The Scope of Axiomatic Set Theory
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I am currently studying ZF set theory in terms of first-order logic and I am having trouble understanding the motivation behind this axiomatic formulation of set theory.
ZF set theory is a first-order language with the proper axioms of extensionality, foundation, specification, union, pairing, replacement, power set, well- ordering and infinity.
But why?
Is ZF set theory just a model for universal set theory in the same manner that Lagrangian Classical Mechanics is just a model for the universe. I.e neither are perfect?
Otherwise, what is the justification in choosing these axioms, what 'logic' are we basing our choices on?
For when we extend the theory to include the axiom of choice, we have added a new axiom.
What is the justification of adding the axiom of choice, and not other statements which we cannot prove (have been proved to be unprovable in set theory); such as the continuum hypotheses, provided they are consistent with the current axioms?
Why can't we add every statement that has been proved to be unprovable within the system, why do we only add the axiom of choice?
In this light, what is the scope of ZF(C) set theory?
logic set-theory axiom-of-choice foundations
edited Feb 15 at 16:57
YuiTo Cheng
2,0322636
asked Feb 15 at 13:40
MatthewMatthew
338112
$endgroup$
|
show 2 more comments
$begingroup$
I am currently studying ZF set theory in terms of first-order logic and I am having trouble understanding the motivation behind this axiomatic formulation of set theory.
ZF set theory is a first-order language with the proper axioms of extensionality, foundation, specification, union, pairing, replacement, power set, well- ordering and infinity.
But why?
Is ZF set theory just a model for universal set theory in the same manner that Lagrangian Classical Mechanics is just a model for the universe. I.e neither are perfect?
Otherwise, what is the justification in choosing these axioms, what 'logic' are we basing our choices on?
For when we extend the theory to include the axiom of choice, we have added a new axiom.
What is the justification of adding the axiom of choice, and not other statements which we cannot prove (have been proved to be unprovable in set theory); such as the continuum hypotheses, provided they are consistent with the current axioms?
Why can't we add every statement that has been proved to be unprovable within the system, why do we only add the axiom of choice?
In this light, what is the scope of ZF(C) set theory?
logic set-theory axiom-of-choice foundations
edited Feb 15 at 16:57
YuiTo Cheng
2,0322636
asked Feb 15 at 13:40
MatthewMatthew
338112
$endgroup$
3
$begingroup$
Gödel tackles much of what you're asking in his paper titled "What is Cantor's Continuum Problem?". I uploaded it to here (it will be autodeleted in thirty days). The takeaway is this: we use these axioms because they are true (from a realist point of view) and because they work - mathematics comes out nicely out of them.
$endgroup$
– Git Gud
Feb 15 at 14:05
1
$begingroup$
You can't add the axiom of choice if you already have well-ordering. I mean, you can, but it won't change anything.
$endgroup$
– Asaf Karagila♦
Feb 15 at 14:11
3
$begingroup$
There seem to be several different questions here: (1) Why do we prefer the system ZFC (as opposed to, say, ZF or ZFC+CH)? (2) Does the inability of ZFC to answer "natural" questions constitute a "fundamental flaw" in ZFC? (3) Is the program of "keep adding axioms whenever you hit an undecidable problem" a good one (and what criteria do we use to judge that)? (4) Why do we look to set theories for foundations of mathematics in the first place (even before settling on ZFC specifically - that is, why sets as opposed to, I don't know, fleens)?
$endgroup$
– Noah Schweber
Feb 15 at 14:31
3
$begingroup$
And possibly more. Can you narrow the scope of this question? I think in the absence of a more concentrated focus, it will be difficult to answer this satisfyingly.
$endgroup$
– Noah Schweber
Feb 15 at 14:32
3
$begingroup$
I removed the last sentence since it is nonsense and risks derailing an otherwise reasonable question.
$endgroup$
– Andrés E. Caicedo
Feb 15 at 16:07
|
show 2 more comments
$begingroup$
I am currently studying ZF set theory in terms of first-order logic and I am having trouble understanding the motivation behind this axiomatic formulation of set theory.
ZF set theory is a first-order language with the proper axioms of extensionality, foundation, specification, union, pairing, replacement, power set, well- ordering and infinity.
But why?
Is ZF set theory just a model for universal set theory in the same manner that Lagrangian Classical Mechanics is just a model for the universe. I.e neither are perfect?
Otherwise, what is the justification in choosing these axioms, what 'logic' are we basing our choices on?
For when we extend the theory to include the axiom of choice, we have added a new axiom.
