category of measurable spaces
- notation:
- objects: measurable spaces
- morphisms: measurable maps
- nLab Link
Limits and colimits can be constructed in the same way as for topological spaces.
Properties
Properties from the database
- is locally small
- is well-powered
- is well-copowered
- is complete
- is cocomplete
- is infinitary distributive
- has a generator
- has a cogenerator
- has disjoint coproducts
Deduced properties
is locally essentially small
Since it is locally small, we deduce that it is locally essentially small.is finitely complete
Since it is complete, we deduce that it is finitely complete.has filtered limits
Since it is complete, we deduce that it has filtered limits.has wide pullbacks
Since it is complete, we deduce that it has wide pullbacks.has connected limits
Since it is complete, we deduce that it has connected limits.has products
Since it is complete, we deduce that it has products.has equalizers
Since it is complete, we deduce that it has equalizers.has finite products
Since it is finitely complete, we deduce that it has finite products.has countable products
Since it has products, we deduce that it has countable products.has a terminal object
Since it has finite products, we deduce that it has a terminal object.has binary products
Since it has finite products, we deduce that it has binary products.has pullbacks
Since it has binary products and has equalizers, we deduce that it has pullbacks.is connected
Since it has a terminal object, we deduce that it is connected.has coproducts
Since it has disjoint coproducts, we deduce that it has coproducts.has disjoint finite coproducts
Since it has disjoint coproducts, we deduce that it has disjoint finite coproducts.has finite coproducts
Since it has disjoint finite coproducts, we deduce that it has finite coproducts.is Cauchy complete
Since it has equalizers, we deduce that it is Cauchy complete.has sequential limits
Since it has equalizers and has countable products, we deduce that it has sequential limits.is distributive
Since it is infinitary distributive, we deduce that it is distributive.has a strict initial object
Since it is distributive, we deduce that it has a strict initial object.is inhabited
Since it is connected, we deduce that it is inhabited.is finitely cocomplete
Since it is cocomplete, we deduce that it is finitely cocomplete.has filtered colimits
Since it is cocomplete, we deduce that it has filtered colimits.has wide pushouts
Since it is cocomplete, we deduce that it has wide pushouts.has connected colimits
Since it is cocomplete, we deduce that it has connected colimits.has coequalizers
Since it is cocomplete, we deduce that it has coequalizers.has countable coproducts
Since it has coproducts, we deduce that it has countable coproducts.has an initial object
Since it has finite coproducts, we deduce that it has an initial object.has binary coproducts
Since it has finite coproducts, we deduce that it has binary coproducts.has pushouts
Since it has binary coproducts and has coequalizers, we deduce that it has pushouts.has sequential colimits
Since it has coequalizers and has countable coproducts, we deduce that it has sequential colimits.
Non-Properties
Non-Properties from the database
- is not cartesian closed
- does not have a strict terminal object
- is not balanced
is not skeletal
trivial- is not Malcev
Deduced Non-Properties*
is not small
Assume for a contradiction that it is small. Since it is small, we deduce that it is essentially small. Since it is essentially small and has products, we deduce that it is thin. Since it is essentially small and is thin and is complete and is infinitary distributive, we deduce that it is cartesian closed. This is a contradiction since we already know that it is not cartesian closed.does not have zero morphisms
Assume for a contradiction that it has zero morphisms. Since it has zero morphisms and has an initial object, we deduce that it is pointed. Since it has a strict initial object and is pointed, we deduce that it is trivial. Since it is trivial, we deduce that it is split abelian. Since it is split abelian, we deduce that it is abelian. Since it is abelian, we deduce that it is Malcev. This is a contradiction since we already know that it is not Malcev.is not preadditive
Assume for a contradiction that it is preadditive. Since it is preadditive, we deduce that it has zero morphisms. This is a contradiction since we already know that it does not have zero morphisms.is not additive
Assume for a contradiction that it is additive. Since it is additive, we deduce that it is preadditive. This is a contradiction since we already know that it is not preadditive.is not abelian
Assume for a contradiction that it is abelian. Since it is abelian, we deduce that it is additive. This is a contradiction since we already know that it is not additive.is not pointed
