKA the empty product).
– Najib Idrissi
Oct 1 '15 at 14:17
2
2
You need to consider degenerate cases like nullary meets/joins. $omega$ does not even have finite meets: it does not have a maximum element.
– Zhen Lin
Oct 1 '15 at 10:07
You need to consider degenerate cases like nullary meets/joins. $omega$ does not even have finite meets: it does not have a maximum element.
– Zhen Lin
Oct 1 '15 at 10:07
In fact, a category has finite products iff it has binary products and a terminal object (AKA the empty product).
– Najib Idrissi
Oct 1 '15 at 14:17
In fact, a category has finite products iff it has binary products and a terminal object (AKA the empty product).
– Najib Idrissi
Oct 1 '15 at 14:17
add a comment |Â
2 Answers
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oldest
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accepted
In a preordered set $(X,le)$, a maximum element (which is the same thing as a terminal object in the associated category) can be described either as the infimum of the empty system or as the supremum of all elements of $X$. This shows that $omega$, which lacks a maximum element, is neither complete nor cocomplete as a category.
This is due to the observation, that when checking universal properties in thin categories, uniqueness of the induced arrows is for free. That is, a terminal object $t$ is characterized by the existence of some arrow $xlongrightarrow t$ for all objects $x$. Both, $t$ being the product of the empty family, and $t$ being the coproduct of all objects, ensure this condition.
answered Oct 1 '15 at 10:03
Jakob Werner
1,4681919
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up vote
0
down vote
It might be interesting to say that in Completeness almost implies cocompleteness Adamek et al. proved the following.
Any cocomplete weakly co-wellpowered category with a weakly colimit-dense small subcategory is complete.
A (kind of-)dual statement holds for completeness.
answered Aug 7 at 18:04
Ivan Di Liberti
2,29311122
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
In a preordered set $(X,le)$, a maximum element (which is the same thing as a terminal object in the associated category) can be described either as the infimum of the empty system or as the supremum of all elements of $X$. This shows that $omega$, which lacks a maximum element, is neither complete nor cocomplete as a category.
This is due to the observation, that when checking universal properties in thin categories, uniqueness of the induced arrows is for free. That is, a terminal object $t$ is characterized by the existence of some arrow $xlongrightarrow t$ for all objects $x$. Both, $t$ being the product of the empty family, and $t$ being the coproduct of all objects, ensure this condition.
answered Oct 1 '15 at 10:03
Jakob Werner
1,4681919
add a comment |Â
up vote
4
down vote
accepted
In a preordered set $(X,le)$, a maximum element (which is the same thing as a terminal object in the associated category) can be described either as the infimum of the empty system or as the supremum of all elements of $X$. This shows that $omega$, which lacks a maximum element, is neither complete nor cocomplete as a category.
This is due to the observation, that when checking universal properties in thin categories, uniqueness of the induced arrows is for free. That is, a terminal object $t$ is characterized by the existence of some arrow $xlongrightarrow t$ for all objects $x$. Both, $t$ being the product of the empty family, and $t$ being the coproduct of all objects, ensure this condition.
answered Oct 1 '15 at 10:03
Jakob Werner
1,4681919
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
In a preordered set $(X,le)$, a maximum element (which is the same thing as a terminal object in the associated category) can be described either as the infimum of the empty system or as the supremum of all elements of $X$. This shows that $omega$, which lacks a maximum element, is neither complete nor cocomplete as a category.
This is due to the observation, that when checking universal properties in thin categories, uniqueness of the induced arrows is for free. That is, a terminal object $t$ is characterized by the existence of some arrow $xlongrightarrow t$ for all objects $x$. Both, $t$ being the product of the empty family, and $t$ being the coproduct of all objects, ensure this condition.
answered Oct 1 '15 at 10:03
Jakob Werner
1,4681919
In a preordered set $(X,le)$, a maximum element (which is the same thing as a terminal object in the associated category) can be described either as the infimum of the empty system or as the supremum of all elements of $X$. This shows that $omega$, which lacks a maximum element, is neither complete nor cocomplete as a category.
This is due to the observation, that when checking universal properties in thin categories, uniqueness of the induced arrows is for free. That is, a terminal object $t$ is characterized by the existence of some arrow $xlongrightarrow t$ for all objects $x$. Both, $t$ being the product of the empty family, and $t$ being the coproduct of all objects, ensure this condition.
answered Oct 1 '15 at 10:03
Jakob Werner
1,4681919
edited Oct 1 '15 at 10:09
answered Oct 1 '15 at 10:03
Jakob Werner
1,4681919
answered Oct 1 '15 at 10:03
Jakob Werner
1,4681919
answered Oct 1 '15 at 10:03
Jakob Werner
1,4681919
1,4681919
add a comment |Â
add a comment |Â
up vote
0
down vote
It might be interesting to say that in Completeness almost implies cocompleteness Adamek et al. proved the following.
Any cocomplete weakly co-wellpowered category with a weakly colimit-dense small subcategory is complete.
A (kind of-)dual statement holds for completeness.
answered Aug 7 at 18:04
Ivan Di Liberti
2,29311122
add a comment |Â
up vote
0
down vote
It might be interesting to say that in Completeness almost implies cocompleteness Adamek et al. proved the following.
Any cocomplete weakly co-wellpowered category with a weakly colimit-dense small subcategory is complete.
A (kind of-)dual statement holds for completeness.
answered Aug 7 at 18:04
Ivan Di Liberti
2,29311122
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It might be interesting to say that in Completeness almost implies cocompleteness Adamek et al. proved the following.
Any cocomplete weakly co-wellpowered category with a weakly colimit-dense small subcategory is complete.
A (kind of-)dual statement holds for completeness.
answered Aug 7 at 18:04
Ivan Di Liberti
2,29311122
It might be interesting to say that in Completeness almost implies cocompleteness Adamek et al. proved the following.
Any cocomplete weakly co-wellpowered category with a weakly colimit-dense small subcategory is complete.
A (kind of-)dual statement holds for completeness.
answered Aug 7 at 18:04
Ivan Di Liberti
2,29311122
edited Aug 7 at 18:12
answered Aug 7 at 18:04
Ivan Di Liberti
2,29311122
answered Aug 7 at 18:04
Ivan Di Liberti
2,29311122
answered Aug 7 at 18:04
Ivan Di Liberti
2,29311122
2,29311122
add a comment |Â
add a comment |Â
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2
You need to consider degenerate cases like nullary meets/joins. $omega$ does not even have finite meets: it does not have a maximum element.
– Zhen Lin
Oct 1 '15 at 10:07
In fact, a category has finite products iff it has binary products and a terminal object (AKA the empty product).
– Najib Idrissi
Oct 1 '15 at 14:17