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Given a enough strong formal theory capable to form Continuum hypothesis. from law of excluded middle and Noncontradiction, one and only one of CH and negative CH should be true. but the consistent and independence of Continuum in ZFC tell me either CH or nagative can be true in ZFC, determined by which one selected as new axiom, So it infer that we can't form the sentence of CH in ZFC. if ZFC is weak to can't even express CH, How could it be? What is wrong in the above literature.

@asaf: you maybe right, what i really concerned is if a sentence can form, then the sentence should be true or false, no matter which new axiom selected. mathematical sentence should be true or false by its self, New axiom add to formal theory can't change the truth/falsity of the sentence. So there is no independence for every mathematical sentence if the sentence is grammar right.

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I don't fully understand the choice of tags here. –  Asaf Karagila May 20 '13 at 0:30
No. This is exactly what I am trying to explain in my answer. Even if you can write a sentence, it doesn't have to be provable. It is true or false in one particular model, not in all models. Provability is equivalent to being true in all models, not just some. $\sf CH$ is true in some models, and false in others, so it's not provable. Being true or false requires us to choose a particular model and asking whether $\sf CH$ is true in that model or not. –  Asaf Karagila May 20 '13 at 0:45
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3 Answers

You confuse true with provable, and provable and expressible.

In a particular model of a theory, statements are true or false. Having a provable statement means that we can write a proof from a certain theory. The incompleteness theorem, for example, shows that not every statement which is true in a particular model is also provable. On the other hand, if a statement is true in every model, then it is provable. The continuum hypothesis is not provable, which means that there are models where it is true, and other models where it is false. Remember that if $\sf ZFC$ have a model, then it has many different models, and not all have the same true statements.

What I mean to say is that if we assume that $\sf ZFC$ is consistent, then it is consistent that $\sf ZFC+CH$ holds, and it is consistent that $\sf ZFC+\lnot CH$ holds. It shows that $\sf ZFC$ is too weak to prove $\sf CH$. It can certainly express it.

For example, one can express it by stating what is a function, and what is an injective function. We can certainly express $\omega$ which is a countably infinite set. Then we can say "Every set $A$ which is a subset of the power set of $\omega$ either has an injection into $\omega$, or there is an injection from the power set of $\omega$ into $A$."

The point is, and I can't stress this enough, that "true" or "false" is a semantic property of a sentence. It depends on the interpretation, the model. But provability is a syntactical property which says that we can write a proof from some theory to some sentence. Syntax and semantics are related, but they are not the same.

There are theories which can prove anything which is true in a particular model, but those theories do not include $\sf ZFC$.


Let's try another example. Let's assume we have a language with $0$ and $+$, then we can write that $+$ is associative and commutative and that $0$ is the identity element.

Now consider the sentence, $\forall x\exists y(x+y=0)$. This sentence says that every element has an additive inverse. Is this sentence true in $\Bbb N$? Is it true in $\Bbb Z$? Being true is a relationship between a sentence and a model, not a sentence and a theory.

And so, if we are given a particular model of $\sf ZFC$ it has meaning to ask "Is $\sf CH$ true or false in this model?", but there is no meaning to the question "Is $\sf CH$ true in $\sf ZFC$?".

Also relevant:

  1. Why is the Continuum Hypothesis (not) true?
  2. Impossible to prove vs neither true nor false
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Well, CH just might be provable, if ZF is inconsistent. –  tomasz May 20 '13 at 0:31
Well, pardon me for assuming that my foundational theory is consistent! ;-) –  Asaf Karagila May 20 '13 at 0:32
@user18921: It is pedagogically wrong to start a course in set theory and say "Okay, $\sf ZFC$ might be inconsistent.", much like it is pedagogically wrong to tell people on their first day in a math degree "There's a reasonable chance that everything we teach you is wrong, and you have wasted your time.". No, first you learn to understand the theory, its intricacies and its beauty, then you learn about the deeper points regarding consistency. I do agree that some remark is mandatory, but it should be left for mid-semester if anything. –  Asaf Karagila May 20 '13 at 0:54
@AsafKaragila: Thanks for your answer, Affer considered your informative post, I think I need form a new question to show you the formal theory of ZFC as a foundation of math is different than that of usual system such as a language with 0 and + which you mentioned. –  logician May 20 '13 at 2:12
@logician: Also note that the choice of foundation is a philosophical issue, rather than purely mathematical. Set theory has advantages, which is why it is considered a good foundational theory. Of course you argue that its disadvantages are far greater, and it should not be a foundational theory, but this is not a mathematical argument per se. And you certainly can't exclaim "Oh, $\sf CH$ is not provable, so the whole theory is inconsistent!" and decide this is a good reason for abandoning $\sf ZFC$. It's a wrong reason. –  Asaf Karagila May 20 '13 at 2:19
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Consider a formal theory $T$ whose language consists of logical operators and symbols $\langle=,0,1\rangle$, and having no axioms except tautologies. Let us assign to symbols $\langle=,0,1\rangle$ their usual meaning in arithmetic.

The sentence $0=1$ can be expressed in the language of $T$, and it is semantically false, but it neither can be proved nor disproved in $T$. But the sentence $(1=0)\lor\neg(1=0)$ is a tautology and so can be proved.

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$0=1$ is not semantically false if you don't introduce an axiom that $0\neq 1$ or something to that effect (or a model in which you could check semantic truth). –  tomasz May 20 '13 at 1:20
@tomasz Does 1 equal to 0? –  Vladimir Reshetnikov May 20 '13 at 1:41
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ZFC can easily express the sentences, "CH is true," and "CH is false." Furthermore, ZFC is able to prove that in every model of ZFC, either CH is true or CH is false (as you observe, this is a consequence of the law of the excluded middle, which is valid in classical logic). In fact, this can be proved in much weaker systems. For instance, Peano Arithmetic is able to prove that in every model of ZFC, either CH is true or CH is false.

However, it's possible that ZFC may have no models at all! Or it may have many models. If it has many models, then some satisfy CH and some don't. Thus we get at the heart of why ZFC cannot decide CH. ZFC can prove a sentence iff all its models satisfy that sentence. Thus, if it has many models, then it cannot prove "CH is true" because it has models in which "CH is false," and it cannot prove "CH is false" because it has models in which "CH is true."

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@user18921: It's possible that someone knows that ZFC is inconsistent, but won't tell for some reason. Like NSA knows that P=NP, but won't tell. ;) –  tomasz May 20 '13 at 1:18
@tomasz: Shelah knows. He knows of a decreasing sequence of ordinals... –  Asaf Karagila May 20 '13 at 1:23
@user18921: who is NSA? –  logician May 20 '13 at 1:40
These last few comments have turned this into a comedy of errors. –  Andres Caicedo May 20 '13 at 2:21
@logician: I think that one has to learn about logic and set theory, and try to look at the mathematical world through that lens. If one also takes into account all the problems that non-rigorous mathematics had, one finds logic and the firm arguments based on set theory to be quite a comfort. Perhaps not $\sf ZFC$, but nonetheless, when you think about it, $\sf ZFC$ does have a reasonable grasp on what properties sets should have. It's very easy to say that something is wrong, until you try and replace it and realize that it wasn't all bad to begin with, and probably quite reasonable. –  Asaf Karagila May 20 '13 at 4:45
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