# statement that is consistent in ZFC but the negation of it can be both consistent and inconsistent in ZFC or vice versa

Is there any known case where ZFC system is known to be consistent with a statement, but is also consistent with the negation of the statement? Or vice versa.

Also, when we say ZFC is consistent with something, are we saying that one particular model of ZFC allows something as being consistent?

Edit: Phrasing the question this way, I see I was somehow mixed up. Answers to two questions are obvious... For some checks, can anyone answer the second one? - though I think the answer is obviously.. (as some other models can behave differently)

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It seems that you may have misunderstood the concept of consistency. A statement cannot be both consistent and inconsistent; also, a statement is usually not said to be consistent in ZFC, but with ZFC. You may be thinking of the kind of theorem that asserts that an axiom system is consistent both with a statement and with its negation; but that's different from what you're saying. The problem might be due to the grammatical error in the first sentence in the use of "allow"; perhaps if you fix the grammar the sentence might become clearer. – joriki Sep 25 '12 at 9:32
I modified the question. – Alphalama Sep 25 '12 at 10:00

Let's say you have a first-order theory $T$ and a sentence $\phi$. If $T$ is both consistent with $\phi$ and consistent with $\neg \phi$ then $\phi$ is independent of $T$, which means that if $T$ has a model, then it has models in which $\phi$ is satisfied and models in which $\neg \phi$ is satisfied.

Here we're considering $T=\text{ZFC}$. Let our $\phi$ be $\text{CH}$, which is the assertion that $2^{\aleph_0}=\aleph_1$ (this is known as the Continuum Hypothesis). Then there are methods which can be used to prove that

$$\text{ZFC is consistent}\ \Leftrightarrow\ \text{ZFC+CH is consistent}\ \Leftrightarrow\ \text{ZFC+(}\neg\text{CH) is consistent}$$

That is, if there is a model of ZFC, then there are models in which the continuum hypothesis holds and models in which its negation holds. (And, in fact, if there is a model in which CH holds then there is a model in which the negation of CH holds.)

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By the way, by '$T$ is consistent with $\phi$' what I mean is $T \cup \{ \phi \} \nvdash \bot$, i.e. $T \cup \{ \phi \}$ is consistent. And 'consistent' $\Leftrightarrow$ 'has a model' by completeness and soundness. I kind of glossed over this in my answer. – Clive Newstead Sep 25 '12 at 13:08

For the first $\varphi$ can be many different sentences which are independent of ZFC, namely if ZFC is consistent then ZFC+$\varphi$ and ZFC+$\lnot\varphi$ are both consistent as well.

Note that all these results are of relative consistency, namely we have to assume that ZFC is consistent to begin with. We add this assumption because ZFC cannot prove its own consistency. Commonly we do our mathematics in a universe of sets which satisfies ZFC to begin with, and therefore when we talk about consistency results we need to assume that ZFC is consistent. If our ambient theory is a stronger one (e.g. ZFC+large cardinals) then the consistency of ZFC is provable, and then the implications are simply true.

One example is the assertion "Every set is constructible" (in the Godel sense) also known as the axiom $V=L$. If ZFC is consistent then ZFC+$V=L$ is consistent and Cohen's work show that if ZFC+$V=L$ is consistent then ZFC+$V\neq L$ is consistent as well. Therefore $V=L$ is independent of ZFC.

For the second question this is an application of the completeness theorem. If $T$ is consistent then it has a model (and vice versa), so assuming ZFC is consistent to begin with, to show that $\varphi$ is independent we take a model of ZFC (guaranteed to exist by our assumption) and we construct two models, one which satisfies $\varphi$ and which satisfies its negation. Now since we have a model of these theories they are both consistent.

Using the example above, Godel showed that if we have a model of ZF then we can create (within the model, actually!) a model of ZFC+$V=L$. There are fine points which I will not go into here, about internal and external definability. Let us just summarize by saying that if ZF is consistent then ZFC+$V=L$ is consistent as well, simply by the fact we exhibited a model.

Cohen, two decades later invented the method of forcing and showed that we can add new sets which are not constructible, and his method began with a model of ZFC+$V=L$ and added sets which contradicted $V=L$, so the new model is a model of ZFC+$V\neq L$. Therefore this theory is consistent as well (assuming ZFC is consistent to begin with of course).

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I'm no logician, but I think the continuum hypothesis and its negation are both consistent with ZFC. – Stefan Smith Sep 27 '12 at 0:45
@bogus: Yes, and? – Asaf Karagila Sep 27 '12 at 5:54
I thought the OP might like a concrete example. – Stefan Smith Sep 28 '12 at 21:08
@bogus: I gave one. The assertion "$V=L$" is also independent of ZFC. – Asaf Karagila Sep 28 '12 at 21:11
@bogus Although the statement of $V=L$ is more complex, the proof of its independence is simpler. A proof of Con(V=L) is properly contained in Goedel's proof of Con(CH), and a proof of V$\ne$L is properly contained in Cohen's proof of Con($\neg$CH). – Trevor Wilson Oct 4 '12 at 20:53