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

Question

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

Answer 1

Your guess is right.

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.