2
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