Reining in the Axiom of Power Set in ZF
3
$begingroup$
Given the powerset operator $mathit P$, we have the following mapping
$tag 1 mathcal Phi: mathbb N to Phi(mathbb N) $
$quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad , n mapsto mathit P^n(mathbb N)$
What happens if we take away the Axiom of Power Set in $ZFpm C$ and replace
it with $text{(1)}$? Would this contradict the other axioms?
set-theory axioms
asked Jan 9 at 11:50
CopyPasteItCopyPasteIt
4,3771828
$endgroup$
|
show 3 more comments
3
$begingroup$
Given the powerset operator $mathit P$, we have the following mapping
$tag 1 mathcal Phi: mathbb N to Phi(mathbb N) $
$quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad , n mapsto mathit P^n(mathbb N)$
What happens if we take away the Axiom of Power Set in $ZFpm C$ and replace
it with $text{(1)}$? Would this contradict the other axioms?
set-theory axioms
asked Jan 9 at 11:50
CopyPasteItCopyPasteIt
4,3771828
$endgroup$
1
$begingroup$
Incidentally, there are some very interesting results on the "amount of powerset" (or really, "powerset-along-replacement") we need in various situations. Most famously, Harvey Friedman showed that more and more iterations of powerset are needed to prove determinacy further and further up the Borel hierarchy.
$endgroup$
– Noah Schweber
Jan 11 at 22:08
1
$begingroup$
On formatting: You can use pairs of L& R dollar-signs to "display" a line instead of typing quad quad quad ..., and to put un-formatted text (like "because") in a displayed line, enclose it within text {...}, like text { because }.
$endgroup$
– DanielWainfleet
Jan 14 at 0:47
$begingroup$
@DanielWainfleet I've felt silly at times trying to (perfectly) align stuff since the final result might depend on the display device. But anyway, not exactly sure what you mean. Please share some links with useful mathstackexchange formatting technique so I can copy n paste it in the future.
$endgroup$
– CopyPasteIt
Jan 14 at 1:27
$begingroup$
@NoahSchweber Another interesting paper, 'Large irredundant sets in operator algebras' arxiv.org/abs/1808.01511?context=math I don't know the formulation for the $text{♢-Axiom}$ (stronger than the continuum hypothesis), but my guess is all modern day physics can be expressed/modeled in $V_{omega+omega} + text{♢}$. I wonder how this holds up: "...every open set of Minkowski spacetime is associated with a C*-algebra" /wikipedia.
$endgroup$
– CopyPasteIt
Jan 14 at 14:55
$begingroup$
Did something not work with the Ask Question page that you felt the need to ask a completely disjoint question on a page where you already accepted one answer?
$endgroup$
– Asaf Karagila♦
Jan 14 at 15:44
|
show 3 more comments
3
3
3
$begingroup$
Given the powerset operator $mathit P$, we have the following mapping
$tag 1 mathcal Phi: mathbb N to Phi(mathbb N) $
$quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad , n mapsto mathit P^n(mathbb N)$
What happens if we take away the Axiom of Power Set in $ZFpm C$ and replace
it with $text{(1)}$? Would this contradict the other axioms?
set-theory axioms
asked Jan 9 at 11:50
CopyPasteItCopyPasteIt
4,3771828
$endgroup$
Given the powerset operator $mathit P$, we have the following mapping
$tag 1 mathcal Phi: mathbb N to Phi(mathbb N) $
$quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad , n mapsto mathit P^n(mathbb N)$
What happens if we take away the Axiom of Power Set in $ZFpm C$ and replace
it with $text{(1)}$? Would this contradict the other axioms?
set-theory axioms
set-theory axioms
asked Jan 9 at 11:50
CopyPasteItCopyPasteIt
4,3771828
asked Jan 9 at 11:50
CopyPasteItCopyPasteIt
4,3771828
edited Jan 14 at 17:15
CopyPasteIt
asked Jan 9 at 11:50
CopyPasteItCopyPasteIt
4,3771828
asked Jan 9 at 11:50
CopyPasteItCopyPasteIt
4,3771828
asked Jan 9 at 11:50
CopyPasteItCopyPasteIt
4,3771828
4,3771828
1
$begingroup$
Incidentally, there are some very interesting results on the "amount of powerset" (or really, "powerset-along-replacement") we need in various situations. Most famously, Harvey Friedman showed that more and more iterations of powerset are needed to prove determinacy further and further up the Borel hierarchy.
