Talk:T1 space

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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.

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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)

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(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)

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In the examples section, can't we choose another notation for the complements of finite sets? I kept switching "O_A" to "A'" in my mind's eye.

[edit] proof

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.