Math 486 Homework 4 Notes
Paul Shafer
[email protected]
February 26, 2008
Problem I.6.8
We give two proofs that a countable graph is 4colorable if every finite subgraph is 4colorable.
Proof via K¨onig’s lemma:
Let
G
= (
{
a
0
, a
1
, . . .
}
, E
) be a countable graph.
Let
G
n
denote the
subgraph induced by
{
a
0
, . . . , a
n

1
}
. Define a tree
T
⊆
4
<ω
(= finite strings over
{
0
,
1
,
2
,
3
}
) as
follows:
σ
∈
T
⇔ ∀
i, j <

σ

((
a
i
, a
j
)
∈
E
→
σ
(
i
)
6
=
σ
(
j
))
.
That is,
σ
is in
T
if and only if
σ
represents a 4coloring of
G

σ

. To prove that
T
is a tree, we need
to show that if
σ
∈
T
and
τ
⊆
σ
then
τ
∈
T
. This is clear because if
∀
i, j <

σ

((
a
i
, a
j
)
∈
E
→
σ
(
i
)
6
=
σ
(
j
)) holds for
σ
then it holds for any substring of
σ
.
T
is finitely branching because it a
subtree of 4
<ω
which is finitely branching. We prove that
T
is infinite by showing that for each
n
there is a
σ
in
T
of length
n
.
G
n
is a finite subgraph of
G
and so is 4colorable by assumption.
Let
c
:
{
a
0
, . . . , a
n

1
} → {
0
,
1
,
2
,
3
}
be a 4coloring of
G
n
.
Let
σ
be the corresponding string.
That is, define
σ
of length
n
by
σ
(
i
) =
c
(
a
i
) for
i
= 0
, . . . , n

1. Then
σ
∈
T
.
T
is an infinite,
finitely branching tree and so has an infinite path
P
=
{
σ
n

n
∈
ω
}
, where

σ
n

=
n
. Define the
coloring
c
:
{
a
0
, a
1
, . . .
} → {
0
,
1
,
2
,
3
}
by
c
(
a
i
) =
σ
n
(
i
) for any (or all)
n > i
, and note that
c
is
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 DORAIS
 Math, Logic

Click to edit the document details