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B
(
W
)=
H
∈W
H
J
→
(
W
)=
{
B
i
For
W ⊆ B,
put
,
i
:
(
j
)
∈ W
for some
j
}
and
J
←
(
W
)=
{
B
i
B
i
)=
B
i
i
:
(
j
)
∈ W
for some
j
}.
Now
(
j
(
1
)
so
.
B
i
B
i
|B
(
W
)
|
=
(
1
)
∪
(
1
)
i
∈
J
→
(
W
)
i
∈
J
←
(
W
)
Note that
J
→
(
W
)
∩
J
←
(
W
)=
0 implies
|W |≤|B|/
2
.
We will divide the analy-
sis into several cases and for each case we will show that
|B
(
W
)
|≥|W |
; the cases
are expressed in terms of
J
→
(
W
)
but the corresponding cases for
J
←
(
W
)
follow
by symmetry and we will often use this fact without explicitly mentioning it in the
arguments below.
Case 1.
Suppose
J
→
(
W
)
contains
i
1
and
i
2
satisfying
|
i
1
−
i
2
| >
1
.
Because
b
≥
a
,
at least one of
A
i
)
∈ B
i
1
(
and
A
i
)
∈ B
i
2
(
for each
i
and
j
,
(
j
1
)
(
j
1
)
holds and at
least one of
A
i
)
∈B
i
1
(
and
A
i
)
∈ B
i
2
(
(
j
1
)
(
j
1
)
holds. Hence we have
B
(
W
)=
Y
→
∪Y
←
=
Y
|B
(
W
)
|
=
|B|≥|W |.
Thus we can assume that
and
J
→
(
W
)
|≤
J
→
(
W
)
|
=
|
2
.
Furthermore, if
|
2
,
then
J
→
(
W
)=
{
|Y
→
|−
(
k
.
k
+
1
}
for some integer
k
so that
B
(
W
)
contains at least
s
−
Y
→
.
This follows because
A
i
A
i
λ
s
+
λ
q
−
q
)
members of
(
j
)
∈B
(
W
)
if
kb
∈
(
j
)
.
|Y
←
|−
(
Y
←
.
Similarly
B
(
W
)
contains at least
s
−
λ
s
)
members of
Case 2.
Suppose
J
→
(
W
)
B
i
(
W
)
⊇ Y
←
and
contains an
i
≤
q
.
We t h e n h ave
holds when
J
→
(
W
)
∩
J
←
(
W
)=
|B
(
W
)
|≥|B|/
2
.
Hence
|B
(
W
)
|≥|W |
/0or
J
←
(
W
)
≤
.
contains a
k
q
Thus, by Case
1
, we only have to consider the case
J
→
(
W
)=
{
J
←
(
W
)
and
J
←
(
W
)
∩{
,
+
},
+
∈
,
,...,
}
=
.
q
q
1
q
1
1
2
q
0
, B
q
+
1
(
Y
→
so
=
)
−
λ
+
λ
−
|B
(
W
)
|≥
If
λ
0
j
contains at least
s
q
members of
s
s
q
J
→
(
W
)
∩
J
←
(
W
)
|
=
|B|/
2
+
s
−
λ
s
+
λ
q
−
q
≥|W |
because
|
1
.
Now
A
i
int
A
i
If
λ
s
=
0
,
s
≥
3 because 2
b
<
1
.
(
j
)
∈ B
(
W
)
if
qb
∈
(
j
)
.
There
Y
→
having
qb
as an interior point so
are at most
s
−
λ
s
members of
B
(
W
)
contains
Y
→
.
at least
(
s
λ
q
−
q
λ
s
)
−
(
s
−
λ
s
)=
s
λ
q
−
s
members of
Now
s
≤
q
+
λ
q
because
(
q
+
λ
q
)
b
≥
qb
+
λ
q
a
≥
1
>
(
s
−
1
)
b
.
Thus
|B
(
W
)
|≥|B|/
2
+
s
λ
q
−
s
≥|B|/
2
+
because
J
→
(
W
)
∩
J
←
(
W
)=
{
(
s
−
1
)
λ
q
−
q
≥|B|/
2
+
λ
q
−
q
≥|W |
q
+
1
}.
By Cases
1
and
2
we can now assume
J
→
(
W
)
|≤
J
←
(
W
)
|≤
|
2
,
and
|
2
(9.7)
and
J
→
(
W
)
∪
J
←
(
W
)
i
∈
implies
i
>
q
.
(9.8)
Thus
J
→
(
W
)
|
+
|
J
←
(
W
)
|
)
.
|W |≤
(
λ
q
−
q
)(
|
(9.9)
Case 3.
Suppose
J
→
(
W
)=
{
k
,
k
+
1
}
where
k
>
q
.
We have already shown that,
Y
←
and at least
for this case,
B
(
W
)
contains at least
s
λ
q
−
q
λ
s
−
(
s
−
λ
s
)
of the
Y
→
.
J
←
(
W
)
|
=
s
λ
q
−
q
λ
s
−
(
s
−
λ
s
+
λ
q
−
q
)
of the
Furthermore, if
|
2
, B
(
W
)
Y
→
.
contains at least
s
λ
q
−
q
λ
s
−
(
s
−
λ
s
)
members of
Thus
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