Math 6a  Solution Set 7
Problem # 1
Liu 5.15
Let
T
1
and
T
2
be two spanning trees of a connected graph
G.
Let
a
be an edge that is in
T
1
but not
T
2
.
Prove that there is an edge
b
in
T
2
but not
T
1
such that both (
T
1
 {
a
}
)
∪ {
b
}
and (
T
2
 {
a
}
)
∪ {
b
}
are spanning trees of
G.
Solution.
Note that if the edge
a
=
{
x,y
}
,
since spanning trees are minimal connected graphs, it
follows that the removal of
a
from
T
1
splits the vertices into two subsets
A
and
B
(without
loss of generality,
x
∈
A
and
y
∈
B
) such that
A
and
B
are connected, and
a
is the only
edge between
A
and
B.
Also, note that
T
2
being a spanning tree and
a
not in
T
2
implies
that if we take
T
2
∪{
a
}
we obtain a graph with a circuit. Clearly that circuit contains
a
and
thus it contains the two points
x
and
y.
So, since it contains the edge
a
it must contain at
least one more edge with one endpoint in
A
and one in
B,
since if it does not, without loss
of generality all of its other edges have both endpoints in
A
which is a contradiction since
there must be another edge with endpoint
y
which is in
B.
Thus, let
b
=
{
x
0
,y
0
}
be an edge
in
T
2
that is in that circuit and has endpoint
x
0
∈
A
and the other endpoint
y
0
∈
B.
Note
that
b
cannot be in
T
1
,
since if it were, we would have a circuit in
T
1
,
simply by taking the
path from
x
to
x
0
in
A,
the edge
b
from
x
0
to
y
0
,
then the path
y
0
to
y
in
B,
followed by the
edge
a
from
y
back to
x.
Note that (
T
1
{
a
}
)
∪{
b
}
and (
T
2
{
b
}
)
∪{
a
}
both have
v

1

1+1 =
v

1 edges (where
v
is the number of vertices in
G
), so if we show both are connected, it will follow that both
are spanning trees of
G
(since if not, we could remove edges from one to get a spanning tree
with less than
v

1 edges, a contradiction). But note that in
T
1
, A
and
B
are connected,
so in (
T
1
 {
a
}
)
∪ {
b
}
they are both connected, and there exists a path from any
z
1
in
A
to
x
0
to
y
0
(by
b
) to any
z
2
in
B.
Thus (
T
1
 {
a
}
)
∪ {
b
}
is connected, and is a spanning tree of
G.
Also, since
T
2
∪ {
a
}
contains a circuit containing the edge
b,
the removal of
b
from that
circuit will not disconnect it and thus (
T
2
{
b
}
)
∪{
a
}
is connected and thus also a spanning