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This question is about an answer in question

Are there any proofs the undecidability of the halting problem that does not depend on self-referencing or diagonalization ?

Bjørn Kjos-Hanssen answer states that:

You can first show that the halting problem (the set $0'$) can be used to compute a set $G\subseteq\mathbb N$ that is 1-generic meaning that in a sense each $\Sigma^0_1$ fact about $G$ is decided by a finite prefix of $G$. Then it is easy to prove that such a set $G$ cannot be computable (i.e., decidable).

I have trouble understanding this sentence as I am new to forcing. For example, what does "finite prefix" mean here? Can someone explain this sentence? (I know what $\Sigma^0_1$ means.)

Can someone explain the general idea behind forcing? I am also looking for easy to read introductions articles and books about forcing.

edit: So, how can we prove that each $\Sigma_1^0$ fact about $G$ being decided by a finite prefix leads to the fact that $G$ cannot be computable?

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  • $\begingroup$ I edited the question to make it more interesting. I hope that I kept the original intention. Feel free to roll back or edit. $\endgroup$ – Kaveh Mar 2 '13 at 8:02
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A finite prefix of $G$ is $G \cap \{0,1,\ldots n\}$ for some $n$. It is easier if you look at the characteristic function of $G$ which can we viewed as an infinite word in $2^\omega$. Then a finite prefix is a finite initial part of $G$.

For forcing, there are many resources, depends on what you want to learn. A point to start is the Wikipedia pages and the references listed there: Forcing (set theory), Forcing (recursion theory). Timothy Chow's article A Beginner's Guide to Forcing is a good introduction.

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  • $\begingroup$ While the links are relevant, they certainly do not give enough information to complete the proof outlined by Bjorn. In particular the page on Forcing and recursion theory is atrocious. $\endgroup$ – cody Apr 3 '13 at 14:13
  • $\begingroup$ @cody, that question was added after my answer. I answered what was originally asked. And I am not sure it is a good idea to give an introduction to forcing here when Tim has already written a very good and readable introduction to it. As I see it the OP in place of reading the easy to read reference he asked for wants us to explain it even simpler. Feel free to do so, but I am not sure I will. $\endgroup$ – Kaveh Apr 3 '13 at 14:45
  • $\begingroup$ I really like Tim's introduction! However he makes no no mention of 1-genericity or give any mention to computability, which is what I meant by "not enough information to complete the proof outline". $\endgroup$ – cody Apr 3 '13 at 15:24
  • $\begingroup$ @cody, OK, I will try to address that (I don't have much free time right now, and I am hoping that Tim himself or someone else (Josh, Andrej, Emil, ...) will answer it before I find time :). $\endgroup$ – Kaveh Apr 3 '13 at 16:36
  • $\begingroup$ Thanks! I'll admit that I am interested in the answer to this question as well. $\endgroup$ – cody Apr 3 '13 at 18:49

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