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The Complete Idiot’s Guide to the Independence of the Continuum Hypothesis

141 points by headalgorithm 6y ago

H8crilA 6y ago
Whenever I get around to studying modern mathematical logic I have this feeling of awe, as if I managed to get a glimpse of the potential of the "true nature of the world". Take the Löwenheim–Skolem theorem (mentioned in the post). It essentially tries to answer "what is the potential of illusion" within the first order logic, and shows that the potential is pretty damn big. You can be mind-blowingly confused about the "true nature of your reality". As you, a first-order logic creature, walk around, explore and interact with your space you can fool yourself into believing all kinds of marvellous things (for example that real numbers are "continuous", that they're inherently more complicated than naturals), you can even prove such things like our good friend Cantor did, never knowing that you were in fact always in a boring, aleph-zero universe. A universe built out of strings, or maybe natural numbers, nothing more than that. It's just that the daemon (engineer) ruling your world is incredibly clever and is adeptly laying out the bricks as you walk around, always putting them down just in time before you manage to turn your head around. And you'll never be able to catch that daemon and unravel the whole scheme. I am looking forward to the next parts! I have also always wanted to understand forcing.
- eximius 6y ago
  In the context of the Löwenheim–Skolem theorem and "Skolem's Paradox", what does it mean for 'the domain of a model to be countable'? My only guess is that the number of 'objects and/or relations' in the model forms a set of countable size. Contrast this with the size of the objects in the model. e.g., if you have a set of countably infinite cardinality where the elements might be sets of uncountably infinite size.
  - lowdanie 6y ago
    Indeed, the statement is that for any list of axioms there exists a countable set of objects satisfying them. For example, you could write down axioms for the real numbers by specifying that there should be relations called + and x with the standard properties such as commutativity, as well as an ordering relation < such that for all elements x and y there is an element z for which: x < z < y. Clearly the real numbers are a model for these axioms. But as it turns out the countable set of rational numbers is a model as well.
    - pdonis 6y ago
      > Clearly the real numbers are a model for these axioms. But as it turns out the countable set of rational numbers is a model as well. You missed the crucial property that rules out the rationals (more precisely, the rationals with their standard ordering): one way of stating it is that every sequence that has an upper bound in the set, must have a least upper bound in the set. The rationals do not satisfy this property (for example, consider the sequence of successive decimal expansions, each one to one more decimal place, of sqrt(2)), but the reals do. The challenge for me is to understand how there can still be countable sets that also satisfy that property of the reals. (Obviously any countable set can be put into one-to-one correspondence with the rationals, but for a countable set that satisfies the least upper bound property of the reals, such a correspondence with the rationals would put an ordering on the rationals that was not the standard one.)
      - zzless 6y ago
        In fact, he did not miss anything. Using the language he started with (variables range over 'numbers', and the relations are <, >, +, and *), the reals and the rationals indeed have the same properties (elementary theory as logicians would put it). The reason things like \sqrt2 present no problems is that it is simply impossible to define such 'sequences of numbers' in this theory (you are only allowed to 'refer' to numbers by your variables, not ordinary sets and the usual language for sets is missing). If I remember right, the fact he was referring to was proved by Tarsky.
        saithound 6y ago
        The reals and the rationals do not have the same elementary theory over the language (>,+,*).
        zzless 6y ago
        You are correct, one would have to exclude the multiplication for that.
        pdonis 6y ago
  - tprice7 6y ago
    I'm rusty with this stuff but I'm pretty sure your guess is correct, a countable model is one with countably many objects. For other readers who might not be familiar, I'll mention that Skolem's paradox is about how there are countable models of set theory, and yet it is a theorem of set theory that uncountable sets exist, so these countable models must contain sets that are uncountable according to the model. I think it seems less paradoxical if you think of it like this: in order for a set to be countable, there needs to exist an injection from that set to the natural numbers. So a countable model can have a set that internally looks uncountable: there is in fact an injection from that set to the natural numbers, it's just that the injection isn't included in the model.
  - deleted 6y ago
    [deleted]
  - dwohnitmok 6y ago