What is the justification of adding the axiom of choice, and not other statements which we cannot prove (have been proved to be unprovable in set theory); such as the continuum hypotheses, provided they are consistent with the current axioms?
Why can't we add every statement that has been proved to be unprovable within the system, why do we only add the axiom of choice?
In this light, what is the scope of ZF(C) set theory?
logic set-theory axiom-of-choice foundations
edited Feb 15 at 16:57
YuiTo Cheng
2,0322636
asked Feb 15 at 13:40
MatthewMatthew
338112
$endgroup$
I am currently studying ZF set theory in terms of first-order logic and I am having trouble understanding the motivation behind this axiomatic formulation of set theory.
ZF set theory is a first-order language with the proper axioms of extensionality, foundation, specification, union, pairing, replacement, power set, well- ordering and infinity.
But why?
Is ZF set theory just a model for universal set theory in the same manner that Lagrangian Classical Mechanics is just a model for the universe. I.e neither are perfect?
Otherwise, what is the justification in choosing these axioms, what 'logic' are we basing our choices on?
For when we extend the theory to include the axiom of choice, we have added a new axiom.
What is the justification of adding the axiom of choice, and not other statements which we cannot prove (have been proved to be unprovable in set theory); such as the continuum hypotheses, provided they are consistent with the current axioms?
Why can't we add every statement that has been proved to be unprovable within the system, why do we only add the axiom of choice?
In this light, what is the scope of ZF(C) set theory?
logic set-theory axiom-of-choice foundations
logic set-theory axiom-of-choice foundations
edited Feb 15 at 16:57
YuiTo Cheng
2,0322636
asked Feb 15 at 13:40
MatthewMatthew
338112
edited Feb 15 at 16:57
YuiTo Cheng
2,0322636
asked Feb 15 at 13:40
MatthewMatthew
338112
edited Feb 15 at 16:57
YuiTo Cheng
2,0322636
edited Feb 15 at 16:57
YuiTo Cheng
2,0322636
edited Feb 15 at 16:57
YuiTo Cheng
2,0322636
2,0322636
asked Feb 15 at 13:40
MatthewMatthew
338112
asked Feb 15 at 13:40
MatthewMatthew
338112
asked Feb 15 at 13:40
MatthewMatthew
338112
338112
3
$begingroup$
Gödel tackles much of what you're asking in his paper titled "What is Cantor's Continuum Problem?". I uploaded it to here (it will be autodeleted in thirty days). The takeaway is this: we use these axioms because they are true (from a realist point of view) and because they work - mathematics comes out nicely out of them.
$endgroup$
– Git Gud
Feb 15 at 14:05
1
$begingroup$
You can't add the axiom of choice if you already have well-ordering. I mean, you can, but it won't change anything.
$endgroup$
– Asaf Karagila♦
Feb 15 at 14:11
3
$begingroup$
There seem to be several different questions here: (1) Why do we prefer the system ZFC (as opposed to, say, ZF or ZFC+CH)? (2) Does the inability of ZFC to answer "natural" questions constitute a "fundamental flaw" in ZFC? (3) Is the program of "keep adding axioms whenever you hit an undecidable problem" a good one (and what criteria do we use to judge that)? (4) Why do we look to set theories for foundations of mathematics in the first place (even before settling on ZFC specifically - that is, why sets as opposed to, I don't know, fleens)?
$endgroup$
– Noah Schweber
Feb 15 at 14:31
3
$begingroup$
And possibly more. Can you narrow the scope of this question? I think in the absence of a more concentrated focus, it will be difficult to answer this satisfyingly.
$endgroup$
– Noah Schweber
Feb 15 at 14:32
3
$begingroup$
I removed the last sentence since it is nonsense and risks derailing an otherwise reasonable question.
$endgroup$
– Andrés E. Caicedo
Feb 15 at 16:07
|
show 2 more comments
3
$begingroup$
Gödel tackles much of what you're asking in his paper titled "What is Cantor's Continuum Problem?". I uploaded it to here (it will be autodeleted in thirty days). The takeaway is this: we use these axioms because they are true (from a realist point of view) and because they work - mathematics comes out nicely out of them.
$endgroup$
– Git Gud
Feb 15 at 14:05
1
$begingroup$
You can't add the axiom of choice if you already have well-ordering. I mean, you can, but it won't change anything.
$endgroup$
– Asaf Karagila♦
Feb 15 at 14:11
3
$begingroup$
There seem to be several different questions here: (1) Why do we prefer the system ZFC (as opposed to, say, ZF or ZFC+CH)? (2) Does the inability of ZFC to answer "natural" questions constitute a "fundamental flaw" in ZFC? (3) Is the program of "keep adding axioms whenever you hit an undecidable problem" a good one (and what criteria do we use to judge that)? (4) Why do we look to set theories for foundations of mathematics in the first place (even before settling on ZFC specifically - that is, why sets as opposed to, I don't know, fleens)?
