Lower limit topology is not second countable

First of all, the remarks about the countability of can be seen here.

A topological space is said to be second countable if there is a countable basis. That is, there exists a countable collection such that, given any open set where .

Recall that a set is dense in if [that is, every point of is either in or a limit point of . If has a countable dense subset, then is said to be separable.

Note: , finite, in the usual topology is separable as the set , [the rational numbers] is both countable and dense.

Theorem If is metric and separable, then is second countable.

Proof: Let be the countable dense subset and then the set is a countable collection of open basis sets. Now if is any open set and then there exists some where . Now there is some where so .

It follows that if is separable but not second countable, there is no metric which produces the topology for ; that is, with that metric is not metrizable.

Example: , the real line with the lower limit topology, is not metrizable.

The proof: Note that is separable as is a countable, dense subset as every open set of the form contains a point of . But is NOT second countable. For if at least one of the . But there are an uncountable number of real numbers.

However does meet an countability condition of a sort. To see this, well use a Lemma:

Lemma: a collection of disjoint open sets in a second countable space is countable. This is easy to see after a moments thought, as there cant be more disjoint open sets than there are basis elements.

Now let be any open cover of . Let consist of all points not contained in the usual topology interior of a . If is empty then is covered by the interiors of these open sets; hence a countable subcollection covers all of . Otherwise, note that is an open set in the usual topology which means that is closed. Now for all there is some where since any other does not lie in an interior. That means that these open intervals are disjoint and so there can only be a countable number of these. On the other hand is covered by open [in the usual topology] intervals and so a countable subcollection suffices for this subset.

Therefore any open cover of has a countable subcover. A space that has this property is called Lindelöf.

A topological space is locally compact if for each there is a compact set where .

If is Hausdorff, then we can say that each is contained in some open set where is compact.

Examples.

1. Any compact topological space is locally compact.

2. finite, usual topology, is locally compact but not compact. Reason: is the required compact set whose interior contains .

3. [countable copies of ] is not locally compact. Reason: any open set has an infinite number of factors and is therefore contained in no compact set.

4. in the lower limit topology is NOT locally compact. Here is why: first remember that compact implies limit point compact [every infinite set has a limit point] and the basis for open sets is . Now let be a sequence in which . That is, the from a monotone increasing sequence whose least upper bound is strictly less than . Note: things like monotone and least upper bound are irrelevant to the topology in question. In fact, let .

Claim: the set has no limit point. Proof of the claim: remember that is open in the lower limit topology. Now no is a limit point as for and . Clearly, no is a limit point [clearly no is a limit point]. Also, no is a limit point as this entire open set misses the sequence. The only candidate for a limit point left is but which is open in and misses the sequence.

Now if is any set in which contains , cannot be compact as it has an infinite set which has no limit point. So is not compact and therefore with the lower limit topology is not locally compact.

Video liên quan