Burali-Forti paradox for cardinals
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It is well-known that in $mathsf{ZF}$ there cannot exist a set containing all of the ordinals (transitive sets of transitive sets).
Is the same true for cardinals? (ordinals satisfying $(forall alphainkappa)[alphalnsimkappa]$)
Of course the answer is "yes" assuming the axiom of choice because we can define successor cardinals, but what about a setting without choice?
set-theory cardinals ordinals
asked Nov 17 at 9:21
Alberto Takase
1,663414
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up vote
3
down vote
favorite
It is well-known that in $mathsf{ZF}$ there cannot exist a set containing all of the ordinals (transitive sets of transitive sets).
Is the same true for cardinals? (ordinals satisfying $(forall alphainkappa)[alphalnsimkappa]$)
Of course the answer is "yes" assuming the axiom of choice because we can define successor cardinals, but what about a setting without choice?
set-theory cardinals ordinals
asked Nov 17 at 9:21
Alberto Takase
1,663414
If $A$ is a set, the cardinality of $mathcal P(bigcup A)$ is strictly larger than the cardinality of any set in $A$.
– Andrés E. Caicedo
Nov 17 at 16:17
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
It is well-known that in $mathsf{ZF}$ there cannot exist a set containing all of the ordinals (transitive sets of transitive sets).
Is the same true for cardinals? (ordinals satisfying $(forall alphainkappa)[alphalnsimkappa]$)
Of course the answer is "yes" assuming the axiom of choice because we can define successor cardinals, but what about a setting without choice?
set-theory cardinals ordinals
asked Nov 17 at 9:21
Alberto Takase
1,663414
It is well-known that in $mathsf{ZF}$ there cannot exist a set containing all of the ordinals (transitive sets of transitive sets).
Is the same true for cardinals? (ordinals satisfying $(forall alphainkappa)[alphalnsimkappa]$)
Of course the answer is "yes" assuming the axiom of choice because we can define successor cardinals, but what about a setting without choice?
set-theory cardinals ordinals
set-theory cardinals ordinals
asked Nov 17 at 9:21
Alberto Takase
1,663414
asked Nov 17 at 9:21
Alberto Takase
1,663414
asked Nov 17 at 9:21
Alberto Takase
1,663414
asked Nov 17 at 9:21
Alberto Takase
1,663414
asked Nov 17 at 9:21
Alberto Takase
1,663414
1,663414
If $A$ is a set, the cardinality of $mathcal P(bigcup A)$ is strictly larger than the cardinality of any set in $A$.
– Andrés E. Caicedo
Nov 17 at 16:17
add a comment |
If $A$ is a set, the cardinality of $mathcal P(bigcup A)$ is strictly larger than the cardinality of any set in $A$.
– Andrés E. Caicedo
Nov 17 at 16:17
If $A$ is a set, the cardinality of $mathcal P(bigcup A)$ is strictly larger than the cardinality of any set in $A$.
– Andrés E. Caicedo
Nov 17 at 16:17
If $A$ is a set, the cardinality of $mathcal P(bigcup A)$ is strictly larger than the cardinality of any set in $A$.
– Andrés E. Caicedo
Nov 17 at 16:17
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Here's a simple proof that the class of cardinals $(dagger)$ (in $mathrm{ZF}$) is a proper class:
We already now that the class of ordinals $mathrm{Ord}$ is a proper class and as such unbounded in the von Neumann hierarchy. Now just note that for every $alpha$ there is a cardinal $alpha^+ > alpha$ (this doesn't use any form of choice -- see Hartogs number). Hence the class of cardinals is unbounded in the von Neumann hierarchy (as it is $in$-cofinal in $mathrm{Ord}$) and as such it must be a proper class.
$(dagger)$ This shows a bit more: The class of wellordered cardinals (or alephs) is a proper class.
answered Nov 17 at 9:56
Stefan Mesken
14.4k32046
1
Hartogs number! I notice that the axiom of replacement is crucial here because $V_{omega+omega}$ has very few cardinals.
– Alberto Takase
Nov 17 at 10:07
2
@AlbertoTakase Replacement certainly plays a role here but notice that the powerset axiom is also important: Let $kappa$ be in infinite cardinal. Then (assuming $mathrm{ZFC}$ in our background universe) $(H_{kappa+}; in) models mathrm{ZFC}^-$, that is $mathrm{ZFC}$ without the powerset axiom and $(H_{kappa^+}; in) models forall x exists f colon kappa to x text{ surjective}$. In particular $(H_{kappa^+}; in) models kappa text{ is the largest cardinal}$.