$endgroup$
– Noah Schweber
Feb 15 at 14:31
3
$begingroup$
And possibly more. Can you narrow the scope of this question? I think in the absence of a more concentrated focus, it will be difficult to answer this satisfyingly.
$endgroup$
– Noah Schweber
Feb 15 at 14:32
3
$begingroup$
I removed the last sentence since it is nonsense and risks derailing an otherwise reasonable question.
$endgroup$
– Andrés E. Caicedo
Feb 15 at 16:07
3
3
$begingroup$
Gödel tackles much of what you're asking in his paper titled "What is Cantor's Continuum Problem?". I uploaded it to here (it will be autodeleted in thirty days). The takeaway is this: we use these axioms because they are true (from a realist point of view) and because they work - mathematics comes out nicely out of them.
$endgroup$
– Git Gud
Feb 15 at 14:05
$begingroup$
Gödel tackles much of what you're asking in his paper titled "What is Cantor's Continuum Problem?". I uploaded it to here (it will be autodeleted in thirty days). The takeaway is this: we use these axioms because they are true (from a realist point of view) and because they work - mathematics comes out nicely out of them.
$endgroup$
– Git Gud
Feb 15 at 14:05
1
1
$begingroup$
You can't add the axiom of choice if you already have well-ordering. I mean, you can, but it won't change anything.
$endgroup$
– Asaf Karagila♦
Feb 15 at 14:11
$begingroup$
You can't add the axiom of choice if you already have well-ordering. I mean, you can, but it won't change anything.
$endgroup$
– Asaf Karagila♦
Feb 15 at 14:11
3
3
$begingroup$
There seem to be several different questions here: (1) Why do we prefer the system ZFC (as opposed to, say, ZF or ZFC+CH)? (2) Does the inability of ZFC to answer "natural" questions constitute a "fundamental flaw" in ZFC? (3) Is the program of "keep adding axioms whenever you hit an undecidable problem" a good one (and what criteria do we use to judge that)? (4) Why do we look to set theories for foundations of mathematics in the first place (even before settling on ZFC specifically - that is, why sets as opposed to, I don't know, fleens)?
$endgroup$
– Noah Schweber
Feb 15 at 14:31
$begingroup$
There seem to be several different questions here: (1) Why do we prefer the system ZFC (as opposed to, say, ZF or ZFC+CH)? (2) Does the inability of ZFC to answer "natural" questions constitute a "fundamental flaw" in ZFC? (3) Is the program of "keep adding axioms whenever you hit an undecidable problem" a good one (and what criteria do we use to judge that)? (4) Why do we look to set theories for foundations of mathematics in the first place (even before settling on ZFC specifically - that is, why sets as opposed to, I don't know, fleens)?
$endgroup$
– Noah Schweber
Feb 15 at 14:31
3
3
$begingroup$
And possibly more. Can you narrow the scope of this question? I think in the absence of a more concentrated focus, it will be difficult to answer this satisfyingly.
$endgroup$
– Noah Schweber
Feb 15 at 14:32
$begingroup$
And possibly more. Can you narrow the scope of this question? I think in the absence of a more concentrated focus, it will be difficult to answer this satisfyingly.
$endgroup$
– Noah Schweber
Feb 15 at 14:32
3
3
$begingroup$
I removed the last sentence since it is nonsense and risks derailing an otherwise reasonable question.
$endgroup$
– Andrés E. Caicedo
Feb 15 at 16:07
$begingroup$
I removed the last sentence since it is nonsense and risks derailing an otherwise reasonable question.
$endgroup$
– Andrés E. Caicedo
Feb 15 at 16:07
|
show 2 more comments
2 Answers
2
active
oldest
votes
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If we add to ZFC every statement that is proved to be unprovable in ZFC, we would obtain an inconsistent system, because there are sentences that are both not-provable and not-disprovable in ZFC.
Take the particular case of the continuum hypothesis, CH. Both CH and its negation are unprovable in ZFC. So we can only add one of them to ZFC at a time. But then we have to decide which one. Of course, we can assume either CH or its negation on a temporary basis, but if we wanted to add one "once and for all" we would have to decide whether there is a consensus to assume CH, or a consensus to assume the negation. Because there is not a widely accepted argument for either of these options, we don't add either axiom to ZFC on a permanent basis.
