T 1 room
Let X be a topological space . X is called T 1 -space if for any two points each has a neighborhood in which the other does not lie. To delimit: In a T Raum-room only one of the two points must have such an environment, in a T₂-room the two environments must be disjoint. It is also said that a T 1 space has a Fréchet topology . In this context, the term Fréchet space , which is a term from functional analysis , should be avoided .
Let X be a topological space. The following statements are equivalent:
- X is a T 1 space.
- X is a Kolmogoroff space and an R 0 space .
- All single point sets in X are closed .
- Every finite set is closed.
- Any set with finite complement is open.
- Every elementary filter for any x only converges to x .
- For every subset S of X it holds that an element x from X is an accumulation point of S if and only if every open neighborhood of x contains an infinite number of elements.
The following implications always apply in topological spaces
- separated ⇒ topologically distinguishable ⇒ disjoint
If the first arrow can be reversed, it is an R 0 space , and precisely in a T 0 space this also applies to the second implication. This shows that a topological space satisfies T 1 if and only if it is both an R 0 -space and a T 0 -space.
The Zariski topology on an algebraic variety (in the classical sense) is T 1 . To see this we consider a point with a local coordinate . The associated one-point set is the set of zeros of the polynomials . The point is thus closed.
For another example, consider the cofinite topology on a countable set, such as the set of integers . As an open set, we define exactly the empty set and the sets with finite complement. So you have all the shape with a finite set A . Now let x and y be two different points. The set is an open set that contains x and y does not. On the other hand, the element y does not contain x . It is therefore actually a T 1 space. But this can also be deduced from the fact that one-element sets are closed. However, this space is not a T 2 space. Because for two finite sets A and B holds what can never be empty. Furthermore, the set of even numbers is compact but not closed, which can never be the case in a T 2 space.
More generally, for every topological space that satisfies the T 1 axiom, its topology already includes the cofinite topology. The co-finite topology is thus the coarsest T 1 topology on a set.