$endgroup$
– Noah Schweber
Jan 11 at 22:08
1
$begingroup$
On formatting: You can use pairs of L& R dollar-signs to "display" a line instead of typing quad quad quad ..., and to put un-formatted text (like "because") in a displayed line, enclose it within text {...}, like text { because }.
$endgroup$
– DanielWainfleet
Jan 14 at 0:47
$begingroup$
@DanielWainfleet I've felt silly at times trying to (perfectly) align stuff since the final result might depend on the display device. But anyway, not exactly sure what you mean. Please share some links with useful mathstackexchange formatting technique so I can copy n paste it in the future.
$endgroup$
– CopyPasteIt
Jan 14 at 1:27
$begingroup$
@NoahSchweber Another interesting paper, 'Large irredundant sets in operator algebras' arxiv.org/abs/1808.01511?context=math I don't know the formulation for the $text{♢-Axiom}$ (stronger than the continuum hypothesis), but my guess is all modern day physics can be expressed/modeled in $V_{omega+omega} + text{♢}$. I wonder how this holds up: "...every open set of Minkowski spacetime is associated with a C*-algebra" /wikipedia.
$endgroup$
– CopyPasteIt
Jan 14 at 14:55
$begingroup$
Did something not work with the Ask Question page that you felt the need to ask a completely disjoint question on a page where you already accepted one answer?
$endgroup$
– Asaf Karagila♦
Jan 14 at 15:44
|
show 3 more comments
1
$begingroup$
Incidentally, there are some very interesting results on the "amount of powerset" (or really, "powerset-along-replacement") we need in various situations. Most famously, Harvey Friedman showed that more and more iterations of powerset are needed to prove determinacy further and further up the Borel hierarchy.
$endgroup$
– Noah Schweber
Jan 11 at 22:08
1
$begingroup$
On formatting: You can use pairs of L& R dollar-signs to "display" a line instead of typing quad quad quad ..., and to put un-formatted text (like "because") in a displayed line, enclose it within text {...}, like text { because }.
$endgroup$
– DanielWainfleet
Jan 14 at 0:47
$begingroup$
@DanielWainfleet I've felt silly at times trying to (perfectly) align stuff since the final result might depend on the display device. But anyway, not exactly sure what you mean. Please share some links with useful mathstackexchange formatting technique so I can copy n paste it in the future.
$endgroup$
– CopyPasteIt
Jan 14 at 1:27
$begingroup$
@NoahSchweber Another interesting paper, 'Large irredundant sets in operator algebras' arxiv.org/abs/1808.01511?context=math I don't know the formulation for the $text{♢-Axiom}$ (stronger than the continuum hypothesis), but my guess is all modern day physics can be expressed/modeled in $V_{omega+omega} + text{♢}$. I wonder how this holds up: "...every open set of Minkowski spacetime is associated with a C*-algebra" /wikipedia.
$endgroup$
– CopyPasteIt
Jan 14 at 14:55
$begingroup$
Did something not work with the Ask Question page that you felt the need to ask a completely disjoint question on a page where you already accepted one answer?
$endgroup$
– Asaf Karagila♦
Jan 14 at 15:44
1
1
$begingroup$
Incidentally, there are some very interesting results on the "amount of powerset" (or really, "powerset-along-replacement") we need in various situations. Most famously, Harvey Friedman showed that more and more iterations of powerset are needed to prove determinacy further and further up the Borel hierarchy.
$endgroup$
– Noah Schweber
Jan 11 at 22:08
$begingroup$
Incidentally, there are some very interesting results on the "amount of powerset" (or really, "powerset-along-replacement") we need in various situations. Most famously, Harvey Friedman showed that more and more iterations of powerset are needed to prove determinacy further and further up the Borel hierarchy.