Assume for a contradiction that it is pointed. Since it is pointed, we deduce that it has zero morphisms. This is a contradiction since we already know that it does not have zero morphisms.is not an elementary topos
Assume for a contradiction that it is an elementary topos. Since it is an elementary topos, we deduce that it is cartesian closed. This is a contradiction since we already know that it is not cartesian closed.is not a Grothendieck topos
Assume for a contradiction that it is a Grothendieck topos. Since it is a Grothendieck topos, we deduce that it is an elementary topos. This is a contradiction since we already know that it is not an elementary topos.is not self-dual
Assume for a contradiction that it is self-dual. Since it is self-dual and has a strict initial object, we deduce that it has a strict terminal object. This is a contradiction since we already know that it does not have a strict terminal object.is not a groupoid
Assume for a contradiction that it is a groupoid. Since it is a groupoid, we deduce that it is self-dual. This is a contradiction since we already know that it is not self-dual.is not essentially small
Assume for a contradiction that it is essentially small. Since it is essentially small and has products, we deduce that it is thin. Since it is essentially small and is thin and is complete and is infinitary distributive, we deduce that it is cartesian closed. This is a contradiction since we already know that it is not cartesian closed.is not thin
Assume for a contradiction that it is thin. Since it is thin and is finitely complete, we deduce that it is Malcev. This is a contradiction since we already know that it is not Malcev.is not discrete
Assume for a contradiction that it is discrete. Since it is discrete, we deduce that it is skeletal. This is a contradiction since we already know that it is not skeletal.is not essentially discrete
Assume for a contradiction that it is essentially discrete. Since it is essentially discrete, we deduce that it is thin. This is a contradiction since we already know that it is not thin.is not finite
Assume for a contradiction that it is finite. Since it is finite, we deduce that it is small. This is a contradiction since we already know that it is not small.is not essentially finite
Assume for a contradiction that it is essentially finite. Since it is essentially finite and has finite products, we deduce that it is thin. This is a contradiction since we already know that it is not thin.is not trivial
Assume for a contradiction that it is trivial. Since it is trivial, we deduce that it is a Grothendieck topos. This is a contradiction since we already know that it is not a Grothendieck topos.does not have a subobject classifier
Assume for a contradiction that it has a subobject classifier. Since it has a subobject classifier, we deduce that it is mono-regular. Since it is mono-regular, we deduce that it is balanced. This is a contradiction since we already know that it is not balanced.is not Grothendieck abelian
Assume for a contradiction that it is Grothendieck abelian. Since it is Grothendieck abelian, we deduce that it is abelian. This is a contradiction since we already know that it is not abelian.is not left cancellative
Assume for a contradiction that it is left cancellative. Since it is left cancellative and has coequalizers, we deduce that it is thin. This is a contradiction since we already know that it is not thin.is not right cancellative
Assume for a contradiction that it is right cancellative. Since it is right cancellative and has equalizers, we deduce that it is thin. This is a contradiction since we already know that it is not thin.is not split abelian
Assume for a contradiction that it is split abelian. Since it is split abelian, we deduce that it is abelian. This is a contradiction since we already know that it is not abelian.is not mono-regular
Assume for a contradiction that it is mono-regular. Since it is mono-regular, we deduce that it is balanced. This is a contradiction since we already know that it is not balanced.is not epi-regular
Assume for a contradiction that it is epi-regular. Since it is epi-regular, we deduce that it is balanced. This is a contradiction since we already know that it is not balanced.
*This also uses the deduced properties.
Unknown properties
For these properties the database currently doesn't have an answer if they are satisfied or not. Please help to complete the data!
Special morphisms
Isomorphisms: bijective measurable maps that map measurable sets to measurable sets
easy- Monomorphisms: injective measurable maps
Epimorphisms: surjective measurable maps
Use the same proof as for sets, where is endowed with the trivial -algebra.