# Bourbaki General Topology I: Exercise 20, sec. 8 ch.1

I have added a picture with the complete exercise. I'm interested only in c), I think I have proved a) and b).

In c) we are given a topological space $$X_0$$ which is semi-regular (i.e. there exists a base for the topology whose open set satisfy the relation $$\mathring{\overline{U}}=U$$) and another topological space $$X$$ (the underlying set is always the same). Then, we have to prove that $$X^*=X_0$$ if and only if there exists a family $$\mathfrak M$$ of dense subsets of $$X$$ such that every finite intersection of sets of $$\mathfrak M$$ belongs to $$\mathfrak M$$ and such that the topology on $$X$$ is generated by the union of $$\mathfrak M$$ and the open sets of $$X_0$$ (here $$X^*$$ is the topology whose base are the regular open sets in $$X$$). Hence, we have to manage to construct examples of topologies finer than regular Hausdorff spaces which are not regular.

To prove the result, the exercise gives us a hint: We should consider the dense open subsets of $$X$$ and notice that every open set in $$X$$ can be written as the intersection of a dense open set in $$X$$ with an open set in $$X_0$$.

I'm not sure how to start with. I know that, if $$D$$ is a dense subset, then $$\overline U = \overline{U\cap D}$$, but I don't think this helps at all.

Any hint will be grateful. Thanks.

EDIT:

The axiom $$\mathrm{O_{III}}$$ is the condition of regularity: For each closed set $$F$$ and each point $$x\in X\setminus F$$, there are disjoint open sets containing $$x$$ and $$F$$, respectively.

CONTEXT:

My goal is th goal of the exercise: I want to give finer topologies than regular Hausdorff that aren't regular Hausdorff. I think it is interesting because finer topologies than $$T_0$$, $$T_1$$, $$T_2$$, $$T_{21/2}$$ or completely Hausdorff are $$T_0$$, $$T_1\dots$$, resp. But it doesn't happen for $$T_3$$, and I would like to know why. My guess is that making finer the topology may appear new closed sets that doesn't verify the $$\mathrm{O_{III}}$$ axiom; namely, we have made the topology finer, but not enough, so we have create new closed sets but not enough open sets to separatd them from points.

Here is the complete exercise

• Your first sentence has a wrong definition for a semi regular space. – Henno Brandsma Mar 7 at 5:02
• What is Bourbaki's Axiom O$_{III}$? – DanielWainfleet Mar 7 at 6:54
• @HennoBrandsma Yes, that's true. Only the sets of a certain base for the topology of $X_0$ satisfy that. Thanks. I hope this doesn't matter for my previous work. I guess no but... – Dog_69 Mar 7 at 6:58
• @DanielWainfleet It is regularity: For each closed set $F$ and each $x\in X\setminus F$ there exists disjoint open sets containing $F$ and $x$. – Dog_69 Mar 7 at 7:00
• The common modern notation for regular space is $T_3$ space. – DanielWainfleet Mar 7 at 7:31