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Thm 2.1.16 (Completeness of the Function Domains) Let D 1 ; D 2 be domains with cpo's vD1 and vD2 , respectively. Then hD 1
 

Summary: Thm 2.1.16 (Completeness of the Function Domains) Let D 1 ; D 2 be domains with
cpo's vD1 and vD2 , respectively. Then hD 1
! D 2
i is the corresponding function domain
and vD1!D2 is a cpo on hD 1
! D 2
i.
Proof. The smallest element of hD 1
! D 2
i is ?D1!D2 . This function is indeed continuous
(since every constant function is continuous).
Now let S be a chain in hD 1
! D 2
i. Since vD1!D2 is complete on the set of all functions
from D 1 to D 2 (Thm. 2.1.13 (c)), there exists a least upper bound tS which is a function
from D 1
to D 2
. We have to show that this function is continuous (this means that it is in
hD 1 ! D 2 i).
So we have to show that (tS)(tT ) = t(tS)(T ) holds for every chain T from D 1

  

Source: Ábrahám, Erika - Fachgruppe Informatik, Rheinisch Westfälische Technische Hochschule Aachen (RWTH)

 

Collections: Computer Technologies and Information Sciences