    In short: They both rely on a "background model" that you are implicitly working in. The "background model" and the model in discussion then disagree on which sets are countable. That is e.g. you fix some model (perhaps a model that satisfies ZFC). Then you work within that model to create a submodel .You can then describe the submodel using properties of the outer model. In fact it turns out that you can have submodels of ZFC that are proper sets in the outer model, i.e. a single set can contain an entire subuniverse of sets that themselves satisfy the entirety of the ZFC axioms. Using the outer model you can then talk about global properties of the submodel. In the case of the Löwenheim–Skolem theorem it turns out that you can have a submodel satisfying the axioms of ZFC that can have arbitrary infinite cardinality in the outer model. In particular you can have a submodel of ZFC that has only a countably infinite number of sets as measured by the outer model. And in fact every element of that set can also be a set of countably infinite size according to the outer model. According to the submodel there is no notion of the cardinality of itself, since ZFC does not have a set of all sets. Likewise the submodel "thinks" many sets within it are countably infinite. This is how the submodel is able to still satisfy ZFC. However, the outer model disagrees with the submodel and instead thinks that the submodel is "impoverished." The submodel is missing the functions that it needs to "realize" that bijections exist between certain sets. These functions exist in the outer model.
    - dwohnitmok 6y ago
      > Likewise the submodel "thinks" many sets within it are countably infinite. I meant uncountably infinite.
- pdonis 6y ago
  > You can be mind-blowingly confused about the "true nature of your reality". Only if you restrict yourself to first-order logic. So the real answer to the problems you pose is "don't do that".
  - bitdizzy 6y ago
    Which formal logic do you suggest to replace it?
    - pdonis 6y ago
      Second-order logic. (Not really to "replace", but to augment--to deal with those situations where first-order logic creates issues.) But I understand that not everyone would agree with that (apparently Scott Aaronson himself falls into this category, based on some of his comments in the comment section of the article).
      - dwohnitmok 6y ago
        The traditional reason that logicians do not trust second-order logic is that usually attempts to try to formalize it start to lean back on first-order logic, e.g. by depending on a background theory of ZFC which is a first-order theory or via Henkin semantics which are also essentially first-order logic semantics and share all the same paradoxes of first-order logic. Hence the solutions that second-order logic presents to the issues of first-order logic are really just smoke and mirrors that fall apart and turn into the same problems of first-order logic on closer inspection.
        andi999 6y ago
        Can you point me to (or explain briefly) what is second order logic?
        yaantc 6y ago
        In first order logic one can only quantify over objects of the domain: in "for all x such that ..." the variable "x" represent an object. In second order logic, one can quantify over relations (and functions, sets...): "for all P such that ...". The wikipaedia page [1] has more details and is quite readable. Be careful about falling into a rabbit hole ;) [1] https://en.wikipedia.org/wiki/Second-order_logic https://en.wikipedia.org/wiki/Second-order_logic
- cjfd 6y ago
  But is this actually 'fooling yourself'. It looks more like an approximation. What is space happens to be discrete but the discrete points are very small compared to every day life? Then for most purposes approximating a space coordinate with real numbers is a pretty good approximation. Never mind that the possible state space of the universe could actually be smaller than such a single real number. The thing is, a human brain can presumably only grasp finitely many things but actually having to reason about finitely many things is difficult. Note that theorems about numbers are easier when they are about 'natural numbers' than about 'all numbers smaller than 2^32' because all theorems then will have preconditions that limit their size in order to exclude integer overflow and hence get somewhat hairy. So as a cognitive trick working with infinite sets works pretty well. However, complicated reasonings about the nature of these infinities turn into navel gazing pretty quickly. The thing to strive for, I think, is extending the finite and the discrete with the infinite and the continuous but but not having the illusion that we can actually say very much about the nature of the infinte.
mcphage 6y ago
The story told to me by my set theory teacher back in college was that a Paul Cohen was an Analyst who attended a symposium on Set Theory and said "hey, this is all the set theorist do? This is easy!". He then switched his field to Set Theory, developed forcing, and proved a bunch of important open questions in the field. I don’t know if it’s true, but I’d like to believe.
- wannabebarista 6y ago
  I've heard a version of this too. With the additional details that after the result was announced, he struggled and took quite a bit of time to get it ready for publication.
- cwzwarich 6y ago
  I don’t really think this is true. See this historical article: http://math.bu.edu/people/aki/14.pdf http://math.bu.edu/people/aki/14.pdf He was drawn to logic by discussions with Feferman and worked unsuccessfully on other problems in logic before moving to independence questions in set theory.
  - mcphage 6y ago
    Alright then. I’m not sure if that makes me feel better or worse.
skybrian 6y ago
> Yes, there’s a “self-hating theory,” F+Not(Con(F)), which believes in its own inconsistency. And yes, by Gödel, this self-hating theory is consistent if F itself is—which means that it has a model (involving “nonstandard integers,” formal artifacts that effectively promise a proof of F’s inconsistency without ever actually delivering it). But this self-hating theory can’t be sound: I mean, just look at it! It’s either unsound because F is consistent, or else it’s unsound because F is inconsistent. I’m not following. What does “unsound” mean here? How can a theory be consistent but unsound?
- pdonis 6y ago
