1
2.2. EBRT, IBRT in A,fA.
This section is intended to be a particularly gentle
introduction to BRT classification theory. It is wholly
subsumed by section 2.3.
Recall the five main BRT settings introduced at the
beginning of this Chapter: (SD,INF), (ELG,INF), (MF,INF),
(ELG
∩
SD,INF), (EVSD,INF).
We begin with the BRT fragments
α
=
EBRT in A,fA on these five BRT settings.
As discussed in sections 1.1 and 2.1, classification of
these BRT fragments amounts to a determination of the true
α
assertions, which take the form
1) (
∀
f
∈
V)(
∃
A
∈
K)(
ϕ
)
where
ϕ
is an
α
equation (since we are in the environment
EBRT).
As discussed in sections 1.1 and 2.1, we work,
equivalently, with the
α
statements, which take the form
1’) (
∀
f
∈
V)(
∃
A
∈
K)(S)
where S is an
α
format, interpreted conjunctively.
Recall that in EBRT, S is correct if and only if 1') holds.
S is incorrect if and only if 1') fails.
In this case of EBRT in A,fA, the number of elementary
inclusions is 4, and the number of formats is 16.
Since 16 is so small, we might as well list all of the
formats S. It is most convenient to list the formats S in
increasing order of their cardinality – which is 0–4.
The four A,fA elementary inclusions are as follows. See
Definition 1.1.36. (These do not depend on the BRT
environment or BRT setting).
A
∩
fA =
∅
.
A
∪
fA = U.
A
⊆
fA.
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fA
⊆
A.
According to Definition 1.1.13 of the universal set U in
BRT settings, we see that on our five BRT settings, U is N.
Before beginning this tabular EBRT classification, we
organize the nontrivial mathematical facts that we will
use.
THEOREM 2.2.1. Let f
∈
EVSD and E
⊆
A
⊆
N, where E is
finite, A is infinite, and E
∩
fE =
∅
. Also let D
⊆
N be
infinite. There exists infinite B such that E
⊆
B
⊆
A, B
∩
fB =
∅
, and neither A nor D are subsets of B
∪
fB.
Moreover, this is provable in RCA
0
.
Proof: Let f,E,A,D be as given. Let n
∈
D be such that
n > max(E
∪
fE), and x
≥
n
→
f(x) > x. Let t > n, t
∈
A. We define an infinite strictly increasing sequence n
1
<
n
2
... by induction as follows.
Define n
1
= min{m
∈
A: m > t}. Suppose n
1
< .
.. < n
k
have
been defined, k
≥
1. Define n
k+1
to be the least element of A
that is greater than n
k
and all elements of f(E
∪
{n
1
,...,n
k
}).
Let B = E
∪
{n
1
,n
2
,...}
⊆
A. Clearly B
∩
fB =
∅
. Also n,t
∉
B, and so A,D are not subsets of B
∪
fB. QED
In the applications of Theorem 2.2.1 to the tabular EBRT
classification below, we can ignore E,A,D. We just use that
for all f
∈
EVSD, there exists infinite B
⊆
N such that B
∩
fB =
∅
.
Here is the other fact that we need.
COMPLEMENTATION THEOREM. For all f
∈
SD there exists A
∈
INF such that fA = N\A.
We proved the Complementation Theorem in section 1.3 within
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 Fall '08
 JOSHUA
 Math, Set Theory, UCI race classifications, A,fA, IBRT

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