– Stefan Mesken
Nov 17 at 10:16
1
@Alberto: Insisting that cardinals are ordinals is missing the point of "cardinals".
– Asaf Karagila♦
Nov 17 at 18:22
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Here's a simple proof that the class of cardinals $(dagger)$ (in $mathrm{ZF}$) is a proper class:
We already now that the class of ordinals $mathrm{Ord}$ is a proper class and as such unbounded in the von Neumann hierarchy. Now just note that for every $alpha$ there is a cardinal $alpha^+ > alpha$ (this doesn't use any form of choice -- see Hartogs number). Hence the class of cardinals is unbounded in the von Neumann hierarchy (as it is $in$-cofinal in $mathrm{Ord}$) and as such it must be a proper class.
$(dagger)$ This shows a bit more: The class of wellordered cardinals (or alephs) is a proper class.
answered Nov 17 at 9:56
Stefan Mesken
14.4k32046
1
Hartogs number! I notice that the axiom of replacement is crucial here because $V_{omega+omega}$ has very few cardinals.
– Alberto Takase
Nov 17 at 10:07
2
@AlbertoTakase Replacement certainly plays a role here but notice that the powerset axiom is also important: Let $kappa$ be in infinite cardinal. Then (assuming $mathrm{ZFC}$ in our background universe) $(H_{kappa+}; in) models mathrm{ZFC}^-$, that is $mathrm{ZFC}$ without the powerset axiom and $(H_{kappa^+}; in) models forall x exists f colon kappa to x text{ surjective}$. In particular $(H_{kappa^+}; in) models kappa text{ is the largest cardinal}$.
– Stefan Mesken
Nov 17 at 10:16
1
@Alberto: Insisting that cardinals are ordinals is missing the point of "cardinals".
– Asaf Karagila♦
Nov 17 at 18:22
add a comment |
up vote
2
down vote
accepted
Here's a simple proof that the class of cardinals $(dagger)$ (in $mathrm{ZF}$) is a proper class:
We already now that the class of ordinals $mathrm{Ord}$ is a proper class and as such unbounded in the von Neumann hierarchy. Now just note that for every $alpha$ there is a cardinal $alpha^+ > alpha$ (this doesn't use any form of choice -- see Hartogs number). Hence the class of cardinals is unbounded in the von Neumann hierarchy (as it is $in$-cofinal in $mathrm{Ord}$) and as such it must be a proper class.
$(dagger)$ This shows a bit more: The class of wellordered cardinals (or alephs) is a proper class.
answered Nov 17 at 9:56
Stefan Mesken
14.4k32046
1
Hartogs number! I notice that the axiom of replacement is crucial here because $V_{omega+omega}$ has very few cardinals.
– Alberto Takase
Nov 17 at 10:07
2
@AlbertoTakase Replacement certainly plays a role here but notice that the powerset axiom is also important: Let $kappa$ be in infinite cardinal. Then (assuming $mathrm{ZFC}$ in our background universe) $(H_{kappa+}; in) models mathrm{ZFC}^-$, that is $mathrm{ZFC}$ without the powerset axiom and $(H_{kappa^+}; in) models forall x exists f colon kappa to x text{ surjective}$. In particular $(H_{kappa^+}; in) models kappa text{ is the largest cardinal}$.
– Stefan Mesken
Nov 17 at 10:16
1
@Alberto: Insisting that cardinals are ordinals is missing the point of "cardinals".
– Asaf Karagila♦
Nov 17 at 18:22
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Here's a simple proof that the class of cardinals $(dagger)$ (in $mathrm{ZF}$) is a proper class:
We already now that the class of ordinals $mathrm{Ord}$ is a proper class and as such unbounded in the von Neumann hierarchy. Now just note that for every $alpha$ there is a cardinal $alpha^+ > alpha$ (this doesn't use any form of choice -- see Hartogs number). Hence the class of cardinals is unbounded in the von Neumann hierarchy (as it is $in$-cofinal in $mathrm{Ord}$) and as such it must be a proper class.
$(dagger)$ This shows a bit more: The class of wellordered cardinals (or alephs) is a proper class.
answered Nov 17 at 9:56
Stefan Mesken
14.4k32046
Here's a simple proof that the class of cardinals $(dagger)$ (in $mathrm{ZF}$) is a proper class:
We already now that the class of ordinals $mathrm{Ord}$ is a proper class and as such unbounded in the von Neumann hierarchy. Now just note that for every $alpha$ there is a cardinal $alpha^+ > alpha$ (this doesn't use any form of choice -- see Hartogs number). Hence the class of cardinals is unbounded in the von Neumann hierarchy (as it is $in$-cofinal in $mathrm{Ord}$) and as such it must be a proper class.