$endgroup$
– Noah Schweber
Jan 11 at 22:08
1
1
$begingroup$
On formatting: You can use pairs of L& R dollar-signs to "display" a line instead of typing quad quad quad ..., and to put un-formatted text (like "because") in a displayed line, enclose it within text {...}, like text { because }.
$endgroup$
– DanielWainfleet
Jan 14 at 0:47
$begingroup$
On formatting: You can use pairs of L& R dollar-signs to "display" a line instead of typing quad quad quad ..., and to put un-formatted text (like "because") in a displayed line, enclose it within text {...}, like text { because }.
$endgroup$
– DanielWainfleet
Jan 14 at 0:47
$begingroup$
@DanielWainfleet I've felt silly at times trying to (perfectly) align stuff since the final result might depend on the display device. But anyway, not exactly sure what you mean. Please share some links with useful mathstackexchange formatting technique so I can copy n paste it in the future.
$endgroup$
– CopyPasteIt
Jan 14 at 1:27
$begingroup$
@DanielWainfleet I've felt silly at times trying to (perfectly) align stuff since the final result might depend on the display device. But anyway, not exactly sure what you mean. Please share some links with useful mathstackexchange formatting technique so I can copy n paste it in the future.
$endgroup$
– CopyPasteIt
Jan 14 at 1:27
$begingroup$
@NoahSchweber Another interesting paper, 'Large irredundant sets in operator algebras' arxiv.org/abs/1808.01511?context=math I don't know the formulation for the $text{♢-Axiom}$ (stronger than the continuum hypothesis), but my guess is all modern day physics can be expressed/modeled in $V_{omega+omega} + text{♢}$. I wonder how this holds up: "...every open set of Minkowski spacetime is associated with a C*-algebra" /wikipedia.
$endgroup$
– CopyPasteIt
Jan 14 at 14:55
$begingroup$
@NoahSchweber Another interesting paper, 'Large irredundant sets in operator algebras' arxiv.org/abs/1808.01511?context=math I don't know the formulation for the $text{♢-Axiom}$ (stronger than the continuum hypothesis), but my guess is all modern day physics can be expressed/modeled in $V_{omega+omega} + text{♢}$. I wonder how this holds up: "...every open set of Minkowski spacetime is associated with a C*-algebra" /wikipedia.
$endgroup$
– CopyPasteIt
Jan 14 at 14:55
$begingroup$
Did something not work with the Ask Question page that you felt the need to ask a completely disjoint question on a page where you already accepted one answer?
$endgroup$
– Asaf Karagila♦
Jan 14 at 15:44
$begingroup$
Did something not work with the Ask Question page that you felt the need to ask a completely disjoint question on a page where you already accepted one answer?
$endgroup$
– Asaf Karagila♦
Jan 14 at 15:44
|
show 3 more comments
1 Answer
1
active
oldest
votes
2
$begingroup$
I assume that by (1) you mean the existence of the operator $Phi$ that you defined above. As you say, the existence of such a $Phi$ is a consequence of the Powerset Axiom and of the other axioms of $ZF$, so if (1) contradicts the other axioms, the Powerset Axiom does as well.
Moreover, it seems to me that $V_{omega_1}$ is a model of $ZFC-P+(1)$, so the consistency strength of this theory is strictly less that that of $ZFC$.
answered Jan 9 at 12:11
Leo163Leo163
1,795512
$endgroup$
5
$begingroup$
Just $V_{omega+omega}$ is already enough.
$endgroup$
– Asaf Karagila♦
Jan 9 at 12:18
1
$begingroup$
@AsafKaragila Thanks for your comment. Since I am not extremely familiar with this topic and I wanted to be sure I opted to go for a regular cardinal, in order to be "safer". If I may add something to this, my answer would be wrong if we considered collection instead of replacement, right? What should we use as a model in that case?