  > How can a theory be consistent but unsound? "Consistent" means that there is no proposition P such that both P and not-P have proofs within the theory. If F itself is consistent, then F+Not(Con(F)) will also be consistent, since Con(F) will have no proof in this theory. (Conversely, if F itself is inconsistent, then Con(F) will have a proof in this theory (since any proposition will have a proof in an inconsistent theory), so F+Not(Con(F)) will also be inconsistent.) "Sound" means that there exists some semantic model of the theory that the theorems of the theory accurately describe. This cannot true of the theory F+Not(Con(F)), since there are only two possibilities: (1) F itself is consistent--which means that, while F+Not(Con(F)) is also consistent, it can't be sound, because the theorem (by virtue of being an axiom) Not(Con(F)) cannot accurately describe any semantic model of the theory; or (2) F itself is inconsistent--which means that the theory F+Not(Con(F)) has no possible semantic model at all (since F, a subset of the theory, cannot--no inconsistent theory can have any semantic model), and therefore cannot be sound (since to be sound a theory must have some semantic model). Possibility #1 above answers your question.
  - dwohnitmok 6y ago
    1. is not true. There are semantic models of `F + Not(Con(F))` that accurately describe the theory. The difference is that these models contain nonstandard natural numbers (these nonstandard numbers become the Godel numberings of the proofs of `Not(Con(F))`) if `F` is "truly" consistent (according to the standard model of PA). The fact that there must be semantic models of any consistent theory is a cornerstone of first-order logic and is e.g. not true of second-order logic. This follows from Godel's Completeness Theorem (not Incompleteness Theorems).
RyanShook 6y ago
Idiot’s tldr: There are problems that math, as we currently understand it, cannot solve. “Even though Cantor proved that there are uncountably many real numbers, that only means there are uncountably many reals for us. We can’t rule out the possibly that God, looking down on our universe, would see countably many reals.”
- kibwen 6y ago
  An important second part of that quote is also that, from god's perspective, there would be impossible numbers that only a meta-god could access, and so on and so on. And it's not so much that math as we currently understand it cannot solve some problems, but more along the lines that math has used math to prove that there are some problems that math cannot solve.
IngoBlechschmid 6y ago
It's great that Scott is having his stab at solving the exposition problem for the Continuum Hypothesis. I'm looking forward to the future parts of the series! The Continuum Hypothesis is also very interesting because it is a poster child in a discussion between adherents of the "traditional dream solution" of the Continuum Hypothesis and the intriguing "multiverse philosophy of set theory". In short, the former is the attempt to devise new axioms which would settle the hypothesis. The latter is the position to embrace all possible mathematical universes. I tried to give an exposition of that circle of ideas here: https://iblech.gitlab.io/bb/multiverse.html https://iblech.gitlab.io/bb/multiverse.html Comments and questions are very welcome.
im3w1l 6y ago
So what I took away was that the Continuum hypothesis is boring. But this Löwenheim Skolem on the other hand, that sounds like some pretty crazy stuff.
framecowbird 6y ago
Set theory was definitely my favorite course that I took in pure mathematics. It blew my mind how you can study “infinity” and find much structure; which I had previously thought was a fairly vague concept. It doesn’t seem surprising to me now, but definitely did then. It might also have had something to do with the professor teaching the course, who was incredible.
zzless 6y ago
I am looking forward to the next part but he has already said something that will be a a problem later on: he said that according to L-S theorem there are countable sets that are models of set theory (ZFC to be precise). This is not really true since it contradicts Goedel's incompleteness theorem. What the method of forcing uses, however, is a pair of theorems by Mostowski: the reflection and the collapse theorem that let one find a model of any finite(ly axiomatizable) portion of ZFC. Finite axiomatizations of set theory (say, GB) have similar tricks. His intuition is right on though.
- Recursing 6y ago
  > he said that according to L-S theorem there are countable sets that are models of set theory (ZFC to be precise). This is not really true since it contradicts Goedel's incompleteness theorem Could you clarify this a bit? I think Scott's very open to suggestions and corrections
  - zzless 6y ago
    I should have said that one cannot prove this in ZFC (of course there may be models of ZFC in which this is true) since that would meant one can prove Con ZFC in ZFC. Mostowski's theorems however are provable in ZFC and are enough for the arithmetic proofs that are produced by the method of forcing.
zzless 6y ago
I have also found that students have a much easier time with forcing after learning about ultraproducts (and Los theorem) and boolean valued forcing (even though technically, the modern version using partially ordered sets of 'conditions' is much more convenient). Then you can think of forcing as a way of taking a 'model' where properties of elements are described by 'boolean valued' probabilities and collapsing it to a bona fide set theoretic universe.
bikenaga 6y ago
Paul Cohen wrote an exposition of his proof. It was republished by Dover: https://store.doverpublications.com/0486469212.html https://store.doverpublications.com/0486469212.html It starts with background material in logic and set theory (fast-paced, but reasonably easy to follow and fairly complete in what it covers). Contents: 1. General background in logic 2. Zermelo-Fraenkel set theory 3. The consistency of the continuum hypothesis and the axiom of choice 4. The independence of the continuum hypothesis and the axiom of choice I haven't gone through the whole thing, but what I've read is pretty clear. And the whole book is only 150 pages. This is not a textbook - no exercises. (Disclaimer: I'm a mathematician, but not a logician or a set theorist.)

