Computability and Logic

304 NONSTANDARD MODELS
Note that a’s block is infinite in both directions if a is nonstandard, and is ordered
like the integers (negative, zero, and positive).
Suppose that a is LESS THAN b and that a and b are in different blocks. Then since
a † is LESS THAN or equal to b, and a and a † are in the same block, a † is LESS THAN b.
Similarly, a is LESS THAN b ‡ . It follows that if there is even one member of a block A
that is LESS THAN some member of a block B, then every member of A is LESS THAN
every member of B. If this is the case, we’ll say that block A is LESS THAN block B.
A block is nonstandard if and only if it contains some nonstandard number. The
standard block is the LEAST block.
There is no LEAST nonstandard block, however. For suppose that b is a nonstandard
NUMBER. Then there is an a LESS THAN b such that either a ⊕ a = b or a ⊕ a ⊕ I = b.
[Why? Because for any (natural) number b greater than zero there is an a less than b
such that either a + a = b or a + a + 1 = b.] Let’s suppose a ⊕ a = b. (The other
case is similar.) If a is standard, so is a ⊕ a.Soa is nonstandard. And a is not in the
same block as b: for if a ⊕ c = b for some standard c, then a ⊕ c = a ⊕ a, whence
c = a, contradicting the fact that a is nonstandard. (The laws of addition that hold in
N hold in M.) So a’s block is LESS THAN b’s block. Similarly, there is no GREATEST
block.
Finally, if one block A is LESS THAN another block C, then there is a third block
B that A is LESS THAN, and that is LESS THAN C. For suppose a is in A and c is in C,
and a is LESS THAN c. There there is an b such that a is LESS THAN b, b is LESS THAN
c, and either a ⊕ c = b ⊕ b or a ⊕ c ⊕ I = b ⊕ b. (Averages, to within a margin of
error of one-half, always exist in N ; b is the AVERAGE in M of a and c.) Suppose
a ⊕ c = b ⊕ b. (The argument is similar in the other case.) If b is in A, then b = a ⊕ d
for some standard d, and so a ⊕ c = a ⊕ d ⊕ a ⊕ d, and so c = a ⊕ d ⊕ d (laws of
addition), from which it follows, as d ⊕ d is standard, that c is in A.Sob is not in A,
and, similarly not in C either. We may thus take as the desired B the block of b.
To sum up: the elements of the domain of any nonstandard model M of arithmetic
are going to be linearly ordered by LESS THAN. This ordering will have an initial
segment that is isomorphic to the usual ordering of natural numbers, followed by a
sequence of blocks, each of which is isomorphic to the usual ordering of the integers
(negative, zero, and positive). There is neither an earliest nor a latest block, and
between any two blocks there lies a third. Thus the ordering of the blocks is what
was called in the problems at the end of Chapter 12 a dense linear ordering without
endpoints, and so, as shown there, it is isomorphic to the usual ordering of the rational
numbers. This analysis gives us the following result.
25.1a Theorem. The order relations on any two enumerable nonstandard models of
arithmetic are isomorphic.
Proof: Let K be the set consisting of all natural numbers together with all pairs
(q, a) where q is a rational number and a and integer. Let < K be the order on K in
which the natural numbers come first, in their usual order, and the pairs afterward,
ordered as follows: (q, a) < K (r, b) if and only if q < r in the usual order on rational
numbers, or (q = r and a < b in the usual order on integers). Then what we
have shown above is that the order relation in any enumerable nonstandard model of