The axioms that are in ZFC are all consistent with a particular understanding of sets as forming a cumulative hierarchy. Indeed, ZFC as a theory is set up for studying the cumulative hierarchy, rather than any sets that might exist outside of it. But our understanding of the cumulative hierarchy is not complete, so that we have no idea whether CH (for example) should hold.
There has been a lot written on whether new axioms should be added as "basic axioms" in mathematics, extending ZFC. For example, see
Feferman, S., Friedman, H., Maddy, P., & Steel, J. (2000). Does Mathematics Need New Axioms? Bulletin of Symbolic Logic, 6(4), 401-446. doi:10.2307/420965
answered Feb 15 at 14:12
Carl MummertCarl Mummert
67.5k7133251
$endgroup$
1
$begingroup$
As an interesting point, in the fourth section of the article I mentioned in the comments, Gödel says that "certain facts (not known at Cantor's time) (...) seem to indicate that Cantor's conjecture [i.e. the CH] will turn out to be wrong".
$endgroup$
– Git Gud
Feb 15 at 15:08
add a comment |
$begingroup$
I'm no historian, so I can't say anything about why Zermelo and Fraenkel came down to that exact list of axioms. However, I do believe that they felt it was a list which embodied the most important properties of "naive" set theory while still avoiding the paradoxes that had begun to shake the foundations of mathematics.
Could they have made a different list? Almost certainly. And set theory as we know it today might have been slightly different. But not too different: remember that they (presumably) wanted to retain as much as possible of naive set theory, so whatever axiom list they made had to result in a theory close to what we have now.
answered Feb 15 at 14:15
ArthurArthur
117k7116200
$endgroup$
6
$begingroup$
Historically, Zermelo came up with a weaker set of axioms first (en.wikipedia.org/wiki/Zermelo_set_theory). Zermelo's 1908 paper was written before our current understanding of first-order logic was very developed. His axioms were refined and polished, based on ideas proposed independently by Fraenkel and Skolem (e.g. the replacement axiom, and using first-order logic to phrase the comprehension scheme). For whatever reason, Skolem's name did not get attached to the theory. The axiom of regularity was added separately, first proposed by Mirimanoff.
$endgroup$
– Carl Mummert
Feb 15 at 14:18
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@CarlMummert. K. Kunen, in Set Theory: An Introduction to Independence Proofs, includes another axiom: Existence: $exists x;(x=x). $ He first shows the axiomatic development of much of ZF and ZFC that can be done without using the axiom of Infinity. Without Infinity and without Existence you have axioms that all begin with "$forall$" and you can't prove any sentence that begins with "$exists$".
$endgroup$
– DanielWainfleet
Feb 28 at 12:09
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@DanielWainfleet: in modern logic the axiom of existence is normally considered part of the underlying first-order logic; it follows from the rules of the proof system even if no nonlogical axioms are assumed. Kunen's mention of it as an axiom is a little idosyncratic.
$endgroup$
– Carl Mummert
Feb 28 at 12:42
add a comment |
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2 Answers
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If we add to ZFC every statement that is proved to be unprovable in ZFC, we would obtain an inconsistent system, because there are sentences that are both not-provable and not-disprovable in ZFC.
Take the particular case of the continuum hypothesis, CH. Both CH and its negation are unprovable in ZFC. So we can only add one of them to ZFC at a time. But then we have to decide which one. Of course, we can assume either CH or its negation on a temporary basis, but if we wanted to add one "once and for all" we would have to decide whether there is a consensus to assume CH, or a consensus to assume the negation. Because there is not a widely accepted argument for either of these options, we don't add either axiom to ZFC on a permanent basis.
The axioms that are in ZFC are all consistent with a particular understanding of sets as forming a cumulative hierarchy. Indeed, ZFC as a theory is set up for studying the cumulative hierarchy, rather than any sets that might exist outside of it. But our understanding of the cumulative hierarchy is not complete, so that we have no idea whether CH (for example) should hold.
There has been a lot written on whether new axioms should be added as "basic axioms" in mathematics, extending ZFC. For example, see
Feferman, S., Friedman, H., Maddy, P., & Steel, J. (2000). Does Mathematics Need New Axioms? Bulletin of Symbolic Logic, 6(4), 401-446. doi:10.2307/420965
answered Feb 15 at 14:12
Carl MummertCarl Mummert
67.5k7133251
$endgroup$
1
$begingroup$
As an interesting point, in the fourth section of the article I mentioned in the comments, Gödel says that "certain facts (not known at Cantor's time) (...) seem to indicate that Cantor's conjecture [i.e. the CH] will turn out to be wrong".