141 points by headalgorithm 6y ago

Whenever I get around to studying modern mathematical logic I have this feeling of awe, as if I managed to get a glimpse of the potential of the "true nature of the world". Take the Löwenheim–Skolem theorem (mentioned in the post). It essentially tries to answer "what is the potential of illusion" within the first order logic, and shows that the potential is pretty damn big. You can be mind-blowingly confused about the "true nature of your reality". As you, a first-order logic creature, walk around, explore and interact with your space you can fool yourself into believing all kinds of marvellous things (for example that real numbers are "continuous", that they're inherently more complicated than naturals), you can even prove such things like our good friend Cantor did, never knowing that you were in fact always in a boring, aleph-zero universe. A universe built out of strings, or maybe natural numbers, nothing more than that. It's just that the daemon (engineer) ruling your world is incredibly clever and is adeptly laying out the bricks as you walk around, always putting them down just in time before you manage to turn your head around. And you'll never be able to catch that daemon and unravel the whole scheme. I am looking forward to the next parts! I have also always wanted to understand forcing.

eximius 6y ago
In the context of the Löwenheim–Skolem theorem and "Skolem's Paradox", what does it mean for 'the domain of a model to be countable'? My only guess is that the number of 'objects and/or relations' in the model forms a set of countable size. Contrast this with the size of the objects in the model. e.g., if you have a set of countably infinite cardinality where the elements might be sets of uncountably infinite size.
- lowdanie 6y ago
  Indeed, the statement is that for any list of axioms there exists a countable set of objects satisfying them. For example, you could write down axioms for the real numbers by specifying that there should be relations called + and x with the standard properties such as commutativity, as well as an ordering relation < such that for all elements x and y there is an element z for which: x < z < y. Clearly the real numbers are a model for these axioms. But as it turns out the countable set of rational numbers is a model as well.
  - pdonis 6y ago
    > Clearly the real numbers are a model for these axioms. But as it turns out the countable set of rational numbers is a model as well. You missed the crucial property that rules out the rationals (more precisely, the rationals with their standard ordering): one way of stating it is that every sequence that has an upper bound in the set, must have a least upper bound in the set. The rationals do not satisfy this property (for example, consider the sequence of successive decimal expansions, each one to one more decimal place, of sqrt(2)), but the reals do. The challenge for me is to understand how there can still be countable sets that also satisfy that property of the reals. (Obviously any countable set can be put into one-to-one correspondence with the rationals, but for a countable set that satisfies the least upper bound property of the reals, such a correspondence with the rationals would put an ordering on the rationals that was not the standard one.)
    - zzless 6y ago
      In fact, he did not miss anything. Using the language he started with (variables range over 'numbers', and the relations are <, >, +, and *), the reals and the rationals indeed have the same properties (elementary theory as logicians would put it). The reason things like \sqrt2 present no problems is that it is simply impossible to define such 'sequences of numbers' in this theory (you are only allowed to 'refer' to numbers by your variables, not ordinary sets and the usual language for sets is missing). If I remember right, the fact he was referring to was proved by Tarsky.
      - saithound 6y ago
        The reals and the rationals do not have the same elementary theory over the language (>,+,*).
        zzless 6y ago
        You are correct, one would have to exclude the multiplication for that.
      - pdonis 6y ago
- tprice7 6y ago
  I'm rusty with this stuff but I'm pretty sure your guess is correct, a countable model is one with countably many objects. For other readers who might not be familiar, I'll mention that Skolem's paradox is about how there are countable models of set theory, and yet it is a theorem of set theory that uncountable sets exist, so these countable models must contain sets that are uncountable according to the model. I think it seems less paradoxical if you think of it like this: in order for a set to be countable, there needs to exist an injection from that set to the natural numbers. So a countable model can have a set that internally looks uncountable: there is in fact an injection from that set to the natural numbers, it's just that the injection isn't included in the model.
