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