2.4.A,B,fA,fBSD062310

2.4.A,B,fA,fBSD062310 - 1 2.4. EBRT in A,B,fA,fB, on...

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1 2.4. EBRT in A,B,fA,fB, on (SD,INF). In this section, we use the tree methodology described in section 2.1 to classify EBRT in A,B,fA,fB, on (SD,INF) and (ELG SD,INF). We handle both BRT settings at once, as they behave the same way for EBRT in A,B,fA,fB, . In particular, we show that they are RCA 0 secure (see Definition 1.1.43). We begin with a list of five Lemmas that we will need for documenting the classification. LEMMA 2.4.1. Let f SD. There exist infinite A B N such that B . fA = N and A = B fB. Proof: By the BRT Fixed Point Theorem, section 1.3, let A be the unique A N such that A = N\fA f(N\fA). Let B = N\fA. Clearly A B and B . fA = N. Also B fB = N\fA f(N\fA) = A. Suppose A is finite. Then N\fA is cofinite and f(N\fA) is infinite. Hence their intersection is infinite, and so A is infinite. So we conclude that A is infinite. QED LEMMA 2.4.2. Let f SD. There exist infinite A B N such that A . fB = N, fA B, and B fB fA. Proof: By the BRT Fixed Point Theorem, section 1.3, let B be the unique B N such that B = N\fB f(N\fB). Let A = N\fB. Then A B, fA B. Now B fB = (N\fB f(N\fB)) fB = f(N\fB) fB fA. Suppose A is finite. Then B = A fA is finite. Hence N\fB = A is infinite, which is a contradiction. Hence A is infinite. Therefore fA,B are infinite. QED The following is a sharpening of the Complementation Theorem. LEMMA 2.4.3. Let f SD and X N. There exists a unique A such that A X A . fA. Proof: We will give a direct argument from scratch. Let f,X be as given. Define membership in A inductively as follows. Suppose membership in A for 0,. ..,n-1 has been defined. Define n A if and only if n X and n fA thus far. The construction is unique. QED
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2 LEMMA 2.4.4. The following is false. For all f ELG SD there exist infinite A B N such that A fB = and fB B. Proof: Let f be given by Lemma 3.2.1, and let A B N, A fB = , and fB B, where A is infinite. Just using fB B, B , we see that fB is cofinite, and hence A is finite. This is the desired contradiction. QED LEMMA 2.4.5. Let f SD. There is no nonempty A N such that A fA. Proof: Let n be the least element of A. Then n fA. QED Note that in the proofs of Lemmas 2.4.1, 2.4.2, 2.4.3, 2.4.5, we never used the fact that f is everywhere defined. Hence these Lemmas hold even for partially defined f. We will use Lemma 2.4.2 for partial f in section 2.5. The 16 A,B,fA,fB pre elementary inclusions are as follows (see Definition 1.1.35). A B fA fB = . A B fA fB = N. A
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2.4.A,B,fA,fBSD062310 - 1 2.4. EBRT in A,B,fA,fB, on...

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