Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 144 / 3 / c

Table of contents
- i c
  - Solution i
- ii c
  - Solution ii

i

Solution

By quantifier elimination, a one-type is determined entirely by the position of

x

relative to the natural-number parameters. Assuming

N = {0, 1, 2, \dots}

, the isolated types and isolating formulas are:

$x = n$ , for each $n \in N$ ;
$x < 0$ ;
$n < x < n + 1$ , for each $n \in N$ .

Each formula fixes every comparison of

x

with every natural number, so it determines a complete type. These are precisely the isolated members of the one-types over the natural numbers in the rational order.

Solved by gpt-5.6-sol high.

ii

Solution

There is exactly one non-isolated type:

p_{\infty} (x) = {n < x : n \in N} .

It is consistent by the compactness theorem, since every finite subset is realized by a sufficiently large rational number. It is complete by quantifier elimination, because it decides every comparison with a parameter from

N

.

No formula isolates it. Any formula belongs to

p_{\infty}

only through finitely many natural-number parameters; after quantifier elimination it holds throughout some final ray. It is consequently also satisfied by a sufficiently large natural number, whose equality type differs from

p_{\infty}

. Thus

p_{\infty}

is non-isolated, and the list in part i exhausts all other cuts of

N

.

Solved by gpt-5.6-sol high.

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