$endgroup$
– Git Gud
Feb 15 at 15:08
add a comment |
$begingroup$
If we add to ZFC every statement that is proved to be unprovable in ZFC, we would obtain an inconsistent system, because there are sentences that are both not-provable and not-disprovable in ZFC.
Take the particular case of the continuum hypothesis, CH. Both CH and its negation are unprovable in ZFC. So we can only add one of them to ZFC at a time. But then we have to decide which one. Of course, we can assume either CH or its negation on a temporary basis, but if we wanted to add one "once and for all" we would have to decide whether there is a consensus to assume CH, or a consensus to assume the negation. Because there is not a widely accepted argument for either of these options, we don't add either axiom to ZFC on a permanent basis.
The axioms that are in ZFC are all consistent with a particular understanding of sets as forming a cumulative hierarchy. Indeed, ZFC as a theory is set up for studying the cumulative hierarchy, rather than any sets that might exist outside of it. But our understanding of the cumulative hierarchy is not complete, so that we have no idea whether CH (for example) should hold.
There has been a lot written on whether new axioms should be added as "basic axioms" in mathematics, extending ZFC. For example, see
Feferman, S., Friedman, H., Maddy, P., & Steel, J. (2000). Does Mathematics Need New Axioms? Bulletin of Symbolic Logic, 6(4), 401-446. doi:10.2307/420965
answered Feb 15 at 14:12
Carl MummertCarl Mummert
67.5k7133251
$endgroup$
1
$begingroup$
As an interesting point, in the fourth section of the article I mentioned in the comments, Gödel says that "certain facts (not known at Cantor's time) (...) seem to indicate that Cantor's conjecture [i.e. the CH] will turn out to be wrong".
$endgroup$
– Git Gud
Feb 15 at 15:08
add a comment |
$begingroup$
If we add to ZFC every statement that is proved to be unprovable in ZFC, we would obtain an inconsistent system, because there are sentences that are both not-provable and not-disprovable in ZFC.
Take the particular case of the continuum hypothesis, CH. Both CH and its negation are unprovable in ZFC. So we can only add one of them to ZFC at a time. But then we have to decide which one. Of course, we can assume either CH or its negation on a temporary basis, but if we wanted to add one "once and for all" we would have to decide whether there is a consensus to assume CH, or a consensus to assume the negation. Because there is not a widely accepted argument for either of these options, we don't add either axiom to ZFC on a permanent basis.
The axioms that are in ZFC are all consistent with a particular understanding of sets as forming a cumulative hierarchy. Indeed, ZFC as a theory is set up for studying the cumulative hierarchy, rather than any sets that might exist outside of it. But our understanding of the cumulative hierarchy is not complete, so that we have no idea whether CH (for example) should hold.
There has been a lot written on whether new axioms should be added as "basic axioms" in mathematics, extending ZFC. For example, see
Feferman, S., Friedman, H., Maddy, P., & Steel, J. (2000). Does Mathematics Need New Axioms? Bulletin of Symbolic Logic, 6(4), 401-446. doi:10.2307/420965
answered Feb 15 at 14:12
Carl MummertCarl Mummert
67.5k7133251
$endgroup$
If we add to ZFC every statement that is proved to be unprovable in ZFC, we would obtain an inconsistent system, because there are sentences that are both not-provable and not-disprovable in ZFC.
Take the particular case of the continuum hypothesis, CH. Both CH and its negation are unprovable in ZFC. So we can only add one of them to ZFC at a time. But then we have to decide which one. Of course, we can assume either CH or its negation on a temporary basis, but if we wanted to add one "once and for all" we would have to decide whether there is a consensus to assume CH, or a consensus to assume the negation. Because there is not a widely accepted argument for either of these options, we don't add either axiom to ZFC on a permanent basis.
The axioms that are in ZFC are all consistent with a particular understanding of sets as forming a cumulative hierarchy. Indeed, ZFC as a theory is set up for studying the cumulative hierarchy, rather than any sets that might exist outside of it. But our understanding of the cumulative hierarchy is not complete, so that we have no idea whether CH (for example) should hold.
There has been a lot written on whether new axioms should be added as "basic axioms" in mathematics, extending ZFC. For example, see
Feferman, S., Friedman, H., Maddy, P., & Steel, J. (2000). Does Mathematics Need New Axioms? Bulletin of Symbolic Logic, 6(4), 401-446. doi:10.2307/420965
answered Feb 15 at 14:12
Carl MummertCarl Mummert
67.5k7133251
answered Feb 15 at 14:12
Carl MummertCarl Mummert
67.5k7133251
answered Feb 15 at 14:12
Carl MummertCarl Mummert
67.5k7133251
answered Feb 15 at 14:12
Carl MummertCarl Mummert
67.5k7133251
67.5k7133251
1
$begingroup$
As an interesting point, in the fourth section of the article I mentioned in the comments, Gödel says that "certain facts (not known at Cantor's time) (...) seem to indicate that Cantor's conjecture [i.e. the CH] will turn out to be wrong".
