Basic Idea: the dual of filters

I was going to write about a topological space being sober. But, for this next post not getting too full of information, I thought it would be nice to talk a little bit about closed sets, first.

This continues a previous post.

Closed sets

Just the same way we have a complete lattice of open sets, we do have a complete lattice given by the closed sets. Everything we did for the open lattice, we can repeat and do exactly the same for the closed lattice. It is a sublattice of $\mathscr{P}(X)$ , because finite unions and intersections of closed sets give closed sets. The infimum also coincides: arbitrary intersections. But the supremum is different: the closure of arbitrary unions. Where the closure is simply the infimum applied to everything that is bigger.

Now… I am certainly nauseating anyone fond of categories. 🤢

In a less pedestrian fashion, if you know a closed set is just the complement of an open one — or maybe it is how you like to define what a “closed set” is —, just notice that you can simply:

Use the complement operation ( $A \mapsto A^c$ ): an order reversing automorphism in $\mathscr{P}(X)$ . When restricted to the open lattice, it gives the (dual) closed lattice.

But, if you like categories… you may just “invert the arrow directions”. 😤

When you simply invert the arrow directions, you get the lattice structure of the closed sets. You do not get the closed sets, because you are not actually talking about sets!

Up to now, we have just been juggling 🤹 stuff from one side to the other. And although I personally do not like treating a closed set simply as the complement of an open one… if you do not want to leave the complete lattice of open sets we have been working with, it is a nice way to do things. Specially if you are not talking about sets!

But… what is the “complement” of an element, then? 🤷

Attention: If all we have is a complete lattice, we do not really know what the closed sets are! Because we are not talking about sets. We are simply representing closed elements using the order dual lattice: the same elements, but with reverse order. For an element $a$ in the complete lattice, we could even use the symbol $a^c$ to talk about the corresponding element in the dual lattice. But, that would be confusing, because we use $\cdot^c$ for set complement. Let’s use $a^*$ for the element $a$ regarded as an element in the dual lattice. And also, things like $\vee_*$ and $\wedge_*$ to refer to the dual lattice operations. In this case, for example,

$\begin{equation} a^* \vee_* b^* = (a \wedge b)^*. \end{equation}$

To make the notation easier to follow, lets agree that if we are treating $a$ as an open element, $a^*$ will be the same element treated as a closed element. And the closed elements will always have a $\cdot^*$ that can be removed to produce an open element. We shall do this even with families of elements: $\mathscr{F}$ is a family of open elements and $\mathscr{C}^*$ is a family of closed elements. The corresponding closed and open families are $\mathscr{F}^*$ and $\mathscr{C}$ .

The only exception shall be $0$ and $1$ . They are supposed to mean the smallest and biggest element for the given order. So, abusing notation a little, $0^* = 1$ , since the smallest element in the open lattice is the biggest element in the closed lattice.

When talking about a real topological space… using sets, when $B$ is a closed set, we shall write $B \sim a^*$ to mean that $B$ is the complement of the open set the element $a$ represents.

Disclaimer: I know almost nothing about categories. Not much beyond having noticed that there are morphisms in different theories. Category theorists… please, forgive my slip-ups! 😇

Closure of sets

Using those blinders… 🙈

We want to talk about closure, but we don’t want to talk about sets! I mean… all we have are the open sets. But we don’t know they are sets. 😱

Closure is easy… it depends a little on what closure means for you. To deal only with open sets, let’s talk about the complement of the closure. It is the biggest open set that does not intersect the original set:

$\begin{equation} \overline{B}^c = \bigvee \{A \in \tau :\, B \cap A = \emptyset\}. \end{equation}$

Or, using $\cdot^*$ …

$\begin{equation} \overline{B} \sim \left(\overline{B}^c\right)^* = {\bigwedge}_* \{A \in \tau :\, B \cap A = \emptyset\}^*. \end{equation}$

Since we are dealing with a complete lattice… we do not see sets… 🙈 let’s deal only with open elements (including $B$ ). Look Mom, no sets:

$\begin{equation} \tag{🚴} \label{closure:no_sets} \overline{B} \sim {\bigwedge}_* \{A \in \tau :\, B \wedge A = 0\}^*. \end{equation}$

Intersecting the Closures

Remember that in a topological space $X$ , we cannot expect much by looking at

$\begin{equation} \bigvee \tau(x) \quad\text{ and }\quad \bigwedge \tau(x). \end{equation}$

The join (of any filter) just gives 1, while the meet frequently is $0$ . In particular, $\bigwedge \mathscr{F}$ usually does not tell us much about a filter $\mathscr{F}$ possibly corresponding or not to the open neighbourhoods of some point.