$endgroup$
– Leo163
Jan 9 at 12:51
4
$begingroup$
Correct, for Replacement we use $H(kappa)$, the set of sets which are hereditarily smaller than $kappa$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 13:18
add a comment |
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1 Answer
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oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
I assume that by (1) you mean the existence of the operator $Phi$ that you defined above. As you say, the existence of such a $Phi$ is a consequence of the Powerset Axiom and of the other axioms of $ZF$, so if (1) contradicts the other axioms, the Powerset Axiom does as well.
Moreover, it seems to me that $V_{omega_1}$ is a model of $ZFC-P+(1)$, so the consistency strength of this theory is strictly less that that of $ZFC$.
answered Jan 9 at 12:11
Leo163Leo163
1,795512
$endgroup$
5
$begingroup$
Just $V_{omega+omega}$ is already enough.
$endgroup$
– Asaf Karagila♦
Jan 9 at 12:18
1
$begingroup$
@AsafKaragila Thanks for your comment. Since I am not extremely familiar with this topic and I wanted to be sure I opted to go for a regular cardinal, in order to be "safer". If I may add something to this, my answer would be wrong if we considered collection instead of replacement, right? What should we use as a model in that case?
$endgroup$
– Leo163
Jan 9 at 12:51
4
$begingroup$
Correct, for Replacement we use $H(kappa)$, the set of sets which are hereditarily smaller than $kappa$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 13:18
add a comment |
2
$begingroup$
I assume that by (1) you mean the existence of the operator $Phi$ that you defined above. As you say, the existence of such a $Phi$ is a consequence of the Powerset Axiom and of the other axioms of $ZF$, so if (1) contradicts the other axioms, the Powerset Axiom does as well.
Moreover, it seems to me that $V_{omega_1}$ is a model of $ZFC-P+(1)$, so the consistency strength of this theory is strictly less that that of $ZFC$.
answered Jan 9 at 12:11
Leo163Leo163
1,795512
$endgroup$
5
$begingroup$
Just $V_{omega+omega}$ is already enough.
$endgroup$
– Asaf Karagila♦
Jan 9 at 12:18
1
$begingroup$
@AsafKaragila Thanks for your comment. Since I am not extremely familiar with this topic and I wanted to be sure I opted to go for a regular cardinal, in order to be "safer". If I may add something to this, my answer would be wrong if we considered collection instead of replacement, right? What should we use as a model in that case?
$endgroup$
– Leo163
Jan 9 at 12:51
4
$begingroup$
Correct, for Replacement we use $H(kappa)$, the set of sets which are hereditarily smaller than $kappa$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 13:18
add a comment |
2
2
2
$begingroup$
I assume that by (1) you mean the existence of the operator $Phi$ that you defined above. As you say, the existence of such a $Phi$ is a consequence of the Powerset Axiom and of the other axioms of $ZF$, so if (1) contradicts the other axioms, the Powerset Axiom does as well.
Moreover, it seems to me that $V_{omega_1}$ is a model of $ZFC-P+(1)$, so the consistency strength of this theory is strictly less that that of $ZFC$.
answered Jan 9 at 12:11
Leo163Leo163
1,795512
$endgroup$
I assume that by (1) you mean the existence of the operator $Phi$ that you defined above. As you say, the existence of such a $Phi$ is a consequence of the Powerset Axiom and of the other axioms of $ZF$, so if (1) contradicts the other axioms, the Powerset Axiom does as well.
Moreover, it seems to me that $V_{omega_1}$ is a model of $ZFC-P+(1)$, so the consistency strength of this theory is strictly less that that of $ZFC$.
answered Jan 9 at 12:11
Leo163Leo163
1,795512
answered Jan 9 at 12:11
Leo163Leo163
1,795512
answered Jan 9 at 12:11
Leo163Leo163
1,795512
answered Jan 9 at 12:11
Leo163Leo163
1,795512
1,795512
5
$begingroup$
Just $V_{omega+omega}$ is already enough.
