subcover In A Sentence
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- Choosing a finite subcover then gives equicontinuity.
- Any finite subcover of this cover must be bounded, because all balls in the subcover are contained in the largest open ball within that subcover.
- Since M is compact, this cover has a finite subcover.
- In other words, every open cover of " X " by sets open relative to " Y " have a finite subcover that covers " X " iff the same is true for open sets relative to " Z ".
- Use Zorn's Lemma to find an open cover without finite subcover that is " maximal " amongst such covers.
- Then is an open cover of " S ", but any finite subcollection of has the form of " C " discussed previously, and thus cannot be an open subcover of " S ".
- This means that ever open cover has a countable subcover which means that in particular the topology on X has a countable subcover and let that countable subcover be the basis and we are done.
- That subcover must be of the form.
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