2
Then, I claim that

A

=

A
0

.
To prove this, note that it is suﬃcient to show that for
any
q,q
0
∈
A
with
q
6
=
q
0
we have
f
(
q
)
6
=
f
(
q
0
)
.
For if this is true, we have a bijective
correspondence between the elements of
A
and
A
0
(simply correspond
q
with
f
(
q
)). So
assume it is not true, ie that we have
q,q
0
∈
A
with
q
6
=
q
0
but
f
(
q
) =
f
(
q
0
)
.
Then, if
q
= (
a,b
) and
q
0
= (
a
0
,b
0
)
,
as
f
(
q
) =
a
and
f
(
q
0
) =
a
0
we have
a
=
a
0
.
Since
q
6
=
q
0
we must
have
b
6
=
b
0
.
But as
b,b
0
∈
L
and
L
is a chain,
b,b
0
must be comparable. So, without loss
of generality,
b < b
0
(relabel
q,q
0
if necessary). But then, we have (
a,b
)
<
(
a,b
0
) and thus
q < q
0
,
contradicting the fact that
A
is an antichain.
Thus,

A

=

A
0

.
So now I claim that
A
0
does not contain a chain of length exceeding 2
.
So again, assume this is false. Then we have
a
1
,a
2
,a
3
∈
A
0
with
a
1
< a
2
< a
3
(since if we
have any chain of length greater than 2 it contains a subchain of length 3). Then, as each
a
i
∈
A
0
we must have
q
1
,q
2
,q
3
∈
A
with
f
(
q
i
) =
a
i
,
for each
i
∈ {
1
,
2
,
3
}
.
So deﬁne
b
i
by
q
i
= (
a
i
,b
i
)
.
Note that
A
is an antichain, so the three elements (
a
1
,b
1
)
,
(
a
2
,b
2