$endgroup$
– Git Gud
Feb 15 at 15:08
add a comment |
1
$begingroup$
As an interesting point, in the fourth section of the article I mentioned in the comments, Gödel says that "certain facts (not known at Cantor's time) (...) seem to indicate that Cantor's conjecture [i.e. the CH] will turn out to be wrong".
$endgroup$
– Git Gud
Feb 15 at 15:08
1
1
$begingroup$
As an interesting point, in the fourth section of the article I mentioned in the comments, Gödel says that "certain facts (not known at Cantor's time) (...) seem to indicate that Cantor's conjecture [i.e. the CH] will turn out to be wrong".
$endgroup$
– Git Gud
Feb 15 at 15:08
$begingroup$
As an interesting point, in the fourth section of the article I mentioned in the comments, Gödel says that "certain facts (not known at Cantor's time) (...) seem to indicate that Cantor's conjecture [i.e. the CH] will turn out to be wrong".
$endgroup$
– Git Gud
Feb 15 at 15:08
add a comment |
$begingroup$
I'm no historian, so I can't say anything about why Zermelo and Fraenkel came down to that exact list of axioms. However, I do believe that they felt it was a list which embodied the most important properties of "naive" set theory while still avoiding the paradoxes that had begun to shake the foundations of mathematics.
Could they have made a different list? Almost certainly. And set theory as we know it today might have been slightly different. But not too different: remember that they (presumably) wanted to retain as much as possible of naive set theory, so whatever axiom list they made had to result in a theory close to what we have now.
answered Feb 15 at 14:15
ArthurArthur
117k7116200
$endgroup$
6
$begingroup$
Historically, Zermelo came up with a weaker set of axioms first (en.wikipedia.org/wiki/Zermelo_set_theory). Zermelo's 1908 paper was written before our current understanding of first-order logic was very developed. His axioms were refined and polished, based on ideas proposed independently by Fraenkel and Skolem (e.g. the replacement axiom, and using first-order logic to phrase the comprehension scheme). For whatever reason, Skolem's name did not get attached to the theory. The axiom of regularity was added separately, first proposed by Mirimanoff.
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– Carl Mummert
Feb 15 at 14:18
$begingroup$
@CarlMummert. K. Kunen, in Set Theory: An Introduction to Independence Proofs, includes another axiom: Existence: $exists x;(x=x). $ He first shows the axiomatic development of much of ZF and ZFC that can be done without using the axiom of Infinity. Without Infinity and without Existence you have axioms that all begin with "$forall$" and you can't prove any sentence that begins with "$exists$".
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– DanielWainfleet
Feb 28 at 12:09
$begingroup$
@DanielWainfleet: in modern logic the axiom of existence is normally considered part of the underlying first-order logic; it follows from the rules of the proof system even if no nonlogical axioms are assumed. Kunen's mention of it as an axiom is a little idosyncratic.
$endgroup$
– Carl Mummert
Feb 28 at 12:42
add a comment |
$begingroup$
I'm no historian, so I can't say anything about why Zermelo and Fraenkel came down to that exact list of axioms. However, I do believe that they felt it was a list which embodied the most important properties of "naive" set theory while still avoiding the paradoxes that had begun to shake the foundations of mathematics.
Could they have made a different list? Almost certainly. And set theory as we know it today might have been slightly different. But not too different: remember that they (presumably) wanted to retain as much as possible of naive set theory, so whatever axiom list they made had to result in a theory close to what we have now.
answered Feb 15 at 14:15
ArthurArthur
117k7116200
$endgroup$
6
$begingroup$
Historically, Zermelo came up with a weaker set of axioms first (en.wikipedia.org/wiki/Zermelo_set_theory). Zermelo's 1908 paper was written before our current understanding of first-order logic was very developed. His axioms were refined and polished, based on ideas proposed independently by Fraenkel and Skolem (e.g. the replacement axiom, and using first-order logic to phrase the comprehension scheme). For whatever reason, Skolem's name did not get attached to the theory. The axiom of regularity was added separately, first proposed by Mirimanoff.
