(c)
Take A={1−n−1:n∈N}∪{2+n−1:n∈N}. Show that 1 and 2 realize the same type over A, but there is no automorphism of Q fixing A and sending 1 to 2.
Answer revised since previous submission (both parts)
(pt1) Take
ϕ∈tp(1/A). Note that
L has no constant or function symbols, so
ϕ has as terms only variables and constant symols from
A. Let
a be those constant symbols and
ϕ′ be
ϕ with
a replaced by free variables, so that
QA⊨ϕ(x)⟺Q⊨ϕ′(x,a)
Then
Q⊨ϕ′(1,a). But DLO admits quantifier elimination, so WLOG
ϕ′ is quantifier-free and thus a boolean combination over atomic sentences of the form
□<□ or
□=□ where each
□ is populated by either the free variable
x or a free variable
a∈a.
Any such atomic sentence produces the same result when evaluated on
x=1 versus
x=2. Inequalities
x<ai are true exactly when
ai>2, and inequalities
ai<x exactly when
ai<1; equalities
x=ai and
ai=x are always false; inequalities
ai<aj and equalities
ai=aj and
x=x likewise do not depend on the choice of
x.
As such,
Q observes the same evaluation for
ϕ′(1,a) and
ϕ′(2,a). Hence by construction of
ϕ, uh,
QA observes the same evaluation for
ϕ(1) and
ϕ(2). Thus the types in question are equal.
(pt2) Let
σ be an automorphism of
Q sending 1 to 2. Let
1<x<2. Then
σ(1)<σ(x)<σ(2), so
2<σ(x) by replacement. Let
2<∈Aa<σ(2); such an
a must exist because elements of
A get arbitrarily close to 2. Then
σ−1(2)<σ−1(a)<σ−1(σ(2)), so by replacement
1<σ−1(a)<2 so
σ−1(a)∈(1,2). But
A∩(1,2)=∅, so
σ−1(a)∈/A, so
σ does not fix
A. Contradiction!