models constructed from constants
The definition of a structure and of the satisfaction relation is
nice, but it raises the following question : how do we get models in
the first place? The most basic construction for models of
first-order theory is the construction that uses constants. Throughout
this entry, L is a fixed first-order language.
Let C be a set of constant symbols of L, and T be a theory in
L. Then we say C is a set of witnesses for T if and only
if for every formula φ with at most one free variable x,
we have T⊢∃x(φ)⇒φ(c) for some c∈C.
Lemma.
Let T is any consistent
set of sentences of L, and C is a set of new symbols
such that |C|=|L|. Let L′=L∪C. Then there is a consistent
set T′⊆L′ extending T and which has C as set of
witnesses.
Lemma.
If T is a consistent theory in L, and C is a set of witnesses
for T in L, then T has a model whose elements are the constants
in C.
Proof:
Let Σ be the signature for L. If T is a consistent set of
sentences of L, then there is a maximal consistent T′⊇T.
Note that T′ and T have the same sets of witnesses. As every
model of T′ is also a model of T, we may assume T is maximal
consistent.
We let the universe of 𝔐 be the set of equivalence classes
C/∼, where a∼b if and only if “a=b”∈T. As T is
maximal consistent, this is an equivalence relation. We interpret the
non-logical symbols as follows :
-
1.
[a]=𝔐[b] if and only if a∼b;
-
2.
Constant symbols are interpreted in the obvious way, i.e. if c∈Σ is a constant symbol, then c𝔐=[c];
-
3.
If R∈Σ is an n-ary relation symbol, then ([a1],…,[an])∈R𝔐 if and only if R(a1,…,an)∈T;
-
4.
If F∈Σ is an n-any function symbol, then F𝔐([a0],…,[an])=[b] if and only if “F(a1,…,an)=b”∈T.
From the fact that T is maximal consistent, and ∼ is an
equivalence relation, we get that the operations are well-defined (it
is not so simple, i’ll write it out later).
The proof that 𝔐⊧ is a straightforward induction
on the
complexity of the formulas of .
Corollary. (The extended completeness theorem) A set of formulas of is consistent if and only if it has a model (regardless of whether or not has witnesses for ).
Proof: First add a set of new constants to , and expand to in such a way that is a set of witnesses for . Then expand to a maximal consistent set . This set has a model consisting of the constants in , and is also a model of .
Corollary. (Compactness theorem) A set of sentences of has a model if and only if every finite subset of has a model.
Proof: Replace “has a model” by “is consistent”, and apply the syntactic compactness theorem.
Corollary. (Gödel’s completeness theorem) Let be a consistent set of formulas of . Then A sentence is a theorem of if and only if it is true in every model of .
Proof: If is not a theorem of , then is consistent with , so has a model , in which cannot be true.
Corollary. (Downward Löwenheim-Skolem theorem) If has a model, then it has a model of power at most .
Proof: If has a model, then it is consistent. The model constructed from constants has power at most (because we must add at most many new constants).
Most of the treatment found in this entry can be read in more details
in Chang and Keisler’s book Model Theory.
Title | models constructed from constants |
---|---|
Canonical name | ModelsConstructedFromConstants |
Date of creation | 2013-03-22 12:44:42 |
Last modified on | 2013-03-22 12:44:42 |
Owner | ratboy (4018) |
Last modified by | ratboy (4018) |
Numerical id | 17 |
Author | ratboy (4018) |
Entry type | Definition |
Classification | msc 03C07 |
Synonym | completeness theorem |
Synonym | Gödel completeness theorem |
Related topic | UpwardsSkolemLowenheimTheorem |
Defines | set of witnesses |