- deleted 6y ago
  [deleted]
- dwohnitmok 6y ago
  In short: They both rely on a "background model" that you are implicitly working in. The "background model" and the model in discussion then disagree on which sets are countable. That is e.g. you fix some model (perhaps a model that satisfies ZFC). Then you work within that model to create a submodel .You can then describe the submodel using properties of the outer model. In fact it turns out that you can have submodels of ZFC that are proper sets in the outer model, i.e. a single set can contain an entire subuniverse of sets that themselves satisfy the entirety of the ZFC axioms. Using the outer model you can then talk about global properties of the submodel. In the case of the Löwenheim–Skolem theorem it turns out that you can have a submodel satisfying the axioms of ZFC that can have arbitrary infinite cardinality in the outer model. In particular you can have a submodel of ZFC that has only a countably infinite number of sets as measured by the outer model. And in fact every element of that set can also be a set of countably infinite size according to the outer model. According to the submodel there is no notion of the cardinality of itself, since ZFC does not have a set of all sets. Likewise the submodel "thinks" many sets within it are countably infinite. This is how the submodel is able to still satisfy ZFC. However, the outer model disagrees with the submodel and instead thinks that the submodel is "impoverished." The submodel is missing the functions that it needs to "realize" that bijections exist between certain sets. These functions exist in the outer model.
  - dwohnitmok 6y ago
    > Likewise the submodel "thinks" many sets within it are countably infinite. I meant uncountably infinite.
pdonis 6y ago
> You can be mind-blowingly confused about the "true nature of your reality". Only if you restrict yourself to first-order logic. So the real answer to the problems you pose is "don't do that".
- bitdizzy 6y ago
  Which formal logic do you suggest to replace it?
  - pdonis 6y ago
    Second-order logic. (Not really to "replace", but to augment--to deal with those situations where first-order logic creates issues.) But I understand that not everyone would agree with that (apparently Scott Aaronson himself falls into this category, based on some of his comments in the comment section of the article).
    - dwohnitmok 6y ago
      The traditional reason that logicians do not trust second-order logic is that usually attempts to try to formalize it start to lean back on first-order logic, e.g. by depending on a background theory of ZFC which is a first-order theory or via Henkin semantics which are also essentially first-order logic semantics and share all the same paradoxes of first-order logic. Hence the solutions that second-order logic presents to the issues of first-order logic are really just smoke and mirrors that fall apart and turn into the same problems of first-order logic on closer inspection.
      - andi999 6y ago
        Can you point me to (or explain briefly) what is second order logic?
        yaantc 6y ago
        In first order logic one can only quantify over objects of the domain: in "for all x such that ..." the variable "x" represent an object. In second order logic, one can quantify over relations (and functions, sets...): "for all P such that ...". The wikipaedia page [1] has more details and is quite readable. Be careful about falling into a rabbit hole ;) [1] https://en.wikipedia.org/wiki/Second-order_logic https://en.wikipedia.org/wiki/Second-order_logic
cjfd 6y ago
But is this actually 'fooling yourself'. It looks more like an approximation. What is space happens to be discrete but the discrete points are very small compared to every day life? Then for most purposes approximating a space coordinate with real numbers is a pretty good approximation. Never mind that the possible state space of the universe could actually be smaller than such a single real number. The thing is, a human brain can presumably only grasp finitely many things but actually having to reason about finitely many things is difficult. Note that theorems about numbers are easier when they are about 'natural numbers' than about 'all numbers smaller than 2^32' because all theorems then will have preconditions that limit their size in order to exclude integer overflow and hence get somewhat hairy. So as a cognitive trick working with infinite sets works pretty well. However, complicated reasonings about the nature of these infinities turn into navel gazing pretty quickly. The thing to strive for, I think, is extending the finite and the discrete with the infinite and the continuous but but not having the illusion that we can actually say very much about the nature of the infinte.

