> This identification is precisely what choice sequences are used to rule out.
But you don't need to rule them out -- Brouwer certainly didn't. He just wanted to rule out decidable sequences. It's the undecidability that's important to his result, not arbitrariness, and that's perfectly compatible with automata. He just didn't know about computational undecidability and thought that undecidability requires lawlessness, but it doesn't. In any event, he didn't think randomness violates finitism. That only finite systems can be the source of axioms has been a constant refrain of his. I see that according to Wikipedia's definition of Finitism, he would be a "classical finitist."
> In particular, without the axioms for choice sequences, you cannot derive the strong counterexamples, because you need classically-false axioms to prove classically-false theorems (like the continuity of all real-valued functions).
The continuity of all real-valued functions can be proven with just computable sequences because of undecidability; you just can't do it if computational undecidability hasn't been discovered yet. In fact, it doesn't even become quite "classically false", but can be "embedded" in classical mathematics by restricting the "all": all computable real functions are continuous is a classical theorem. AFAIK, all constructive theorems can be embedded in classical mathematics by restricting them to computable entities. Do you have an example of any result that relies on uncomputable sequences?
> It's the undecidability that's important to his result, not arbitrariness
Undecidability is arbitrariness - to say that, for example, the axiom of choice is independent of ZF is the same thing as saying that there is a arbitrary choice[0] between ZF+C and ZF+!C. Similarly, the sequence[1] G(PA),G(PA+G(PA)),G(PA+G(PA)+G(PA+G(PA))),... requires (at least) one additional arbitrary bit per item.
0: not asserting anything about what makes that choice, though.
1: where G(X) is the Godel sentence, roughly "X cannot cannot prove G(X)"
What I meant was that Brouwer doesn't need lawless sequences; all the results are valid with lawful sequences only. He just didn't know that because he didn't know that lawful sequences can be undecidable. You can call it arbitrary, but it doesn't have to be random, it just needs to be unknowable, which is the case even if it's chosen by a computational rule. It is true that most of the sequences of the kind you presented are not computable, but I don't think any result of Brouwer's is affected.
But you don't need to rule them out -- Brouwer certainly didn't. He just wanted to rule out decidable sequences. It's the undecidability that's important to his result, not arbitrariness, and that's perfectly compatible with automata. He just didn't know about computational undecidability and thought that undecidability requires lawlessness, but it doesn't. In any event, he didn't think randomness violates finitism. That only finite systems can be the source of axioms has been a constant refrain of his. I see that according to Wikipedia's definition of Finitism, he would be a "classical finitist."
> In particular, without the axioms for choice sequences, you cannot derive the strong counterexamples, because you need classically-false axioms to prove classically-false theorems (like the continuity of all real-valued functions).
The continuity of all real-valued functions can be proven with just computable sequences because of undecidability; you just can't do it if computational undecidability hasn't been discovered yet. In fact, it doesn't even become quite "classically false", but can be "embedded" in classical mathematics by restricting the "all": all computable real functions are continuous is a classical theorem. AFAIK, all constructive theorems can be embedded in classical mathematics by restricting them to computable entities. Do you have an example of any result that relies on uncomputable sequences?