The hierarchy is a way of classifying separation axioms in topology. They describe how well points and sets can be distinguished by open sets.
Kolmogorov Space
Kolmogorov Space
A topological space is called a Kolmogorov space or -space if for any two distinct points in the space, there exists an open set that contains one of the points but not the other.
Fréchet Space
Space
A topological space is called Fréchet or -space if for any two distinct points , in the space, there exists open sets , respectively such that
For , the intersection of all open sets containing , is .
Proof If is , for any singleton set , there are no limit points, so the closure of is , which is closed; Now assume every singleton set is closed, then for any point in the intersection of all open sets containing , if , since is closed, is an open set containing , but which is a contradiction; Lastly, suppose the intersection of all open sets containing is , and , are two distinct points, then and serve as open sets containing and respectively, but not the other.
Hausdorff Space
Hausdorff Space
A topological space is called Hausdorff or if for any with there exist open neighborhoods of and of such that .
e.g. It is clear that all metric spaces are Hausdorff. In fact, most of the spaces we will encounter is Hausdorff.
Proposition
Let be a Hausdorff space and . A point is a limit point of if and only if any neighborhood of contains infinitely many points of .
Proposition
In Hausdorff spaces, limits of sequences are unique if they exist.
Proof Assume a sequence in a Hausdorff space has two distinct limits and . Then there exist neighborhoods of and of such that . However, since and , there exists an integer such that and for . Thus , which is a contradiction.
Proof direction is immediate because any singleton is compact. Conversely, suppose is Hausdorff, we first show that any compact set and singleton set can be separated by disjoint open neighborhoods. For any , we can pick an open neighbourhood of and an open neighborhood of such that . Then forms an open cover of , so it has a finite subcover . Let , then and are disjoint open neighborhoods of and .
Now for any two disjoint compact sets and , for any , we can find open neighborhoods of and of such that . Then similar to the previous case, forms an open cover of , so it has a finite subcover . Let , then and are disjoint open neighborhoods of and .
Proposition
Every finite set in a Hausdorff space is closed. More generally, any compact set in a Hausdorff space is closed.
Proof It suffices to show for any the set is closed. For any , by the Hausdorff property we can find an open set containing but . Thus and hence it is open. Consequently is closed. To show this is true for a compact set , note that super Hausdorff lemma shows every has an open neighborhood that is disjoint from , so is open.
Proof Any metric space is Hausdorff, so is closed. Without loss of generality, we can assume that is nonempty, so we can fix some . Note that the open balls forms an open cover of (thus an open cover of ), so it has a finite subcover . Let , then , so is bounded.
Proposition
Let be a Hausdorff space. If is a finer topology on , then is also a Hausdorff space.
Proof Let with . Since , the open sets in are also open in . Thus there exist open sets such that , and , that is is a Hausdorff space under .
Proof Suppose is Hausdorff, then for every distinct , there are disjoint open neighborhoods , of and respectively. Then is an open neighborhood of , hence is open and is closed. The converse is exactly the same argument in reverse.