mcphage 6y ago

The story told to me by my set theory teacher back in college was that a Paul Cohen was an Analyst who attended a symposium on Set Theory and said "hey, this is all the set theorist do? This is easy!". He then switched his field to Set Theory, developed forcing, and proved a bunch of important open questions in the field. I don’t know if it’s true, but I’d like to believe.

wannabebarista 6y ago
I've heard a version of this too. With the additional details that after the result was announced, he struggled and took quite a bit of time to get it ready for publication.
cwzwarich 6y ago
I don’t really think this is true. See this historical article: http://math.bu.edu/people/aki/14.pdf http://math.bu.edu/people/aki/14.pdf He was drawn to logic by discussions with Feferman and worked unsuccessfully on other problems in logic before moving to independence questions in set theory.
- mcphage 6y ago
  Alright then. I’m not sure if that makes me feel better or worse.

skybrian 6y ago

> Yes, there’s a “self-hating theory,” F+Not(Con(F)), which believes in its own inconsistency. And yes, by Gödel, this self-hating theory is consistent if F itself is—which means that it has a model (involving “nonstandard integers,” formal artifacts that effectively promise a proof of F’s inconsistency without ever actually delivering it). But this self-hating theory can’t be sound: I mean, just look at it! It’s either unsound because F is consistent, or else it’s unsound because F is inconsistent. I’m not following. What does “unsound” mean here? How can a theory be consistent but unsound?

pdonis 6y ago
> How can a theory be consistent but unsound? "Consistent" means that there is no proposition P such that both P and not-P have proofs within the theory. If F itself is consistent, then F+Not(Con(F)) will also be consistent, since Con(F) will have no proof in this theory. (Conversely, if F itself is inconsistent, then Con(F) will have a proof in this theory (since any proposition will have a proof in an inconsistent theory), so F+Not(Con(F)) will also be inconsistent.) "Sound" means that there exists some semantic model of the theory that the theorems of the theory accurately describe. This cannot true of the theory F+Not(Con(F)), since there are only two possibilities: (1) F itself is consistent--which means that, while F+Not(Con(F)) is also consistent, it can't be sound, because the theorem (by virtue of being an axiom) Not(Con(F)) cannot accurately describe any semantic model of the theory; or (2) F itself is inconsistent--which means that the theory F+Not(Con(F)) has no possible semantic model at all (since F, a subset of the theory, cannot--no inconsistent theory can have any semantic model), and therefore cannot be sound (since to be sound a theory must have some semantic model). Possibility #1 above answers your question.
- dwohnitmok 6y ago
  1. is not true. There are semantic models of `F + Not(Con(F))` that accurately describe the theory. The difference is that these models contain nonstandard natural numbers (these nonstandard numbers become the Godel numberings of the proofs of `Not(Con(F))`) if `F` is "truly" consistent (according to the standard model of PA). The fact that there must be semantic models of any consistent theory is a cornerstone of first-order logic and is e.g. not true of second-order logic. This follows from Godel's Completeness Theorem (not Incompleteness Theorems).

