Talk:T1 space

Untitled

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 "O_A" to "A'" in my mind's eye.

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.

Rename article to "Separation classes" or something like it

Since the article covers many other flavors of "separable space" besides $T_{1}$ 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)

Two lists?

Why are there two lists headed by "If X is a topological space then the following conditions are equivalent" under conditions? What is the reason for this split? ~2026-23862-66 (talk) 15:38, 18 April 2026 (UTC)

One list gives characterizations of T1 spaces. The other does the same for R0 spaces. PatrickR2 (talk) 22:19, 19 April 2026 (UTC)

Request to add a proposition of a theorem of T1 spaces being equivalent to the Axiom of Choice

I sincerely hope that this will not fall under Wikipedia:NOT#IINFO, but as I still find this proposition that I have found to be rather interesting, may I add the fact that the statement "For any set X, there is a unique smallest topology J such that (X,J) is a T_1 space" and the AC are equivalent? PicoMath (talk) 16:15, 20 April 2026 (UTC)

Sorry, but this seems dubious. Even if true, it is certainly not mentioned in any standard source. Do you have any specific reference for this? PatrickR2 (talk) 18:11, 20 April 2026 (UTC)

Not for the above proposition per se; I just got one part of the theorem (that statement about unique T_1 spaces) from an exercise in Kelley's General Topology, and thought that it might be nice to prove equivalence with AC.

Now that I think about it, I believe that I shouldn't add this, possibly because it goes against Wikipedia's rules about "original research". PicoMath (talk) 00:34, 21 April 2026 (UTC)

The unique smallest T1 topology is an easy thing, and does not require AC, as far as I can tell. That's why I was curious about a reference for it. If you have doubts, you can ask a question on math stackexchange. But as you said, wikipedia is not the place for "original research". And even if the result were true, it is probably not "notable". PatrickR2 (talk) 04:14, 22 April 2026 (UTC)