Why do some mathematicians feel like mathematical objects are “really out there” in some metaphysically fundamental sense? For example, if you ask mathematicians whether ZFC + not Consistent(ZFC) is consistent, they will say “no, of course not!” But given ZFC is consistent, the statement is in fact consistent due to by Godel’s second incompleteness theorem[1]. Similarly, if we have the Peano axioms without induction, mathematicians will say that induction should be there, but in fact you cannot prove this fact from within Peano, and given induction mathematicians will say transfinite induction should be there.
I argue that an explanation could be from logical induction. In logical induction, fast but possibly wrong sub-processes bet with each other over whether different mathematical facts will be proven true or false by a slow but ground-truth formal system prover. Another example of backstops in learning. But one result of this is that the successful sub-processes are not selected very hard to give null results on unprovable statements, producing spurious generalization and the subjective feeling—as expressed by probabilities for propositions—that some impossible theorems are true.
Of course, the platonist can still claim that this logical induction stuff is very similar to bayesian updating in the sense that both tell you something about the world, even when you can’t directly observe the relevant facts. If a photon exists your lightcone, there is no reason to stop believing the photon exists, even though there is no chance for you to ever encounter it again. Similarly, just because a statement is unprovable, doesn’t mean its right for you to have no opinion on the subject, insofar as the simplest & best internal logical-induction market traders have strong beliefs on the subject, they may very well be picking up on something metaphysically fundamental. Its simply the simplest explanation consistent with the facts.
The argument here is that there are two ways of proving ZFC + not Consistent(ZFC) is inconsistent. Either you prove not Consistent(ZFC) from axioms in ZFC or you contradict an axiom of ZFC from not Consistent(ZFC). The former is impossible by Godel’s second incompleteness theorem. The ladder is equivalent to proving Consistent(ZFC) from an axiom of ZFC (its contrapositive), which is also impossible by Godel.
insofar as the simplest & best internal logical-induction market traders have strong beliefs on the subject, they may very well be picking up on something metaphysically fundamental. Its simply the simplest explanation consistent with the facts.
Theorem 4.6.2 in logical induction says that the “probability” of independent statements does not converge to 1 or 0, but to something in-between.
So even if a mathematician says that some independent statement feels true (eg some objects are “really out there”), logical induction will tell him to feel uncertain about that.
For example, if you ask mathematicians whether ZFC + not Consistent(ZFC) is consistent, they will say “no, of course not!”
Certainly not a mathematician with any background in logic.
Similarly, if we have the Peano axioms without induction, mathematicians will say that induction should be there, but in fact you cannot prove this fact from within Peano
What exactly do you mean here? That the Peano axioms minus induction do not adequately characterize the natural numbers because they have nonstandard models? Why would I then be surprised that induction (which does characterize the natural numbers) can’t be proven from the remaining axioms?
Transfinite induction is a consequence of ZF that makes sense in the context in sets. Yes, it can prove additional statements about the natural numbers (e.g. goodstein sequences converge), but why would it be added as an axiom when the natural numbers are already characterized up to isomorphism by the Peano axioms? How would you even add it as an axiom in the language of natural numbers? (that last question is non-rhetorical).
You are making an error here: ZFC + not Consistent(ZFC) != ZFC.
Assuming ZFC + not Consistent(ZFC) we can prove Consistent(ZFC), because inconsistent systems can prove everything and ZFC + not Consistent(ZFC) + Consistent(ZFC) is, in fact, inconsistent. But it doesn’t say anything about consistency of ZFC itself, because you can freely assume any sufficiently powerful system instead of ZFC. If you assume inconsistent system, then system + not Consistent(system) is still inconsistent, if you assume consistent system, then system + not Consistent(system) is inconsistent for reasoning above, so it can’t prove whether assumed system is consistent or not.
The mistake you are making is assuming that “ZFC is consistent” = Consistent(ZFC) where the ladder is the Godel encoding for “ZFC is consistent” specified within the language of ZFC.
If your logic were valid, it would just as well break the entirety of the second incompleteness theorem. That is, you would say “well of course ZFC can prove Consistent(ZFC) if it is consistent, for either ZFC is consistent, and we’re done, or ZFC is not consistent, but that is a contradiction since ‘ZFC is consistent’ ⇒ Consistent(ZFC)”.
The fact is that ZFC itself cannot recognize that Consistent(ZFC) is equivalent to “ZFC is consistent”.
@Morpheus you too seem confused by this, so tagging you as well.
Completeness theorem states that consistent countable FO theory has a model. Compactness theorem states that FO theory has a model iff every finite subset of FO theory has a model. Both theorems are provable in ZFC.