RyanShook 6y ago

Idiot’s tldr: There are problems that math, as we currently understand it, cannot solve. “Even though Cantor proved that there are uncountably many real numbers, that only means there are uncountably many reals for us. We can’t rule out the possibly that God, looking down on our universe, would see countably many reals.”

kibwen 6y ago
An important second part of that quote is also that, from god's perspective, there would be impossible numbers that only a meta-god could access, and so on and so on. And it's not so much that math as we currently understand it cannot solve some problems, but more along the lines that math has used math to prove that there are some problems that math cannot solve.

IngoBlechschmid 6y ago

It's great that Scott is having his stab at solving the exposition problem for the Continuum Hypothesis. I'm looking forward to the future parts of the series! The Continuum Hypothesis is also very interesting because it is a poster child in a discussion between adherents of the "traditional dream solution" of the Continuum Hypothesis and the intriguing "multiverse philosophy of set theory". In short, the former is the attempt to devise new axioms which would settle the hypothesis. The latter is the position to embrace all possible mathematical universes. I tried to give an exposition of that circle of ideas here: https://iblech.gitlab.io/bb/multiverse.html https://iblech.gitlab.io/bb/multiverse.html Comments and questions are very welcome.

im3w1l 6y ago

So what I took away was that the Continuum hypothesis is boring. But this Löwenheim Skolem on the other hand, that sounds like some pretty crazy stuff.

framecowbird 6y ago

Set theory was definitely my favorite course that I took in pure mathematics. It blew my mind how you can study “infinity” and find much structure; which I had previously thought was a fairly vague concept. It doesn’t seem surprising to me now, but definitely did then. It might also have had something to do with the professor teaching the course, who was incredible.

zzless 6y ago

I am looking forward to the next part but he has already said something that will be a a problem later on: he said that according to L-S theorem there are countable sets that are models of set theory (ZFC to be precise). This is not really true since it contradicts Goedel's incompleteness theorem. What the method of forcing uses, however, is a pair of theorems by Mostowski: the reflection and the collapse theorem that let one find a model of any finite(ly axiomatizable) portion of ZFC. Finite axiomatizations of set theory (say, GB) have similar tricks. His intuition is right on though.

Recursing 6y ago
> he said that according to L-S theorem there are countable sets that are models of set theory (ZFC to be precise). This is not really true since it contradicts Goedel's incompleteness theorem Could you clarify this a bit? I think Scott's very open to suggestions and corrections
- zzless 6y ago
  I should have said that one cannot prove this in ZFC (of course there may be models of ZFC in which this is true) since that would meant one can prove Con ZFC in ZFC. Mostowski's theorems however are provable in ZFC and are enough for the arithmetic proofs that are produced by the method of forcing.

zzless 6y ago

I have also found that students have a much easier time with forcing after learning about ultraproducts (and Los theorem) and boolean valued forcing (even though technically, the modern version using partially ordered sets of 'conditions' is much more convenient). Then you can think of forcing as a way of taking a 'model' where properties of elements are described by 'boolean valued' probabilities and collapsing it to a bona fide set theoretic universe.

bikenaga 6y ago

Paul Cohen wrote an exposition of his proof. It was republished by Dover: https://store.doverpublications.com/0486469212.html https://store.doverpublications.com/0486469212.html It starts with background material in logic and set theory (fast-paced, but reasonably easy to follow and fairly complete in what it covers). Contents: 1. General background in logic 2. Zermelo-Fraenkel set theory 3. The consistency of the continuum hypothesis and the axiom of choice 4. The independence of the continuum hypothesis and the axiom of choice I haven't gone through the whole thing, but what I've read is pretty clear. And the whole book is only 150 pages. This is not a textbook - no exercises. (Disclaimer: I'm a mathematician, but not a logician or a set theorist.)