# 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 $$X$$ [According to the exercise, here should be $$X_0$$ instead of $$X$$. AR] 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. Mar 7, 2019 at 5:02
• What is Bourbaki's Axiom O$_{III}$? Mar 7, 2019 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... Mar 7, 2019 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 $x$. Mar 7, 2019 at 7:00
• The common modern notation for regular space is $T_3$ space. Mar 7, 2019 at 7:31

I didn’t find in [B] a definition of a topology $$\tau$$ generated by a family $$\mathfrak N$$ of subsets of a set $$X$$, so I assume that $$\mathfrak N$$ is a subbase for $$\tau$$.

Hence, we have to manage to construct examples of topologies finer than regular Hausdorff spaces which are not regular. cone topolgies.

Yes, a simple example is when $$X_0$$ is a unit segment $$[0,1]$$ endowed with the natural topology and $$\mathfrak M=\{[0,1]\setminus\{1/n:n\in\Bbb N\}\}$$.

I was acquainted with this exersise from Bourbaki’s book from your answer, but I applied this construction almost twenty years ago and used it to build Haudorff non-regular paratopological groups, see this my answer and Examples 3 and 2 from [Rav]. This construction turned out to be so basic tool to build counterexamples that later I wrote a paper [Rav2] devoted to its applications.

$$X^*=X_0$$ if and only if there exists a family $$\mathfrak M$$ of dense subsets of $$X_0$$ 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 $$X_0$$.

($$\Rightarrow$$) Put $$\mathfrak M=\{Y: Y$$ is open in $$X$$ and $$X\setminus Y$$ is nowhere dense in $$X_0\}$$. Clearly, each set $$Y\in\mathfrak M$$ is dense in $$X_0$$. It is easy to check that every finite intersection of sets of $$\mathfrak M$$ belongs to $$\mathfrak M$$. Now let $$Z$$ be any open set of $$X$$. Let $$\overline{Z}$$ be the closure of $$Z$$ in $$X$$. Since $$X^*=X_0$$, the interior $$Z_0$$ of the set $$\overline{Z}$$ in $$X$$ is open in $$X_0$$ and by b) the set $$\overline{Z}$$ is closed in $$X_0$$. Let $$Y=X\setminus (\overline{Z}\setminus Z)$$. It is easy to check that $$Y\in\mathfrak M$$ and $$Z=Y\cap Z_0$$.

($$\Leftarrow$$) Let $$\tau$$ be the topology of the space $$X_0$$ and $$\sigma$$ be the topology on the set $$X$$ with the subbase (in fact, a base) $$\mathfrak M$$. The topology of the space $$X$$ is a supremum $$\tau\vee\sigma$$ of topologies $$\tau$$ and $$\sigma$$. It is easy to see that the topologies $$\tau$$ and $$\sigma$$ are cowide and the topology $$\sigma$$ is wide, see definitions on [Rav2, p.10]. Since the topology $$\tau$$ is semiregular, $$\tau_r=\tau$$ (see [Rav2, p.11]) and by [Rav2, Lemma 7], $$(\tau\vee\sigma)_r=\tau_r=\tau$$, that is $$X^*=X_0$$.

References

[B] Nicolas Bourbaki, Elements of mathematics. General topology 1, Springer, 1966?.

[Rav] Alex Ravsky, *Pseudocompact paratopological groups , version 5.

[Rav2] Alex Ravsky, Cone topologies of paratopological groups.

• I need some time to understand what I done when traying to solve the exercise and to read and understand your answer. As soon as I make some progress or have some doubt, I'll let you know. Anyway, I thank you for reading my answer and trying to answer it. Jul 5, 2019 at 15:58
• @Dog_69 I added an acknowledgement to you for pointing me this exercise, see p. 28 of the paper “On feebly compact paratopological groups” by Taras Banakh and me. Aug 8, 2019 at 1:46
• you are so kind. I haven't stopped to analysed your answer yet and you are thanking me just to post an answer. Thank you so much. I don't I deserve it. Sep 17, 2019 at 8:30