In the previous post, I discussed the notion of network of a topological space. It was noted that for any space , the network weight (the least cardinality of a network for ) is always the weight (the least cardinality of a base for ). When is compact, the network weight and weight would coincide. Are there other classes of spaces for which network weight = weight? I would like to discuss two other classes of spaces where network weight and weight coincide, namely metrizable spaces and locally compact spaces. The following two theorems are proved. For a basic discussion on network, see the previous post.
Theorem 1. If is metrizable, then .
Proof. For the case of , we have . Now consider the case that is an uncountable cardinal. Based on the Bing-Nagata-Smirnov metrization theorem, any metrizable space has a discrete base. Let be a discrete base for the metrizable space . Now let . Because each is a discrete collection of open sets, any network woulld have cardinality at least as big as for each . If for some , then . If is the least upper bound of , then . Both cases imply . Since always hold, we have .
Theorem 2. If is a locally compact space, then .
Proof. Let be locally compact where . The idea is that we can obtain a base for of cardinality (i.e. ). Let be a network whose cardinality is . Here’s a sketch of the proof. Each point in has an open neighborhood whose closure is compact. For the compact closure of such open neighborhood, the weight would coincide with the network weight. Thus we can find a base of size within such open neighborhood. Because , we only need to consider many such open neighborhoods with compact closure. Thus we can obtain a base for of cardinality . To make this sketch more precise, consider the following three claims.
The collection of all , where is compact, is a cover of the space .
For every compact set , there is an open set such that and is compact.
If is open with compact, then we can obtain a base for the open subspace with .
For each in Claim 1, we can select an open (as in Claim 2)such that and is compact. Let be the union of all the in Claim 3 over all such . There are many in Claim 1. Thus . Note that would form a base for the whole space .
Both Claim 1 and Claim 2 are direct consequence of locally compactness. To see Claim 3, let be open such that is compact. We have (the network weight of a subspace cannot exceed the original network weight). By the result in the previous post, we have . We now have (the weight of a subspace cannot exceed the weight of the space containing it). So the weight of any open subspace with compact closure cannot exceed .
Corollary. For both metrizable spaces and locally compact spaces , .