Separation Axioms in Topological Spaces
In nowadays, an axiom of separation is a property that satisfies certain topological spaces. We are going to describe these axioms that are labeled as but first let’s recall some basic definitions.
Separeted sets
Let be a topological space and let and be points points such that . Let be neighbourhoods of and respectively such that . Then, we say that:

and are distinguishable, if there a neighborhood such that but ,

and are separated by a closed neighborhood, if they have disjoints closed neighborhoods (e.g. and ),

and are separated by a continuous function, if there a map such that and the image and ,

and are precisely separated by a continuous function, if the preimage and .
Axioms

is (or Kolmogorov), the simplest axiom which satisfies that for any two distinct points and from , then and are distinguishable.

is (or Fréchet), if satisfies and if any two dinstinguishable points in are separated.

is (or Hausdorff), if satisfies and if any two dinstinguishable points in are separated by neighbourhoods.

is (or Urysohn), if any two distinct points in are separated by closed neighbourhoods. implies .

is (or regular Hausdorff), if satisfies and if for any and any closed set such that , then and are separated by neighbourhoods.

is (or Tychonoff,), if satisfies and if for any and any closed set such that , then and are separated by a continuous function. implies .

is (or normal Hausdorff), if satisfies and if any two disjoints closed subsets of S are separated by neighbourhoods.

is (or completely normal Hausdorff), if satisfies and if for any two separated sets are separated by neighbourhoods.

is (or perfectly normal Hausdorff), if satisfies and if any two disjoints closed sets are precisely separated by a continuous function.
Concluding:
since satisfies the properties of for a .
Example
Let’s make a very precise example. Let be a topological space and let be any two differrent closed disjoints subsets from such that there a function where is the distance from a point to the subset defined as . Then,

is , if we take any , then for any closed since are disjoints. Then for any , clearly are distinguishable.

is , let be closed subsets from such that and , then are disjoints closed neighborhood from and respectively. Indeed, for any and , then and are separared. Since is , then claim follows.

is , as we described previously that for any an are dinstinguishable points in are separated by neighbourhoods and respectively, then claim follows.

is , this is a little stronger than since follows that for any even if they are not dinstinguishable points, they are separated by closed neighbourhoods. Since already our example fulfills all the requirements about this, then claim follows.

is , if we taky any and we suppose that , then and are separated by neighbourhoods, this is easy to see if we suppose that , since and are disjoints subsets from and since it satisfies then claim follows.

is , this is a little stronger than , since requires that for any is separated by a closed subset , as previously we take and and they are closed disjoints subsets, then claim follows.

is , let and be two disjoints closed subsets from such that and , then we can say that and are separated by neighbourhoods and since satisfies , then claim follows.

is , this is a little stronger than , since it satisfies that two any subsets from are separated even if they are not closed. As we have decomposed in closed subsets and already our example fulfills all the requirements about this, then claim follows.

is , let be a continuous function such that and as we described previously is the distance from a point to the subset defined as , then separates and as closed subsets from . Note that we never divide by zero, that only could be possible if and are both zero, but this is impossible since and are disjoints. Since satisfies , then claim follows.
This post is inspired by the question asked on MathExchange and the example was inspired by the user Akiva about his answer to the question.