Therefore:
Consistent(ZFC) <-> all finite subsets of ZFC have a model ->
not Consistent(ZFC) <-> some finite subsets of ZFC don’t have a model ->
some finite subsets of ZFC + not Consistent(ZFC) don’t have a model <->
Let’s back up here and clarify definitions before invoking any theorems. In the language of set theory, we have a countably infinite set of finite statements. Some statements imply other statements. A subset A of these statements is said to be consistent if they can all be assigned to true such that, when following the basic rules of logic, one does not arrive at a contradiction.
The compactness theorem is helpful when A is an infinite set.ZFC is a finite set of axioms, so let’s ignore everything about finite subsets of A and the compactness theorem; it’s not relevant. [Edit: as indicated by Amalthea’s reaction, this is wrong; some “axioms” in ZF are actually infinite sets of axioms, such as replacement]
I’ll now rewrite your last sentence as:
ZFC + not Consistent(ZFC) has no model <-> not Consistent(ZFC + not Consistent(ZFC))
This is true but irrelevant. Assuming ZFC is consistent, ZFC will not be able to prove its own consistency so [not Consistent(ZFC)] can be added as an axiom without affecting its consistency. This means that ZFC + [not Consistent(ZFC)] would indeed have a model; I forget how this goes but I think it’s something like “start with a model of ZFC, throw in a c that’s treated as a natural number and corresponds to the contradiction found in ZFC, then close”. I think c is automatically treated as greater than every “actual” natural number (and the way to show that this can be added without issue (I think) involves the compactness theorem).
Okay, I kinda understood where I am wrong spiritually-intuitively, but now I don’t understand where I’m wrong formally. Like which inference in chain
not Consistent(ZFC) → some subsets of ZFC don’t have a model → some subsets of ZFC + not Consistent(ZFC) don’t have a model → not Consistent(ZFC + not Consistent(ZFC))
It’s completely valid. And we can simplify it further to:
not Consistent(ZFC) → not Consistent(ZFC + not Consistent(ZFC))
because if a set of axioms is already inconsistent, then it’s inconsistent with anything added. But you still won’t be able to actually derive a contradiction from this.
Embarrassingly, that was a semi-unintended reaction — I would bet a small amount against that statement if someone gave me a resolution method, but am not motivated to figure one out, and realized this a second after making it — that I hadn’t figured out how to remove by the time you made that comment. Sorry.
Why do some mathematicians feel like mathematical objects are “really out there” in some metaphysically fundamental sense? For example, if you ask mathematicians whether
ZFC + not Consistent(ZFC)is consistent, they will say “no, of course not!” But given ZFC is consistent, the statement is in fact consistent due to by Godel’s second incompleteness theorem[1]. Similarly, if we have the Peano axioms without induction, mathematicians will say that induction should be there, but in fact you cannot prove this fact from within Peano, and given induction mathematicians will say transfinite induction should be there.I argue that an explanation could be from logical induction. In logical induction, fast but possibly wrong sub-processes bet with each other over whether different mathematical facts will be proven true or false by a slow but ground-truth formal system prover. Another example of backstops in learning. But one result of this is that the successful sub-processes are not selected very hard to give null results on unprovable statements, producing spurious generalization and the subjective feeling—as expressed by probabilities for propositions—that some impossible theorems are true.
Of course, the platonist can still claim that this logical induction stuff is very similar to bayesian updating in the sense that both tell you something about the world, even when you can’t directly observe the relevant facts. If a photon exists your lightcone, there is no reason to stop believing the photon exists, even though there is no chance for you to ever encounter it again. Similarly, just because a statement is unprovable, doesn’t mean its right for you to have no opinion on the subject, insofar as the simplest & best internal logical-induction market traders have strong beliefs on the subject, they may very well be picking up on something metaphysically fundamental. Its simply the simplest explanation consistent with the facts.
The argument here is that there are two ways of proving
ZFC + not Consistent(ZFC)is inconsistent. Either you provenot Consistent(ZFC)from axioms inZFCor you contradict an axiom ofZFCfromnot Consistent(ZFC). The former is impossible by Godel’s second incompleteness theorem. The ladder is equivalent to provingConsistent(ZFC)from an axiom ofZFC(its contrapositive), which is also impossible by Godel.Theorem 4.6.2 in logical induction says that the “probability” of independent statements does not converge to 1 or 0, but to something in-between. So even if a mathematician says that some independent statement feels true (eg some objects are “really out there”), logical induction will tell him to feel uncertain about that.
Certainly not a mathematician with any background in logic.
What exactly do you mean here? That the Peano axioms minus induction do not adequately characterize the natural numbers because they have nonstandard models? Why would I then be surprised that induction (which does characterize the natural numbers) can’t be proven from the remaining axioms?
