limit of sequence of sets
Recall that and of a sequence of sets denote the and the of , respectively. Please click here (http://planetmath.org/LimitSuperiorOfSets) to see the definitions and here (http://planetmath.org/LimitSuperior) to see the specialized definitions when they are applied to the real numbers.
Theorem. Let be a sequence of sets with . Then
for ranging over all infinite subsets of ,
for ranging over all subsets of with finite compliment,
We need to show, for ranging over all infinite subsets of ,
Let be an element of the LHS, the left hand side of Equation (1). Then for some infinite subset . Certainly, . Now, suppose . Since is infinite, we can find an such that . Being a member of , we have that . By induction, we have for all . Thus is an element of the RHS. This proves one side of the inclusion () in (1).
To show the other inclusion, let be an element of the RHS. So for all In , pick the least element such that . Next, in , pick the least such that . Then the set fulfills the requirement , showing the other inclusion ().
Here we have to show, for ranging over all subsets of with finite,
Suppose first that is an element of the LHS so that for some with finite. Let be a upper bound of the finite set such that for any , . This means that any , we have . Therefore, and is an element of the RHS.
Next, suppose is an element of the RHS so that for some . Then the set is a subset of with finite complement that does the job for the LHS.
The set of all subsets (of ) with finite complement is a subset of the set of all infinite subsets. The third assertion is now clear from the previous two propositions. QED
Corollary. If is a decreasing sequence of sets, then
Similarly, if is an increasing sequence of sets, then
Proof. We shall only show the case when we have a descending chain of sets, since the other case is completely analogous. Let be a descending chain of sets. Set . We shall show that
First, by the definition of of a sequence of sets:
Now, by Assertion 3 of the above Theorem, , so we only need to show that . But this is immediate from the definition of , being the intersection of all with subscripts taking on all values of . Its complement is the empty set, clearly finite. Having shown both the existence and equality of the and of the ’s, we conclude that the limit of ’s exist as well and it is equal to . QED
|Title||limit of sequence of sets|
|Date of creation||2013-03-22 15:00:34|
|Last modified on||2013-03-22 15:00:34|
|Last modified by||CWoo (3771)|