$endgroup$
– Asaf Karagila♦
Jan 9 at 12:18
1
$begingroup$
@AsafKaragila Thanks for your comment. Since I am not extremely familiar with this topic and I wanted to be sure I opted to go for a regular cardinal, in order to be "safer". If I may add something to this, my answer would be wrong if we considered collection instead of replacement, right? What should we use as a model in that case?
$endgroup$
– Leo163
Jan 9 at 12:51
4
$begingroup$
Correct, for Replacement we use $H(kappa)$, the set of sets which are hereditarily smaller than $kappa$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 13:18
add a comment |
5
$begingroup$
Just $V_{omega+omega}$ is already enough.
$endgroup$
– Asaf Karagila♦
Jan 9 at 12:18
1
$begingroup$
@AsafKaragila Thanks for your comment. Since I am not extremely familiar with this topic and I wanted to be sure I opted to go for a regular cardinal, in order to be "safer". If I may add something to this, my answer would be wrong if we considered collection instead of replacement, right? What should we use as a model in that case?
$endgroup$
– Leo163
Jan 9 at 12:51
4
$begingroup$
Correct, for Replacement we use $H(kappa)$, the set of sets which are hereditarily smaller than $kappa$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 13:18
5
5
$begingroup$
Just $V_{omega+omega}$ is already enough.
$endgroup$
– Asaf Karagila♦
Jan 9 at 12:18
$begingroup$
Just $V_{omega+omega}$ is already enough.
$endgroup$
– Asaf Karagila♦
Jan 9 at 12:18
1
1
$begingroup$
@AsafKaragila Thanks for your comment. Since I am not extremely familiar with this topic and I wanted to be sure I opted to go for a regular cardinal, in order to be "safer". If I may add something to this, my answer would be wrong if we considered collection instead of replacement, right? What should we use as a model in that case?
$endgroup$
– Leo163
Jan 9 at 12:51
$begingroup$
@AsafKaragila Thanks for your comment. Since I am not extremely familiar with this topic and I wanted to be sure I opted to go for a regular cardinal, in order to be "safer". If I may add something to this, my answer would be wrong if we considered collection instead of replacement, right? What should we use as a model in that case?
$endgroup$
– Leo163
Jan 9 at 12:51
4
4
$begingroup$
Correct, for Replacement we use $H(kappa)$, the set of sets which are hereditarily smaller than $kappa$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 13:18
$begingroup$
Correct, for Replacement we use $H(kappa)$, the set of sets which are hereditarily smaller than $kappa$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 13:18
add a comment |
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1
$begingroup$
Incidentally, there are some very interesting results on the "amount of powerset" (or really, "powerset-along-replacement") we need in various situations. Most famously, Harvey Friedman showed that more and more iterations of powerset are needed to prove determinacy further and further up the Borel hierarchy.
$endgroup$
– Noah Schweber
Jan 11 at 22:08
1
$begingroup$
On formatting: You can use pairs of L& R dollar-signs to "display" a line instead of typing quad quad quad ..., and to put un-formatted text (like "because") in a displayed line, enclose it within text {...}, like text { because }.
$endgroup$
– DanielWainfleet
Jan 14 at 0:47
$begingroup$
@DanielWainfleet I've felt silly at times trying to (perfectly) align stuff since the final result might depend on the display device. But anyway, not exactly sure what you mean. Please share some links with useful mathstackexchange formatting technique so I can copy n paste it in the future.
$endgroup$
– CopyPasteIt
Jan 14 at 1:27
$begingroup$
@NoahSchweber Another interesting paper, 'Large irredundant sets in operator algebras' arxiv.org/abs/1808.01511?context=math I don't know the formulation for the $text{♢-Axiom}$ (stronger than the continuum hypothesis), but my guess is all modern day physics can be expressed/modeled in $V_{omega+omega} + text{♢}$. I wonder how this holds up: "...every open set of Minkowski spacetime is associated with a C*-algebra" /wikipedia.
$endgroup$
– CopyPasteIt
Jan 14 at 14:55
$begingroup$
Did something not work with the Ask Question page that you felt the need to ask a completely disjoint question on a page where you already accepted one answer?
$endgroup$
– Asaf Karagila♦
Jan 14 at 15:44