If we make those sets just a little “fuller-bodied”… we get

$\begin{equation} \tag{$\overline{\text{v}}$} \label{vx:closed_intersection} \overline{\{x\}} \subset \bigcap_{V \in \mathcal{V}(x)} \overline{V}. \end{equation}$

Of course, in $\eqref{vx:closed_intersection}$ , we could have intersected only the closure of the open neighbourhoods and it would give the same closed set. Dually, using “closure with no sets” $\eqref{closure:no_sets}$ , we could do

$\begin{equation} \overline{\{x\}} \subset \bigcup_{V \in \mathcal{V}(x)} \bigcup_{\substack{A \in \tau \\ A \cap V = \emptyset}} A \sim {\bigwedge}_* \mathscr{C}_x^*, \end{equation}$

where

$\begin{equation} \mathscr{C}_x = \{A \in \tau :\, \exists U \in \tau(x),\, U \wedge A = 0\}. \end{equation}$

Notice that $\mathscr{C}_x$ is not a filter. I mean… $\mathscr{C}_x^*$ is a filter in the dual lattice, using the reverse order:

$\begin{gather} p^*, q^* \in \mathscr{C}_x^* \Rightarrow p^* \wedge_* q^* \in \mathscr{C}_x^* \\ a^* \in \tau^*,\; p^* \in \mathscr{C}_x^* \Rightarrow a^* \vee_* p^* \in \mathscr{C}_x^*, \end{gather}$

where the last impllication is the same as $p^* \prec a^* \Rightarrow a^* \in \mathscr{C}_x^*$ .

But in the original open lattice, the family $\mathscr{C}_x$ is the dual of a filter: an ideal. That is,

$\begin{gather} p, q \in \mathscr{C}_x \Rightarrow p \vee q \in \mathscr{C}_x \\ a \in \tau,\; p \in \mathscr{C}_x \Rightarrow a \wedge p \in \mathscr{C}_x. \end{gather}$

In a sense, $\mathscr{C}_x^*$ is a family of sets a little bigger than those in $\tau(x)$ : we took the closure of sets. However, instead of working on the lattice of closed sets, we are using $\mathscr{C}_x$ because we want to stay in the open lattice. Still… I’d like to think that $\mathscr{C}_x$ is a filter (in the dual order) a little “fuller-bodied” than $\tau(x)$ :

$\begin{equation} \tag{💪} \label{fuller-bodied} \bigcap \tau(x) \subset \left(\bigcup \mathscr{C}_x\right)^c \sim {\bigwedge}_* \mathscr{C}_x^* \end{equation}$

And that the right side above corresponds to the intersection of the closed neighbourhoods.

Even Easier 🤹

Let $\mathscr{F} \subsetneq L$ be a filter in a lattice $L$ . We can do exactly as with $\mathscr{C}_x$ , except we do not have the point $x$ . We have $\mathscr{F}$ :

$\begin{equation} \mathscr{C}_{\mathscr{F}} = \{a \in L :\, \exists u \in \mathscr{F},\, u \wedge a = 0\}. \end{equation}$

This is an ideal that corresponds to the idea of closed neighbourhoods of $\mathscr{F}$ . But there is an even “easier” family to work with:

$\begin{equation} \mathscr{F}^c = L \setminus \mathscr{F}. \end{equation}$

Notice that $\mathscr{C}_{\mathscr{F}} \subset \mathscr{F}^c$ , because $\mathscr{F}$ is a proper filter ( $0 \not \in \mathscr{F}$ ).

$\begin{equation} \bigvee \mathscr{C}_{\mathscr{F}} \prec \bigvee \mathscr{F}^c. \end{equation}$

Or… using closed elements…

$\begin{equation} \tag{💪💪} {\bigwedge}_* {\mathscr{F}^c}^* \prec_* {\bigwedge}_* {\mathscr{C}_{\mathscr{F}}^*}. \end{equation}$

The left side representing the intersection of every closed neighbourhood of the point, and the right side representing the intersection of every closed set that contains the point.

Not ideal, though 😟

So, $\mathscr{F}^c$ is a nice family of elements from our complete lattice… although, a family not ideal… 😟.

I mean… even if $\mathscr{F}$ is a proper filter, $\mathscr{F}^c$ might not be an ideal!!!

Well… $\mathscr{F}^c$ is a wannabe ideal. But let me tell you…

If it was an ideal, it would be prime!

$\begin{align} A \wedge B \in \mathscr{F}^c &\Leftrightarrow A \wedge B \not \in \mathscr{F} \\ &\Rightarrow A \not \in \mathscr{F} \text{ or } B \not \in \mathscr{F} \quad\text{(since $\mathscr{F}$ is closed by $\wedge$)} \\ &\Leftrightarrow A \in \mathscr{F}^c \text{ or } B \in \mathscr{F}^c \end{align}$

And, hey! Cheer up!! Not being an ideal is not a bad thing! It turns out that when the lattice we are working with is the topology $\tau$ of a topological space $X$ , and the filter is the open neighbourhood filter $\tau(x)$ for some point $x$ , the family $\tau(x)^c$ is indeed an ideal!!! 🎉 🎊 🥳

In fact, notice that $\tau(x)$ is a prime filter:

$\begin{gather} A \cup B \in \tau(x) \Rightarrow A \in \tau(x) \text{ or } B \in \tau(x) \\\text{equivalently}\\ A, B \not \in \tau(x) \Rightarrow A \cup B \not \in \tau(x). \end{gather}$

And, if $\mathscr{F}$ is a prime filter, $\mathscr{F}^c$ is an ideal. The main argument being:

$\begin{align} A, B \in \mathscr{F}^c &\Leftrightarrow A, B \not \in \mathscr{F} \\ &\Rightarrow A \vee B \not \in \mathscr{F} \\ &\Leftrightarrow A \vee B \in \mathscr{F}^c. \end{align}$

Change Log

2025-06-06: I have just realized that for a filter $\mathscr{F}$ , its complement, $\mathscr{F}^c$ might not be an ideal. So, I have fixed the post. Now, I understand much better the relation between those filters and those ideals in a lattice.

Continues… next post.