$endgroup$
– Carl Mummert
Feb 15 at 14:18
$begingroup$
@CarlMummert. K. Kunen, in Set Theory: An Introduction to Independence Proofs, includes another axiom: Existence: $exists x;(x=x). $ He first shows the axiomatic development of much of ZF and ZFC that can be done without using the axiom of Infinity. Without Infinity and without Existence you have axioms that all begin with "$forall$" and you can't prove any sentence that begins with "$exists$".
$endgroup$
– DanielWainfleet
Feb 28 at 12:09
$begingroup$
@DanielWainfleet: in modern logic the axiom of existence is normally considered part of the underlying first-order logic; it follows from the rules of the proof system even if no nonlogical axioms are assumed. Kunen's mention of it as an axiom is a little idosyncratic.
$endgroup$
– Carl Mummert
Feb 28 at 12:42
add a comment |
$begingroup$
I'm no historian, so I can't say anything about why Zermelo and Fraenkel came down to that exact list of axioms. However, I do believe that they felt it was a list which embodied the most important properties of "naive" set theory while still avoiding the paradoxes that had begun to shake the foundations of mathematics.
Could they have made a different list? Almost certainly. And set theory as we know it today might have been slightly different. But not too different: remember that they (presumably) wanted to retain as much as possible of naive set theory, so whatever axiom list they made had to result in a theory close to what we have now.
answered Feb 15 at 14:15
ArthurArthur
117k7116200
$endgroup$
I'm no historian, so I can't say anything about why Zermelo and Fraenkel came down to that exact list of axioms. However, I do believe that they felt it was a list which embodied the most important properties of "naive" set theory while still avoiding the paradoxes that had begun to shake the foundations of mathematics.
Could they have made a different list? Almost certainly. And set theory as we know it today might have been slightly different. But not too different: remember that they (presumably) wanted to retain as much as possible of naive set theory, so whatever axiom list they made had to result in a theory close to what we have now.
answered Feb 15 at 14:15
ArthurArthur
117k7116200
edited Feb 15 at 14:28
answered Feb 15 at 14:15
ArthurArthur
117k7116200
answered Feb 15 at 14:15
ArthurArthur
117k7116200
answered Feb 15 at 14:15
ArthurArthur
117k7116200
117k7116200
6
$begingroup$
Historically, Zermelo came up with a weaker set of axioms first (en.wikipedia.org/wiki/Zermelo_set_theory). Zermelo's 1908 paper was written before our current understanding of first-order logic was very developed. His axioms were refined and polished, based on ideas proposed independently by Fraenkel and Skolem (e.g. the replacement axiom, and using first-order logic to phrase the comprehension scheme). For whatever reason, Skolem's name did not get attached to the theory. The axiom of regularity was added separately, first proposed by Mirimanoff.
$endgroup$
– Carl Mummert
Feb 15 at 14:18
$begingroup$
@CarlMummert. K. Kunen, in Set Theory: An Introduction to Independence Proofs, includes another axiom: Existence: $exists x;(x=x). $ He first shows the axiomatic development of much of ZF and ZFC that can be done without using the axiom of Infinity. Without Infinity and without Existence you have axioms that all begin with "$forall$" and you can't prove any sentence that begins with "$exists$".
$endgroup$
– DanielWainfleet
Feb 28 at 12:09
$begingroup$
@DanielWainfleet: in modern logic the axiom of existence is normally considered part of the underlying first-order logic; it follows from the rules of the proof system even if no nonlogical axioms are assumed. Kunen's mention of it as an axiom is a little idosyncratic.
$endgroup$
– Carl Mummert
Feb 28 at 12:42
add a comment |
6
$begingroup$
Historically, Zermelo came up with a weaker set of axioms first (en.wikipedia.org/wiki/Zermelo_set_theory). Zermelo's 1908 paper was written before our current understanding of first-order logic was very developed. His axioms were refined and polished, based on ideas proposed independently by Fraenkel and Skolem (e.g. the replacement axiom, and using first-order logic to phrase the comprehension scheme). For whatever reason, Skolem's name did not get attached to the theory. The axiom of regularity was added separately, first proposed by Mirimanoff.
$endgroup$
– Carl Mummert
Feb 15 at 14:18
$begingroup$
@CarlMummert. K. Kunen, in Set Theory: An Introduction to Independence Proofs, includes another axiom: Existence: $exists x;(x=x). $ He first shows the axiomatic development of much of ZF and ZFC that can be done without using the axiom of Infinity. Without Infinity and without Existence you have axioms that all begin with "$forall$" and you can't prove any sentence that begins with "$exists$".
