Talk:Base (topology)
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Dcoetzee:
Overall, I think your changes make the article much better. I just have a couple comments.
- My main concern is that it's still not clear that you can DEFINE a topology by giving a base. For people who are familiar with the concept of "generating objects" (i.e. basis of a vector space, set generating a free group, etc.) the distinction between the generating object being used to understand or analyse a previously given structure vs using that object to define a new structure seems obvious...but to people not familiar with it, I think it's very confusing at first. In the article, you talk in several places about bases "generating" topologies, and while we know that you mean "base defining topology", the reader might not. So, when they read "An example of a collection of open sets which is not a basis is the set S of all semi-infinite intervals of..." my concern is that the reader might ask "not a basis OF WHAT TOPOLOGY"? It was to avoid this confusion in the reader that I explicitly spelled out the two uses of the term. While my presentation was admittedly too formal, I think the distinction still needs to be made explicit.
- A basis of a vector space is far from unique, either. I can't think of many generating objects in math that are completely unique, so I'm not sure what to compare it to, here.
Otherwise, your additions are very nice. Revolver
- Oops. A basis of a vector space isn't unique, but has unique cardinality, at least. This is what I meant to say here. Fixing this.
- As for the "not a basis of what topology", I think it suffices to say "not a basis of any topology on this set". I see how the concept and the two uses could be confusing though, and need to be distinguished, but not as severely as they were in the original article (to say these are the two uses of bases is a bit extreme).
- Derrick Coetzee 01:32, 2 Jan 2004 (UTC)
A compact space with any topology has a finite base (because any base forms an open cover.) Similarly, a Lindelöf space with any topology has a countable base, and so is second-countable.
Is this true? It doesn't seem right. It's true, any base of a compact set is an open cover, but that doesn't mean it's finite. I'm not certain about the Lindelof thing, I think it's true if you throw in another assumption (Hausdorff, probably, but I don't remember.) Revolver
- Oops, this of course isn't true. Every base of a compact space has a finite subset that covers, being an open cover, but it need not be a base itself. Same bad reasoning into the second fact. I shall remove these. I think we need more theorems about bases in general, and maybe an example of a proof. I want to get the idea across of why bases are useful, which in my opinion is because it's generally easier to deal with an arbitrary base element than, say, an arbitrary open set (if the basis is nicely chosen of course).
- Derrick Coetzee 01:32, 2 Jan 2004 (UTC)
