T1 space

 The title of this article is incorrect because of technical limitations. The correct title is T_{1} space.
In topology and related branches of mathematics, T_{1} spaces and R_{0} spaces are particularly nice kinds of topological spaces. The T_{1} and R_{0} properties are examples of separation axioms.
Contents 
Definitions
Let X be a topological space and let x and y be points in X. We say that x and y can be separated if each lies in an open set which does not contain the other point.
 X is a T_{1} space if any two distinct points in X can be separated.
 X is a R_{0} space if any two topologically distinguishable points in X can be separated.
A T_{1} space is also called an accessible space or a Fréchet space and a R_{0} space is also called an symmetric space. (The term Fréchet space also refers to an entirely different notion from functional analysis. For this reason, the term T_{1} space is preferred).
Properties
Let X be a topological space. Then the following conditions are equivalent:
 X is a T_{1} space.
 X is a T_{0} space and an R_{0} space.
 Points are closed in X; i.e. given any x in X, the singleton set {x} is a closed set.
 Every finite set is closed.
 Every cofinite set of X is open.
 The fixed ultrafilter at x converges only to x.
 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.
Let X be a topological space. Then the following conditions are equivalent:
 X is an R_{0} space.
 Given any x in X, the closure of {x} owns only the points that x is topologically indistinguishable from.
 The fixed ultrafilter at x converges only to the points that x is topologically indistinguishable from.
 The Kolmogorov quotient of X (which identifies topologically indistinguishable points) is T_{1}.
In any topological space we have, as properties of any two points, the following implications
 separated ⇒ topologically distinguishable ⇒ distinct
If the first arrow can be reversed the space is R_{0}. If the second arrow can be reversed the space is T_{0}. If the composite arrow can be reversed the space is T_{1}. Clearly, a space is T_{1} if and only if it's both R_{0} and T_{0}.
Note that a finite T_{1} space is necessarily discrete (since every set is closed).
Examples
The Zariski topology on an algebraic variety is T_{1}. To see this, note that a point with local coordinates (c_{1},...,c_{n}) is the zero set of the polynomials x_{1}c_{1}, ..., x_{n}c_{n}. Thus, the point is closed. However, this example is well known as a space that is not Hausdorff (T_{2}).
For a more concrete example, let's look at the cofinite topology on an infinite set. Specifically, let X be the set of integers, and define the open sets O_{A} to be those subsets of X which contain all but a finite subset A of X. Then given distinct integers x and y:
 the open set O_{{x}} contains y but not x, and the open set O_{{y}} contains x and not y;
 equivalently, every singleton set {x} is the complement of the open set O_{{x}}, so it is a closed set;
so the resulting space is T_{1} by each of the definitions above. This space is not T_{2}, because the intersection of any two open sets O_{A} and O_{B} is O_{A∪B}, which is never empty. Alternatively, the set of even integers is compact but not closed, which would be impossible in a Hausdorff space.
We can modify this example slightly to get an R_{0} space that is neither T_{1} nor R_{1}. Let X be the set of integers again, and using the definition of O_{A} from the previous example, define a basis of open sets G_{x} for any integer x to be G_{x} = O_{{x, x+1}} if x is an even number, and G_{x} = O_{{x1, x}} if x is odd. Then the open sets of X are, unions of the basis sets
 <math>U_A := \bigcup_{x \in A} G(x) <math>
The resulting space is not T_{0} (and hence not T_{1}), because the points x and x + 1 (for x even) are topologically indistinguishable; but otherwise it is essentially equivalent to the previous example.
Generalisations to other kinds of spaces
The terms "T_{1}", "R_{0}", and their synonyms can also be applied to such variations of topological spaces as uniform spaces, Cauchy spaces, and convergence spaces. The characteristic that unites the concept in all of these examples is that limits of fixed ultrafilters (or constant nets) are unique (for T_{1} spaces) or unique up to topological indistinguishability (for R_{0} spaces).
As it turns out, uniform spaces, and more generally Cauchy spaces, are always R_{0}, so the T_{1} condition in these cases reduces to the T_{0} condition. But R_{0} alone can be an interesting condition on other sorts of convergence spaces, such as pretopological spaces.
See also
 T_{0} space
 Hausdorff space (T_{2} space)
 separation axiomspl:T1 przestrzeń