# Truth and undecidability

I believe this is more of a philosophical question.

Given a consistent theory T and a statement S independent of T. Can S be true or false in T? (I don't see any contradiction with that)

I read that Godel thinks the Continuum Hypothesis is false (in ZFC ?) even though independent of ZFC. (so a statement could be undecidable and false according to him)

But at the same time his first theorem shows that if statement S is independent of T then T+S AND T+(not S) are both consistent.

If S was true or false in T then we would have T+S OR T+(not S) consistent. (but you couldn't prove the consistency of T+S or T+(not S) otherwise S would be decidable in T)

So to conclude from the above statement I get: if S is true or false in the consistent theory T and independent of T then the consistency of T+S is undecidable (in T) ?

So my questions are, Do my above statements make sense ? Can their exist undecidable statements that are true ? Is it possible to build one ? is this question undecidable ? if yes in what axiomatic system ? or is it just philosophy ? ... is decidability always decidable (in T) ?

Note: Each time I wrote undecidable I tried to write the theory i'm working with, but I am not sure I am working in the right one.

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What does it mean for a statement $S$ to be true in a theory $T$? Does it mean that $S$ is a logical consequence of $T$, i.e. $S$ is satisfied by every model of $T$? Gödel's completeness theorem (en.wikipedia.org/wiki/G%C3%B6del%27s_completeness_theorem) says that a statement is provable from a theory iff it is a logical consequence of that theory. –  LostInMath Oct 10 '11 at 10:10
@LostInMath I'm saying that S can be true but not provable, so I am not talking aboout truth in term of logical consequence of T. For example when I say that all even number > 2 can be written as the sum of 2 prime numbers, It might be true (if I test every single even number) but It might be not provable (If there is no proof and the only way is to test every odd number and testing an infinit number of values isn't a proof). So how would you express this kind of truth ? –  Ricky Bobby Oct 10 '11 at 11:54
I think that truth in your particular example means truth with respect to the model (or structure) of natural numbers. Levon Haykazyan clarifies the matter nicely in the second paragraph of his answer. –  LostInMath Oct 10 '11 at 12:51
@LostInMath yes,thx, I answered your comment before looking at Levon's answer. but Levon's answer perfectly answered my question –  Ricky Bobby Oct 10 '11 at 13:22
Ricky, what Goedel meant is that formally the statement is undecidable in ZFC, but as a Platonist he believed that there exists an ideal universe of mathematics. In that Platonist universe, or so he believed, the continuum hypothesis is false. –  Asaf Karagila Oct 28 '11 at 12:01

Given a theory $T$ and sentence $\phi$, the truth of $\phi$ in $T$ is not well defined (at least there is no universally accepted definition). However, given a structure ${\cal M}$, one can define the truth of $\phi$ in ${\cal M}$. Usually when discussing some theory, one has an intended structure in his/her mind. In this case one may casually call sentences true or false (every sentence is true or false with respect to a given structure). For example, for $PA$ (Peano Arithmetic) the intended structure is the set of natural numbers (with appropriately defined operations). In this example, Godel's sentence is undecidable in $PA$, but true nonetheless.
As you have correctly pointed out, $\phi$ is independent of $T$ if and only if $T + \phi$ and $T + \lnot \phi$ are consistent. A sufficiently strong theory can prove it himself. However, if $T$ is consistent, then for every $\phi$ either $T + \phi$ or $T + \lnot \phi$ is consistent. But the consistency of $T + \phi$ or $T + \lnot \phi$ implies the consistency of $T$. So a sufficiently strong theory $T$ cannot prove the consistency of $T + \phi$ for any $\phi$.