$endgroup$
– DanielWainfleet
Feb 28 at 12:09
$begingroup$
@DanielWainfleet: in modern logic the axiom of existence is normally considered part of the underlying first-order logic; it follows from the rules of the proof system even if no nonlogical axioms are assumed. Kunen's mention of it as an axiom is a little idosyncratic.
$endgroup$
– Carl Mummert
Feb 28 at 12:42
6
6
$begingroup$
Historically, Zermelo came up with a weaker set of axioms first (en.wikipedia.org/wiki/Zermelo_set_theory). Zermelo's 1908 paper was written before our current understanding of first-order logic was very developed. His axioms were refined and polished, based on ideas proposed independently by Fraenkel and Skolem (e.g. the replacement axiom, and using first-order logic to phrase the comprehension scheme). For whatever reason, Skolem's name did not get attached to the theory. The axiom of regularity was added separately, first proposed by Mirimanoff.
$endgroup$
– Carl Mummert
Feb 15 at 14:18
$begingroup$
Historically, Zermelo came up with a weaker set of axioms first (en.wikipedia.org/wiki/Zermelo_set_theory). Zermelo's 1908 paper was written before our current understanding of first-order logic was very developed. His axioms were refined and polished, based on ideas proposed independently by Fraenkel and Skolem (e.g. the replacement axiom, and using first-order logic to phrase the comprehension scheme). For whatever reason, Skolem's name did not get attached to the theory. The axiom of regularity was added separately, first proposed by Mirimanoff.
$endgroup$
– Carl Mummert
Feb 15 at 14:18
$begingroup$
@CarlMummert. K. Kunen, in Set Theory: An Introduction to Independence Proofs, includes another axiom: Existence: $exists x;(x=x). $ He first shows the axiomatic development of much of ZF and ZFC that can be done without using the axiom of Infinity. Without Infinity and without Existence you have axioms that all begin with "$forall$" and you can't prove any sentence that begins with "$exists$".
$endgroup$
– DanielWainfleet
Feb 28 at 12:09
$begingroup$
@CarlMummert. K. Kunen, in Set Theory: An Introduction to Independence Proofs, includes another axiom: Existence: $exists x;(x=x). $ He first shows the axiomatic development of much of ZF and ZFC that can be done without using the axiom of Infinity. Without Infinity and without Existence you have axioms that all begin with "$forall$" and you can't prove any sentence that begins with "$exists$".
$endgroup$
– DanielWainfleet
Feb 28 at 12:09
$begingroup$
@DanielWainfleet: in modern logic the axiom of existence is normally considered part of the underlying first-order logic; it follows from the rules of the proof system even if no nonlogical axioms are assumed. Kunen's mention of it as an axiom is a little idosyncratic.
$endgroup$
– Carl Mummert
Feb 28 at 12:42
$begingroup$
@DanielWainfleet: in modern logic the axiom of existence is normally considered part of the underlying first-order logic; it follows from the rules of the proof system even if no nonlogical axioms are assumed. Kunen's mention of it as an axiom is a little idosyncratic.
$endgroup$
– Carl Mummert
Feb 28 at 12:42
add a comment |
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Gödel tackles much of what you're asking in his paper titled "What is Cantor's Continuum Problem?". I uploaded it to here (it will be autodeleted in thirty days). The takeaway is this: we use these axioms because they are true (from a realist point of view) and because they work - mathematics comes out nicely out of them.
$endgroup$
– Git Gud
Feb 15 at 14:05
1
$begingroup$
You can't add the axiom of choice if you already have well-ordering. I mean, you can, but it won't change anything.
$endgroup$
– Asaf Karagila♦
Feb 15 at 14:11
3
$begingroup$
There seem to be several different questions here: (1) Why do we prefer the system ZFC (as opposed to, say, ZF or ZFC+CH)? (2) Does the inability of ZFC to answer "natural" questions constitute a "fundamental flaw" in ZFC? (3) Is the program of "keep adding axioms whenever you hit an undecidable problem" a good one (and what criteria do we use to judge that)? (4) Why do we look to set theories for foundations of mathematics in the first place (even before settling on ZFC specifically - that is, why sets as opposed to, I don't know, fleens)?
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– Noah Schweber
Feb 15 at 14:31
3
$begingroup$
And possibly more. Can you narrow the scope of this question? I think in the absence of a more concentrated focus, it will be difficult to answer this satisfyingly.
$endgroup$
– Noah Schweber
Feb 15 at 14:32
3
$begingroup$
I removed the last sentence since it is nonsense and risks derailing an otherwise reasonable question.
$endgroup$
– Andrés E. Caicedo
Feb 15 at 16:07