The general structure of topological space is quite nice, but it is often not sufficient in practice. We often have to consider more certain types of topological spaces, that satisfy an additional *separation axiom*. In this section, we will present several such axioms, classified from less to more stringent. They basically tell us how efficient the open sets are in separating points or sets.

Definition 1.

Let be a topological space.

- ( It is called (or Kolmogorov) if, for two points and in with , there exists an open set such that and
orthere exists an open set such that and .- It is called (or Fr\’echet) if, for for two points and in with , there exists an open set such that and
andthere exists an open set such that and .- It is called (or Hausdorff) if, for two points and in with , there exists open sets and such that , , and .
- It is called if it is both and
regular, that is to say, for any closed set and , there exist open sets and such that , and .- It is called (or Tychonoff) if it is both and
completely regular, that is to say, for any closed set and , there exists a continuous function , such that and on .- It is called if it is both and
normal, that is to say, for any closed sets and there exist open sets and such that , , and .

The most important property is y far : in spaces (which I will call Hausdorff spaces from now on), a convergent sequence has a unique limit. A topological space is if, and only if, for all , (the *singleton* containing ) is closed. A moment of reflexion shows that a topological space is (take and ). Clearly, a space is , a space is , and a space is . It is also clear that a space is , but it is not obvious that a space should be . This is indeed the case, and can be proved with the help of the following result.

Lemma 1.(Urysohn’s Lemma) If is normal, if and are subsets of with closed, open, and , then there exists a continuous function such that on and on .

This is very far from obvious. I have just learned about it, and I will try to convey my understanding of the proof (such as it is). Urysohn’s Lemma essentially tells us that we can separate closed sets using a continuous function. Let us try to do it on an example. We take as a topological space the segment , with its topology as a subspace of the real line equipped with the usual topology (the one I described at the beginning of the section on topological spaces). We take our closed set to be and our open set to be (recall that we use the subspace topology). The function defined by then satisfies the three conditions we require: it is continuous, and . To be more general, we have to define this function in such a way that an analogous definition could be found in a general normal space. It is reasonable (though I admit not obvious) to think that a definition of using a countable family of conditions on would be suited to this task.

Let us consider the set

We call it the set of dyadic numbers. Those are the numbers in that can be written with a finite number of binary digits. For any , we set . Now is a family of open sets of such that:

- for all , ;
- for all , ;
- for all and in such that , .

We can now define the function using using the family :

This definition can be generalized.

Lemma 2.Let be a topological space, a closed set and an open set in . Let us assume that is a family of open sets in satisfying:

- for all , ;
- for all , ;
- for all and in such that , ;
- .
Let us define the function by

Then is continous and on .

*Proof.* Since for all and , we have on . Let us note (this will be used repeatedly ) that if , , and if , . It remains to prove that is continuous. According to the statement at the end of the first section, it is enough to show that is continuous at every point of . Let be a point of and a neighborhood of in .

** Case 1:** . In that case . We recall that the intervals (with and ) form a basis for the topological space . According to the definition of the subspace topology, there exists therefore such that . There also exists such that . Then is an open set, containing . For all , we have according to the remark at the beginning, and thus .

**Case 2:** with . With the same arguments than in the preceding case, there exists such that . There exists such that . Then is an open set containing . For , we have , and therefore .

**Case 3:** with . With the same arguments as before, there exists such that . There exist also and in such that

Then is an open set containing such that for all .

We have shown that in all cases there is a neighborhood of , such that $f(N’)\subset N$. This shows that is continuous at and concludes the proof.

Let us turn to the proof of Urysohn’s Lemma. For , we denote by the set

that is to say the set of numbers in that have at most binary digits. We have and . Let us fix an open set and a closed set in . We will construct recursively a family of open set satisfying the hypotheses in Lemma 2.

Since is normal, there exist two open sets and such that , and . In particular, is contained in the closed set and thus . We have “constructed” the open set indexed by the element in .

Let now assume that, for some , we have a family of open set satisfying the hypotheses in Lemma 2. Let us now pick .

**Case 1:** . Then the closed sets and are disjoint, and, proceeding as in the first step, we find an open set such that

and .

**Case 2:** for some . Then the closed sets and are disjoint, and, proceeding as before, we find an open set such that and .

**Case 3:** . Then the closed sets and are disjoint and, proceeding as before, we find an open set such that and .

We can therefore build recursively a family of open sets satisfying the hypotheses of Lemma 2, and thus, by applying Lemma 2, a continuous function such that on and on . This proves Urysohn’s Lemma.

We can now prove that a space is by applying Urysohn’s Lemma to the closed set and the open set .