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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,.
..,n1 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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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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 Fall '08
 JOSHUA
 Math

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