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Lemma 6.34
Let
R S × D sub ( S ) be closed. Then Ch (
R
) is also closed.
Proof
Straightforward.
is also closed.
Corollary 6.10
Let
R
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
S
× D sub ( S ) be closed and convex, and C
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
D sub ( S ))
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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