Talk:Trivial topology
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To AxelBoldt: most of your changes were good, but you changed this line:
- Given a subset S of X, an element of S is a limit point of S if and only if S is not a singleton. In particular, every infinite subset of X has a limit point.
to this:
- If X has more than one element and S is some non-empty subset of X, then each element of X is a limit point of S.
The second is not correct. Given a singleton S={s}, s is not a limit point of S, because it does not lie in the closure of S-{s}, which is empty. It can still be generalized somewhat, however. I changed it to this, maybe the language could be cleaned up some:
Given a subset S of X which is not a singleton, all elements of X are limit points of S. If S is a singleton, every point of X \ S is still a limit point of S. In particular, every infinite subset of X has a limit point.
Derrick Coetzee 15:52, 29 Nov 2003 (UTC)
Yup you're right, except that your version doesn't work when S is empty. I also don't see the point of the last statement about infinite sets: you don't need infinitely many elements, you only need two. AxelBoldt 16:18, 29 Nov 2003 (UTC)
Oops, you're right. In the original statement I spoke of limit points of S in S, which worked when it was empty, but not currently. I added the statement about infinite subsets having limit points because this is the definition of limit point compactness (in some texts), and is an important property of metric compact spaces. If there were a page on limit point compactness I'd link to that instead.
Derrick Coetzee 00:28, 30 Nov 2003 (UTC)
