This post comes as a sequel of On Set Convergence I where we introduced some necessary notions that are useful for studying how sets converge in a topological space. Alongside, we provided two non-topological definitions of convergence, namely the limits inferior and superior. In this post we will delve a bit further into the topological properties of the inner and outer limits of sequences of sets.
We start with an important result which characterises the topological nature of the inner limit:
Proposition 4. Let be a sequence of sets in a Hausdorff topological space . Then
Proof. (1). If then we can find a sequence such that while and is a strictly increasing sequence of indices. For any there is a such that for all it is: ; but also . Thus, . Therefore is in the right-hand side set of equation the first equation in Proposition 4.
(2). For the reverse direction assume that belongs to the right-hand side set of the first equation in Proposition 4. Then, there is a strictly increasing sequence such that for every we can find a Hence, (in the topology ). ♣
Or course for every result regarding the inner limit, there is a corresponding one for its complementary object: the outer limit. Therefore, following similar steps with the ones in the proof of Proposition 4 we can show that:
Proposition 5. Let be a sequence of sets in a Hausdorff
topological space . Then
Instead of arbitrary open sets, when is a normed space we may use open balls, i.e. sets of the form . We denote the unit ball of by . Then This leads us to the following corollary:
Corollary 1. Let be a sequence of sets in a normed space
This corollary yields another equivalent characterization for the inner limit. The predicate “” translates into the intersection over all , the requirement “” will be written as the union over all and then “” corresponds to an intersection. This allows us to restate the corollary
using set elementary operations as follows:
Corollary 2. Let be a sequence of sets in a normed space . Then,
Before we state and prove the following result (Proposition 7) we need to recall the following description of the closure of a set. Firstly, if is a topological space and , its closure is defined as:
Then, we have the following useful proposition:
We may then show the following result:
Proposition 7. Let be a Hausdorff topological space and be a sequence of sets in . Then,
Proof. (1). Let and let . Let be a neighbourhood of . There is a sucht that for all such that :
(2). Assume that . Then, there is an open neighbourhood of , let , such that . Therefore, . This completes the proof. ♣
Proposition 6 reveals a very important property of the inner limit:
Corollary 3. For any sequence of sets in a Hausdorff topological space the limit is closed.
Indeed, the inner limit is given as an arbitrary intersection of closed sets which is closed.
The following corollary is another immediate consequence of Proposition 6.
Corollary 4. Let be a sequence of sets in . Then,
The following result is preliminary to what follows. I put it here just because the proof has some interesting steps. It is expedient to know beforehand that the interior limit of a sequence of sets in normed spaces is described by:
Proposition 8. Let be a normed space and be a sequence of sets in . If then .
Proof. Let . We will construct a sequence and a strictly increasing sequence of indices such that while .
By Proposition 6, if , then for any :
Choose to be the smallest integer such that . Set .
Note: is cofinal for . Then,
and take to the be smallest such that .
Eventually, we construct the aforementioned sequences. We now have that:
Therefore we have proven that , i.e. . ♣
The previous result is extended to provide a particularly important description of the notion of the inner limit on normed spaces. We use the point-to-set distance function to describe the inner limit of a sequence of sets:
Proposition 9. Let be a normed space and be a sequence of sets in . The inner limit of a sequence of sets is:
Proof. (1). We firstly need to show that for any it is . Let us assume that , i.e. there exists an increasing sequence of indices so that . This suggests that there is a such that for all one has that . However, according to Proposition 6,
which contradicts our initial assumption. Hence, , i.e. and this way we have proven that is in the right-hand side set.
(2). Conversely, assume that in the right-hand side of the equation in the statement of Proposition 8, that is . For any ; we can find such that for all . By definition, we have that
thus we can find a such that
Therefore, from which it follows that (see Corollary 1). ♣
Note. For a sequence , in order to show that , it suffices to find a subsequence of it that converges to , i.e. it suffices to determine a strictly increasing sequence such that , i.e. . Using this fact we can prove the following:
Proposition 10. Let be a normed space and be a sequence of sets in . The outer limit of a sequence of sets is: