Talk:T1 space
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[edit]Gathered together bits and pieces from other articles -- I know nothing about this subject, and hope that someone who knows something about this will write a better article here.
Comment on "These conditions are examples of separation axioms." How can a T1 or an R0 space be a condition? Surely they're mathematical objects?
Even then the sentence would not make sense ....
"These mathematical objects are examples of separation axioms." I think some work needs to be done on this sentence. User:David Martland
- The original writer was mixing up the property of being T1, i.e., obeying the T1 axiom, with a T1 space, which is a space having the T1 property. I'll fix it - thanks! Chas zzz brown 09:29 Nov 20, 2002 (UTC)
- (Question: given the above definitions, is any space with X = {x} and open sets {{},X}, a trivial T1 space? What about a T2 space?)
Yup, a space with a single point is T1 and T2. This is an example of a vacuous truth: in this space, it is impossible to pick two different points x and y. AxelBoldt 21:47 Nov 23, 2002 (UTC)
In the examples section, can't we choose another notation for the complements of finite sets? I kept switching "OA" to "A'" in my mind's eye.
proof
[edit]For every point x in X and every subset S of X, x is a limit point of S if and only if every open neighbourhood of x contains infinitely many points of S.
- Proof. Suppose singletons are closed in X. Let S be a subset of X and x a limit point of S. Suppose there is an open neighbourhood U of x that contains only finitely many points of S. Then U \ (S \ {x}) is an open neighbourhood of x that does not contain any points of S other than x. (Here is where we use the fact that singletons are closed.) This contradicts the fact that x is a limit point of S. Thus, every open neighbourhood of x contains infinitely many points of S. Conversely, suppose there is a point x in X such that the singleton {x} is not closed. Then there is a point y ≠ x in the closure of {x}. We claim that any open neighbourhood U of y contains x. For suppose not; then the complement of U in X would be a closed set containing x, and the closure of {x} would be contained in the complement of U. Since y is in the closure of {x}, this would force y not to be in U, contradicting the fact that U is a neighbourhood of y. We have shown that y is a limit point of S = {x}. But it is clear that X is a neighbourhood of y that does not contain infinitely many points of S.
I deleted this proof from the article. See WP:NOT#IINFO.
Rename article to "Separation classes" or something like it
[edit]Since the article covers many other flavors of "separable space" besides it should be renamed for the general concept rather than for this particular one. Maybe separation classes, separability classes, separability axioms separability types, ... And a link to this article should be inserted in the articles on special types, such as separable space, Hausdorff space, etc. Jorge Stolfi (talk) 21:50, 15 November 2023 (UTC)
