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Y where
members of
η (
s
q
2
)
if k
=
s
,
ρ (
k
)=
η (
s
q
1
)
if k
=
s
and so at least
λ q if
ρ (
k
)(
s
1
) ( η +
2
) .
Hence, using the symmetric result if
J ( W ) | =
|
1
,
we have
|B ( W ) |≥|W |
if
ρ (
k
)(
s
1
) ( η +
2
) .
For s
q
+
3
, ρ (
k
)(
s
1
) η (
s
1
)
so
|B ( W ) |≥|W |
for
η
2andfor
η =
1
and s
4
.
However
η =
1and s
<
4gives s
=
3and q
=
0
.
By Remark 1 we only
need to consider the case b
=
1
/
3and a
=
1
/ λ q .
Because
η =
1
, λ q
5
.
It is easy
Y so
to see that
B ( W )
contains at least 6
> λ q members of
|B ( W ) |≥|W |.
Using the Remarks, we have therefore shown that, for all
W ⊆ B
which are
needed for the proof of the theorem
|B ( W ) |≥|W |.
Thus, for all relevant
B, B
has
a set of distinct representatives by Hall's Theorem. Thus B i
) ∈X gives rise to
(
j
a pure Defender strategy with intervals B i
(
j
)
and A where A
∈ Y
is the representa-
B i
; we say that A is the correspondent of B i
tive of
(
j
)
(
j
) .
Pure Defender strate-
gies involving B i
(
j
)
are obtained similarly. Thus we have a total of 2
(
s
λ q
q
λ s )
Y ∪X ∪X occurs in precisely one of
them. Let S denote the Defender strategy which selects one of these pure strate-
gies at random. We show that, for w
pure strategies and every member of
[
,
] ,
(
λ
+ λ
)
0
1
S has at least 2
s
q
pure
s
q
.
strategies which have intervals containing w
Now
λ s
s
B i
A i
(
j
)
(
j
)
j
=
1
,..., λ q
q
i = 1
i = 1
λ q
q
λ s
B i
A i
A i
(
j
+ λ q
q
)
(
j
+ λ q
q
)
(
j
)
j
=
1
,...,
s
λ s
i = 1
i = 1
i
=
1
+ λ s
are coverings of
and so are the above expressions with the arrows reversed.
Denote the set of these coverings by
[
0
,
1
]
Y ∪X ∪X
C.
Note that every member of
occurs in precisely one member of
C .
For w
[
0
,
1
] ,
a covering C
∈ C
has a (unique) first interval I C (
w
)
containing w
,
by which we mean that the left-hand endpoint of I C (
w
)
is strictly less than the left-
hand endpoint of any other interval of C containing w
.
Note that I C (
w
)
starts strictly
to the left of w if w
is in precisely one of the pure strategies of S
for any w so we can define a mapping of
=
0
.
Now I C (
w
)
C
into the pure strategies of S by mapping
C to the pure strategy containing I C (
We show that this mapping is an injection.
Suppose two different coverings C 1 and C 2 map into the same pure strategy. This
pure strategy therefore comprises the intervals I C 1 (
w
) .
w
)
and I C 2 (
w
) ,
one of which is
of length b and the other (its correspondent) is of length a
.
Because an interval of
length b does not contain its correspondent, w
>
0
.
Hence I C 1 (
w
)
and I C 2 (
w
)
have an
interval
(
w
ε ,
w
]
in common for some
ε >
0
.
Thus, by the definition of the
B i (
j
) ,
the interval of length b must be of the form B i
or B i
(
j
)
(
j
)
where i
∈{
q
,
s
}
for
some j and the interval of length a of the form A λ i (
or A λ i (
j 1 )
j 1 )
respectively where
If B i
contains w so must B i
Similarly, if A λ i (
i
∈{
q
,
s
}
for some j 1 .
(
j
)
(
j 1 ) .
j 1 )
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