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Lemma 6.34
Let
R
ↆ
S
×
D
sub
(
S
)
be closed. Then
Ch
(
R
)
is also closed.
Proof
Straightforward.
†
is also closed.
Corollary 6.10
Let
R
ↆ
S
×
D
sub
(
S
)
be closed and convex. Then
R
Proof
For any
ʔ
∈
D
sub
(
S
), we know from Proposition
6.1
that
†
ʔ
·
R
={
Exp
ʔ
(
f
)
|
f
∈
Ch
(
R
)
}
.
The function Exp
ʔ
(
−
) is continuous. By Lemma
6.34
, the set of choice functions of
R
is closed and it is also bounded, thus being compact. Its image is also compact,
thus being closed.
Lemma 6.35
Let
R
ↆ
S
×
D
sub
(
S
)
be closed and convex, and C
ↆ
D
sub
(
S
)
be
†
closed. Then the set
{
ʔ
|
ʔ
·
R
∩
C
=∅}
is also closed.
Proof
=
Exp
ʘ
(
f
), which is obviously continuous. Then we know from the previous lemma
that
Ch
(
First define
E
:
D
sub
(
S
)
×
(
S
ₒ
D
sub
(
S
))
ₒ
D
sub
(
S
)by
E
(
ʘ
,
f
)
R
) is closed. Finally let
E
−
1
(
C
)
=
∩
D
sub
(
S
)
×
R
Z
ˀ
1
(
(
Ch
(
))),
where
ˀ
1
is the projection onto the first component of a pair. We observe that the con-
tinuity of
E
ensures that the inverse image of the closed set
C
is closed. Furthermore,
E
−
1
(
C
)
)) is compact because it is both closed and bounded. Its
image under the continuous function
ˀ
1
is also compact. It follows that
Z
is closed.
But
Z
={
ʔ
|
ʔ
·
R
∩
(
D
sub
(
S
)
×
Ch
(
R
†
∩
C
=∅}
because
∈
E
−
1
(
C
) for some
f
ʔ
∈
Z
iff (
ʔ
,
f
)
∈
Ch
(
R
)
iff
E
(
ʔ
,
f
)
∈
C
for some
f
∈
Ch
(
R
)
iff Exp
ʔ
(
f
)
∈
C
for some
f
∈
Ch
(
R
)
iff
ʔR
†
ʔ
for some
ʔ
∈
C
The reasoning in the last line is an application of Proposition
6.1
, which requires the
convexity of
R
.
An immediate corollary of this last result is:
Corollary 6.11
In a finitary pLTS the following sets are closed:
(i)
{
ʔ
|
ʔ
−ₒ
ʵ
}
A
(ii)
{
ʔ
|
ʔ
⃒
−ₒ}
Proof
By Lemma
6.17
, we see that
⃒
is closed and convex. Therefore, we can
apply the previous lemma with
C
={
ʵ
}
to obtain the first result. To obtain the second
A
we apply it with
C
={
ʘ
|
ʘ
−ₒ}
, which is easily seen to be closed.
The result is also used in the proof of:
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