T1 space

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Topological spaces in
separation axiom
Kolmogorov (T0) version
T0 | T1 | T2 | T | completely T2
T3 | T | T4 | T5 | T6
The correct title of this article is T1 space. It features superscript or subscript characters that are substituted or omitted because of technical limitations.

In topology and related branches of mathematics, T1 spaces and R0 spaces are particular kinds of topological spaces. The T1 and R0 properties are examples of separation axioms.

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[edit] 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 T1 space if any two distinct points in X can be separated.

A T1 space is also called an accessible space or a Fréchet space and a R0 space is also called a symmetric space. (The term Fréchet space also refers to an entirely different notion from functional analysis. For this reason, the term T1 space is preferred. There is also a notion of a Fréchet-Urysohn space as a type of sequential space.)

[edit] Properties

Let X be a topological space. Then the following conditions are equivalent:

  • X is a T1 space.
  • X is a T0 space and an R0 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 yx 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 R0 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 T1.
  • Every open set is the union of closed sets.

In any topological space we have, as properties of any two points, the following implications

separatedtopologically distinguishabledistinct

If the first arrow can be reversed the space is R0. If the second arrow can be reversed the space is T0. If the composite arrow can be reversed the space is T1. Clearly, a space is T1 if and only if it's both R0 and T0.

Note that a finite T1 space is necessarily discrete (since every set is closed).

[edit] Examples

  • The cofinite topology on an infinite set is a simple example of a topology that is T1 but is not Hausdorff (T2). This follows since no two open sets of the cofinite topology are disjoint. Specifically, let X be the set of integers, and define the open sets OA 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 T1 by each of the definitions above. This space is not T2, because the intersection of any two open sets OA and OB is OAB, which is never empty. Alternatively, the set of even integers is compact but not closed, which would be impossible in a Hausdorff space.
  • The above example can be modified slightly to create the double-pointed cofinite topology, which is an example of an R0 space that is neither T1 nor R1. Let X be the set of integers again, and using the definition of OA from the previous example, define a subbase of open sets Gx for any integer x to be Gx = O{x, x+1} if x is an even number, and Gx = O{x-1, x} if x is odd. Then the basis of the topology are given by finite intersections of the subbasis sets: given a finite set A, the open sets of X are
U_A := \bigcap_{x \in A} G_x
The resulting space is not T0 (and hence not T1), because the points x and x + 1 (for x even) are topologically indistinguishable; but otherwise it is essentially equivalent to the previous example.

[edit] Generalisations to other kinds of spaces

The terms "T1", "R0", 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 T1 spaces) or unique up to topological indistinguishability (for R0 spaces).

As it turns out, uniform spaces, and more generally Cauchy spaces, are always R0, so the T1 condition in these cases reduces to the T0 condition. But R0 alone can be an interesting condition on other sorts of convergence spaces, such as pretopological spaces.

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