Transfinite induction is a consequence of ZF that makes sense in the context in sets. Yes, it can prove additional statements about the natural numbers (e.g. goodstein sequences converge), but why would it be added as an axiom when the natural numbers are already characterized up to isomorphism by the Peano axioms? How would you even add it as an axiom in the language of natural numbers? (that last question is non-rhetorical).
You are making an error here:
ZFC + not Consistent(ZFC)!=ZFC.Assuming
ZFC + not Consistent(ZFC)we can proveConsistent(ZFC), because inconsistent systems can prove everything andZFC + not Consistent(ZFC) + Consistent(ZFC)is, in fact, inconsistent. But it doesn’t say anything about consistency of ZFC itself, because you can freely assume any sufficiently powerful system instead of ZFC. If you assume inconsistent system, thensystem + not Consistent(system)is still inconsistent, if you assume consistent system, thensystem + not Consistent(system)is inconsistent for reasoning above, so it can’t prove whether assumed system is consistent or not.The mistake you are making is assuming that “ZFC is consistent” =
Consistent(ZFC)where the ladder is the Godel encoding for “ZFC is consistent” specified within the language of ZFC.If your logic were valid, it would just as well break the entirety of the second incompleteness theorem. That is, you would say “well of course ZFC can prove Consistent(ZFC) if it is consistent, for either ZFC is consistent, and we’re done, or ZFC is not consistent, but that is a contradiction since ‘ZFC is consistent’ ⇒
Consistent(ZFC)”.The fact is that ZFC itself cannot recognize that
Consistent(ZFC)is equivalent to “ZFC is consistent”.@Morpheus you too seem confused by this, so tagging you as well.
Completeness theorem states that consistent countable FO theory has a model. Compactness theorem states that FO theory has a model iff every finite subset of FO theory has a model. Both theorems are provable in ZFC.
Therefore:
Consistent(ZFC) <-> all finite subsets of ZFC have a model ->
not Consistent(ZFC) <-> some finite subsets of ZFC don’t have a model ->
some finite subsets of ZFC + not Consistent(ZFC) don’t have a model <->
not Consistent(ZFC + not Consistent(ZFC)),
proven in ZFC + not Consistent(ZFC)
Let’s back up here and clarify definitions before invoking any theorems. In the language of set theory, we have a countably infinite set of finite statements. Some statements imply other statements. A subset A of these statements is said to be consistent if they can all be assigned to true such that, when following the basic rules of logic, one does not arrive at a contradiction.
The compactness theorem is helpful when A is an infinite set.ZFC is a finite set of axioms, so let’s ignore everything about finite subsets of A and the compactness theorem; it’s not relevant. [Edit: as indicated by Amalthea’s reaction, this is wrong; some “axioms” in ZF are actually infinite sets of axioms, such as replacement]
I’ll now rewrite your last sentence as:
ZFC + not Consistent(ZFC) has no model <-> not Consistent(ZFC + not Consistent(ZFC))
This is true but irrelevant. Assuming ZFC is consistent, ZFC will not be able to prove its own consistency so [not Consistent(ZFC)] can be added as an axiom without affecting its consistency. This means that ZFC + [not Consistent(ZFC)] would indeed have a model; I forget how this goes but I think it’s something like “start with a model of ZFC, throw in a c that’s treated as a natural number and corresponds to the contradiction found in ZFC, then close”. I think c is automatically treated as greater than every “actual” natural number (and the way to show that this can be added without issue (I think) involves the compactness theorem).
Okay, I kinda understood where I am wrong spiritually-intuitively, but now I don’t understand where I’m wrong formally. Like which inference in chain
not Consistent(ZFC) → some subsets of ZFC don’t have a model → some subsets of ZFC + not Consistent(ZFC) don’t have a model → not Consistent(ZFC + not Consistent(ZFC))
is actually invalid?
It’s completely valid. And we can simplify it further to:
not Consistent(ZFC) → not Consistent(ZFC + not Consistent(ZFC))
because if a set of axioms is already inconsistent, then it’s inconsistent with anything added. But you still won’t be able to actually derive a contradiction from this.
Edit: I think the right thing to do here is look at models for PA + not consistent(PA). I can’t find a nice treatment of this at the moment, but here’s a possibly wrong one by someone who was learning the subject at the time: https://angyansheng.github.io/blog/a-theory-that-proves-its-own-inconsistency
I suspect than many people’s intuitive interpretation of “consistent” is ω-consistent, especially if they are not aware of the distinction.
@Nick_Tarleton How much do you want to bet, and what resolution method do you have in mind?
Embarrassingly, that was a semi-unintended reaction — I would bet a small amount against that statement if someone gave me a resolution method, but am not motivated to figure one out, and realized this a second after making it — that I hadn’t figured out how to remove by the time you made that comment. Sorry.
You’re putting quite a lot of weight on what “mathematicians say”. Probably these people just haven’t thought very hard about it?