HW 12, due Monday, May 1: problems 10 and 18 on pp. 110, 111 of the book.
Let me add some clarifications to these problems, and then indicate the solutions.
Problem 10. Here L = {f } with f a unary function symbol. A permutation of
a set A is by definition a bijection b : A → A. In (i) you are asked to produce an
L-sentence σsur such that for all L-structures A = hA; f A i,
A |= σsur ⇐⇒ f A : A → A is surjective.
In (iv) you need to find an L-sentence σ such that (a) and (b) below hold:
(a) for all L-structures A = hA; . . .i, if A |= σ, then A is infinite;
(b) for every infinite set A there is a model A = hA; . . .i of σ.
Solution. Let x, y be distinct variables. In (i), let σsur be ∀y∃x(f (x) = y). In (ii),
let σin be ∀x, y(f (x) = f (y) → x = y). In (iii) we can take σbi to be σsur ∧ σin . As
to (iv), one can take σ to be σin ∧¬σsur . For (v), {σbi , ∀x(f (x) 6= x)} has an infinite
model, namely hZ; f i where f (k) = k + 1 for all k ∈ Z. Thus by Skolem-Löwenheim
it has for each infinite cardinal κ a model hA; f i with |A| = κ. Given any infinite
set B, let κ := |B|, take hA; f i as above, and take a bijection b : A → B. Then
b ◦ f ◦ b−1 : B → B is a permutation of B without fixed points.
Problem 18. Here k is a fixed (but arbitrary) natural number ≥ 1. Let the
language L have just a binary relation symbol R. Then a graph is an L-structure
G = hG; Ri where R is a symmetric irreflexive binary relation on the (nonempty) set
G whose elements are thought of as the vertices of the graph. A subgraph of such G
is a graph G0 = hG0 ; R∩G20 } with G0 ⊆ G. A k-coloring of such a graph G = hG; Ri
is by definition a function c : G → {1, . . . , k} such that c(g) 6= c(h) for all pairs
(g, h) ∈ R. In the statement of the problem the language L is augmented by k extra
unary predicate symbols C1 , . . . , Ck , to give the language Lk = L∪{C1 , . . . , Ck } and
it is observed that G has a k-coloring iff G can be expanded to an Lk -structure that
satisfies the sentences listed there involving the new symbols C1 , . . . , Ck . Please
check for yourself that this is a correct observation, and then solve the problem.
Hint: given G, you might consider extending Lk further by names for the elements
of G.
Solution. Let G = hG; Ri be a graph. Extend the language Lk to Lk (G) by adding
for each g ∈ G a name g. Let the set Σ(G) consist of the following Lk (G)-sentences:
(1) for all elements g, h ∈ G, the sentence g 6= h whenever g 6= h, the sentence
R(g, h) whenever (g, h) ∈ R, and the sentence ¬R(g, h) whenever
(g, h) ∈
/ R.
V
(2) for each g ∈ G the sentences C1 (g)∨· · ·∨Ck (g) and 1≤i<j≤k ¬(Ci (g)∧Cj (g))
(expressing that g gets painted with exactly one of the k colors);
Vk
(3) for all (g, h) ∈ R the sentence i=1 ¬ Ci (g) ∧ Ci (h) (expressing that the
adjacent vertices g and h do not get painted the same color).
Observe that G is k-colorable if and only if Σ(G) has a model.
Now assume that every finite subgraph G0 = hG0 ; · · ·i of G is k-colorable. Then
for every finite nonempty subset G0 of G the set Σ(G0 ) has a model. Since every
finite subset of Σ(G) is contained in Σ(G0 ) for some finite nonempty G0 ⊆ G, it
follows that every finite subset of Σ(G) has a model. Hence by compactness, Σ(G)
has a model. Thus by the observation above, G is k-colorable.
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