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.
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.
Table of contents |
2 Examples 3 Generalisations to other kinds of spaces |
Definitions
A topological space X is T_{1} if and only if either of the following equivalent conditions is satisfied:
- Given any two distinct points x and y in X, each lies in an open set which does not contain the other. In other words, the singleton sets {x} and {y} are separated unless x = y.
- Given any x in X, {x} is a closed set. In other words, 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. This completes the proof.
- 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. This completes the proof.
- Given any two topologically distinguishable points x and y in X, each lies in an open set which does not contain the other. In other words, {x} and {y} are separated unless x and y are topologically indistinguishable.
- Given any x in X, the closure of {x} owns only the points that x is topologically indistinguishable from. In other words, the fixed ultrafilter at x converges only to the points that x is topologically indistinguishable from.
A space is T_{1} if and only if it's both R_{0} and T_{0} (which says that topologically indistinguishable points must be equal). Conversely, a space is R_{0} if and only if its Kolmogorov quotient (which identifies topologially indistinguishable points) is T_{1}.
Do not confuse the term "Fréchet topology", which is equivalent to "T_{1} topology", with the term "Fréchet space" which refers to an entirely different notion from functional analysis.
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;
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_{{x-1, x}} if x is odd. Then the open sets of X are, unionss of the basis sets
- U_{A} := ∪_{x in A} G_{x}.
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 netss) 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.