How to make a space sober 😵‍💫🚿

Two related questions:

1. How to make a space sober?
2. How to make a sober space?

In our main reference,

📚 Jorge Picado & Aleš Pultr Frames and Locales: topology without points. Springer Basel, Series Frontiers in Mathematics, Vol. 28, xx + 400 pp., 2012 (ISBN: 978-3-0348-0153-9).,

there is a note emphasizing that sobriety is a completeness-type property. This suggest to me that we can:

1. Punch holes in a space to make it, if not non-sober, at least a little dizzy. 😵‍💫
2. Complete a non-sober space in order to make it sober.

I shall leave “punching holes” do another post.

How to make a space sober?

What is the glucose shot 💉 for topological spaces?

Actually, to make a space sober, we can inject 💉 some points into it. Or… equivalently, inject it 💉 into a sober space. 🤯

Injecting 💉 one point

Suppose the space $(X, \tau)$ has a completely prime filter $\mathscr{F}$ such that there is no $x \in X$ making

Notice that we are using $\cap$ instead of $\wedge_*$, because we are identifying

and treating ${\mathscr{F}^c}^*$ as the family of all closed sets that contain $x$.

We want to append to $X$ a new point $w \not \in X$, without changing the complete lattice structure for the open sets. This is very easy, we just have to inject $w$ into all sets in $\mathscr{F}$. The new topological space becomes $\hat{X} = X \cup \{w\}$, and the new topology is given by the family of open sets

In this case, it is very obvious that $\mathscr{F}^c$ is the family of open sets from $\hat{X}$ that do not contain $w$. Therefore,

is the biggest open set that does not contain $w$. It is also clear that $X$ has the topology induced by $\hat{X}$.

Injecting all points at once

We actually do not even need the above construction. We did it just to get the idea of how it feels to complete a space, making it a little more sober. It is much easier to build up a sober space from scratch and embed $X$ into it.

We already have the complete lattice structure. Each completely prime filter must correspond to a point.

Since $\tau(x)$ is completely prime for all $x \in X$, we can send $X$ to $\widetilde{X}$:

To say that this map is an injection is exactly the same thing as saying that $X$ is a $T_0$ space.

The open sets must be in bijection with the original $\tau$. It is just like associating each point to its open neighbourhood filter, $\mathscr{F} = \tau(x)$. Except that we might not have the point $x$ itself. But the open sets that contain $x$ are exactly those in $\mathscr{F}$.

Each “original open set” $A \in \mathscr{F}$ is now supposed to correspond to an open neighbourhood of $\mathscr{F}$. So, given $A \in \tau$, the corresponding open set in $\widetilde{X}$ is

This might look a little strange, but recall that for an open set $A$,

This is is basically what is written in $\eqref{complete:open}$:

Finally, although a little confusing 😵‍💫, it is very clear that $X$ has the topology induced from $\widetilde{X}$ through $\iota$. And, in case of a $T_0$ space, $X$ can be seen as a subspace of $\widetilde{X}$, with the subspace topology.

Conclusion

I have found that completing a non-sober space is a process similar to constructing the Stone-Čech compactification using ultrafilters. It is much easier, though. Because we are not generating a new topology. We already have the complete lattice structure for it.

I wonder if, given a complete lattice, this construction always gives a (sober) topology for this lattice.

The question above looks to me equivalent to asking:

In a complete lattice $L$ is every $b \in L$ contained in a completely prime filter $\mathscr{F}$?

In terms of algebra this looks a lot like asking if — in a certain ring — there is a prime ideal $\mathscr{F}^c$ that does not contain the (non-zero) element $b$. Usually, a maximal ideal that does not contain $b$ is actually prime.

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