Platonic Truth and 1st Order Logic - Take 2 As an algebraist, I have some strong intuitions about what it means for an algebraic result to be true.  In particular, my intuition would lead me to believe that if I cannot construct a counter-example to a claim, then the claim must be true.  This is what motivated my previous question, where it became clear that my intuition needed some tweaking.  (In other words, in some contexts it just doesn't hold true!)  The answers given there are excellent, and I recommend people read them before continuing.
So here is a follow-up question to see just how far we can take the intuition.
Let $T_0$ be the theory ${\rm PA}$ over a countable language.  Let $T_1$ be the extended theory obtained by adding as new axioms all $\Pi_1^0$ statement which are independent of ${\rm PA}$.  This new theory is consistent and sound, assuming that the standard model of the natural numbers exists.
I would guess that this new theory $T_1$ is already not effectively computable.  At any rate, suppose now that we let $T_2$ be the extended theory of $T_1$ obtained by adding as new axioms all $\Pi_2^0$ statement which are independent of $T_1$.
Is $T_2$ consistent?  If so, is the theory $T_n$ (defined in the obvious recursive way) consistent?  If so, is $\bigcup_{n\in \mathbb{N}}T_n$ the true theory of the standard model?
If $T_2$ is not consistent, is there some natural way to fix the problem?
Finally, what happens if we repeat these ideas for ${\rm ZFC}$ instead?
 A: Yes, your theory is the same as the true arithmetic. In
particular, yes, it is consistent.
I claim that at stage $n$, your theory $T_n$ consists of PA plus
the collection of true $\Pi^0_n$ sentences (that is, true in the
standard model). This starts out true with $T_0$. If $T_n$ is like
that, then consider $T_{n+1}$. If $\psi$ is a true $\Pi^0_{n+1}$
statement, then in particular, it is consistent with $T_n$, since
both are true in the standard model. So either it is provable from
$T_n$, in which case it is already there, or else it is
independent, in which case you add it at stage $n+1$. Conversely,
if $\psi$ is a $\Pi^0_{n+1}$ statement that is independent of
$T_n$, then in particular, it asserts $\forall x\ \phi(x)$, where
$\phi(x)$ is $\Sigma_n$. Since $\psi$ is true in a model of $T_n$,
it follows that $\phi(m)$ must hold of every standard $m$, since if $\phi(m)$ failed then $\neg\phi(m)$ would be a true $\Pi^0_n$ sentence and hence part of $T_n$, preventing $\psi$ from being consistent with $T_n$. So
$\psi$ is true.
Thus, by induction, you are adding all and only the true $\Pi^0_n$
statements at each stage, and you've got the theory of true
arithmetic.
A: No. It works with T0 and Π01 sentences, as PA refutes any false Π01 sentence. But not so with T1 and Π02 sentences: there are false Π02 sentences which are independent of T1. 
