Math 5020 Handout 1.2
Proofs of Theorems in Handout 1
Proposition
A
⊆
B
iff

A
 ⊆ 
B

and for every constant symbol
c
∈
L
,
n
ary relation symbol
R
∈
L
and
function symbol
f
∈
L
,
c
B
=
c
A
, R
B
∩ 
A

n
=
R
A
, f
B

A

n
=
f
A
.
Proof
(
⇒
) Suppose
A
⊆
B
.
Then by definition

A
 ⊆ 
B

and the identity map
e
:

A
 → 
B

is an
embedding. By the definition of embedding, we have that
•
if
c
is a constant symbol, then
e
(
c
A
) =
c
B
; but
e
(
c
A
) =
c
A
, and so
c
A
=
c
B
.
•
if
f
∈
L
is an
n
ary function symbol, then
f
B
(
e
(
a
1
)
, . . . , e
(
a
n
)) =
e
(
f
A
(
a
1
, . . . , a
n
) if
a
1
, . . . , a
n
∈ 
A

;
since
e
is the identity map, we have
f
B
(
⃗a
) =
f
A
(
⃗a
), and therefore
f
B

A

n
=
f
A
.
•
if
R
∈
L
is an
n
ary relation symbol, then
R
B
(
⃗a
)
⇐⇒
R
A
(
⃗a
) for
⃗a
∈ 
A

n
; thus
R
B
∩ 
A

n
=
R
A
.
(
⇐
) All the steps in the above proof can be reversed.
Lemma
Suppose
A
0
⊆
A
1
⊆ · · · ⊆
A
n
⊆
. . . . . .
Then there is a unique
L
structure
A
such that

A

=
∪
n

A
n

and
A
n
⊆
A
for all
n
∈
N
.
Proof
We tacitly assume that the substructure relation is transitive (see Homework Problem 1.2). To define
the structure
A
we only need to interpret the symbols in
L
.
Let
c
∈
L
be a constant symbol. Note that
c
A
0
=
c
A
1
=
. . .
and we must define
c
A
=
c
A
1
in order for
A
1
⊆
A
. Since they all agree we do have
c
A
n
=
c
A
for all
n
∈
N
.
Next let
f
∈
L
be a
k
ary function symbol. To define
f
A
we consider an arbitrary
k
tuple
a
1
, . . . , a
k
∈

A

=
∪
n

A
n

. Let
n
0
be the smallest natural number such that
a
1
, . . . , a
k
∈ 
A
n
0

. Then for all
n
≥
n
0
,
a
1
, . . . , a
k
∈ 
A
n

, and
f
A
n
(
a
1
, . . . , a
k
) =
f
A
n
0
(
a
1
, . . . , a
k
)
since
A
n
0
⊆
A
n
. In order for
A
n
0
⊆
A
we must define
f
A
(
a
1
, . . . , a
k
) =
f
A
n
0
(
a
1
, . . . , a
k
)
. We do have that for all
n
∈
N
and
k
tuples
a
1
, . . . , a
k
∈ 
A
n

,
f
A
(
a
1
, . . . , a
k
) =
f
A
n
(
a
1
, . . . , a
k
)
.
Finally let
R
∈
L
be a
k
ary relation symbol. To define
R
A
we consider an arbitrary
k
tuple
a
1
, . . . , a
k
∈

A

=
∪
n

A
n

. Let
n
0
be the smallest natural number such that
a
1
, . . . , a
k
∈ 
A
n
0

. Then for all
n
≥
n
0
,
a
1
, . . . , a
k
∈ 
A
n

, and
R
A
n
(
a
1
, . . . , a
k
)
⇐⇒
R
A
n
0
(
a
1
, . . . , a
k
)
since
A
n
0
⊆
A
n
. In order for
A
n
0
⊆
A
we must define
R
A
(
a
1
, . . . , a
k
)
⇐⇒
R
A
n
0
(
a
1
, . . . , a
k
)
. We do have that for all
n
∈
N
and
k
tuples
a
1
, . . . , a
k
∈ 
A
n

,
R
A
(
a
1
, . . . , a
k
)
⇐⇒
R
A
n
(
a
1
, . . . , a
k
)
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
We have by now defined the structure
A
so that

A

=
∪
n

A
n

and checked along the way that
A
n
⊆
A
for all
n
∈
N
. From our definitions and remarks it is clear that the definition is unique.
Tarski’s Elementary Chain Theorem
Suppose
A
0
≺
A
1
≺ · · · ≺
A
n
≺≺ · · · · · ·
Let
A
=
∪
n
A
n
. Then
A
n
≺
A
for all
n
∈
N
.
Proof
By the preceding lemma the structure
A
is well defined. We show the following statement by induction
on the
L
formula
φ
:
For any
L
formula
φ
(
x
1
, . . . , x
k
), any
n
∈
N
and
k
tuple
a
n
1
, . . . , a
n
k
∈ 
A
n

,
A
n

=
φ
[
a
n
1
, . . . , a
n
k
]
⇐⇒
A

=
φ
[
a
n
1
, . . . , a
n
k
]
.
In the following we omit the superscripts for simplicity, and in fact write
⃗a
whenever the arity is not
important.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 staff
 Math, Set Theory, Model theory, A

Click to edit the document details