$(dagger)$ This shows a bit more: The class of wellordered cardinals (or alephs) is a proper class.
answered Nov 17 at 9:56
Stefan Mesken
14.4k32046
edited Nov 17 at 10:12
answered Nov 17 at 9:56
Stefan Mesken
14.4k32046
answered Nov 17 at 9:56
Stefan Mesken
14.4k32046
answered Nov 17 at 9:56
Stefan Mesken
14.4k32046
14.4k32046
1
Hartogs number! I notice that the axiom of replacement is crucial here because $V_{omega+omega}$ has very few cardinals.
– Alberto Takase
Nov 17 at 10:07
2
@AlbertoTakase Replacement certainly plays a role here but notice that the powerset axiom is also important: Let $kappa$ be in infinite cardinal. Then (assuming $mathrm{ZFC}$ in our background universe) $(H_{kappa+}; in) models mathrm{ZFC}^-$, that is $mathrm{ZFC}$ without the powerset axiom and $(H_{kappa^+}; in) models forall x exists f colon kappa to x text{ surjective}$. In particular $(H_{kappa^+}; in) models kappa text{ is the largest cardinal}$.
– Stefan Mesken
Nov 17 at 10:16
1
@Alberto: Insisting that cardinals are ordinals is missing the point of "cardinals".
– Asaf Karagila♦
Nov 17 at 18:22
add a comment |
1
Hartogs number! I notice that the axiom of replacement is crucial here because $V_{omega+omega}$ has very few cardinals.
– Alberto Takase
Nov 17 at 10:07
2
@AlbertoTakase Replacement certainly plays a role here but notice that the powerset axiom is also important: Let $kappa$ be in infinite cardinal. Then (assuming $mathrm{ZFC}$ in our background universe) $(H_{kappa+}; in) models mathrm{ZFC}^-$, that is $mathrm{ZFC}$ without the powerset axiom and $(H_{kappa^+}; in) models forall x exists f colon kappa to x text{ surjective}$. In particular $(H_{kappa^+}; in) models kappa text{ is the largest cardinal}$.
– Stefan Mesken
Nov 17 at 10:16
1
@Alberto: Insisting that cardinals are ordinals is missing the point of "cardinals".
– Asaf Karagila♦
Nov 17 at 18:22
1
1
Hartogs number! I notice that the axiom of replacement is crucial here because $V_{omega+omega}$ has very few cardinals.
– Alberto Takase
Nov 17 at 10:07
Hartogs number! I notice that the axiom of replacement is crucial here because $V_{omega+omega}$ has very few cardinals.
– Alberto Takase
Nov 17 at 10:07
2
2
@AlbertoTakase Replacement certainly plays a role here but notice that the powerset axiom is also important: Let $kappa$ be in infinite cardinal. Then (assuming $mathrm{ZFC}$ in our background universe) $(H_{kappa+}; in) models mathrm{ZFC}^-$, that is $mathrm{ZFC}$ without the powerset axiom and $(H_{kappa^+}; in) models forall x exists f colon kappa to x text{ surjective}$. In particular $(H_{kappa^+}; in) models kappa text{ is the largest cardinal}$.
– Stefan Mesken
Nov 17 at 10:16
@AlbertoTakase Replacement certainly plays a role here but notice that the powerset axiom is also important: Let $kappa$ be in infinite cardinal. Then (assuming $mathrm{ZFC}$ in our background universe) $(H_{kappa+}; in) models mathrm{ZFC}^-$, that is $mathrm{ZFC}$ without the powerset axiom and $(H_{kappa^+}; in) models forall x exists f colon kappa to x text{ surjective}$. In particular $(H_{kappa^+}; in) models kappa text{ is the largest cardinal}$.
– Stefan Mesken
Nov 17 at 10:16
1
1
@Alberto: Insisting that cardinals are ordinals is missing the point of "cardinals".
– Asaf Karagila♦
Nov 17 at 18:22
@Alberto: Insisting that cardinals are ordinals is missing the point of "cardinals".
– Asaf Karagila♦
Nov 17 at 18:22
add a comment |
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If $A$ is a set, the cardinality of $mathcal P(bigcup A)$ is strictly larger than the cardinality of any set in $A$.
– Andrés E. Caicedo
Nov 17 at 16:17