Infinite intersections in filters

Question

A filter on $X$ is defined as a family $F$ of subsets $\subseteq X$ such that:

1. $\emptyset\not\in F$;
2. $X\in F$;
3. $(A\in F\land A\subseteq B\subseteq X)\implies B\in F$;
4. $(A\in F\land B\in F)\implies A\cap B\in F$.

Using property 1 and 4 we can prove the finite intersection property: If we have a finite family $A_1, \dots, A_n\in F$, then their intersection is not empty. This is because if it would be empty, then it couldn't be an element of $F$. But by repeatedly using 4 we can in fact show that $\bigcap A_i\in F$ is in $F$.

Wondering whether this is also true for infinite intersections, two questions came to mind:

1. Is there a special filter $F$ on some set such that there is a family of sets $A_i\in F$ (for each $i\in I$ such a set $A_i$) with $\bigcap A_i \not\in F$? As we saw, in this case $I$ had to be infinite.

2. Similarly, can it happen that we have an infinite family $(A_i)$ whose intersection is empty?

Independent of this "intersection property" topic, I also have this question concerning filters:

3. Is there for each single set $X$ a filter $F$ on $X$?

Answer 1

1 & 2: Consider the filter $F$ on $\mathbb{N}$ (or any other infinite set) consisting of sets with finite complements, i.e. $$F= \{A\subset\mathbb{N}:\; \#(\mathbb{N}\setminus A)<\infty\}.$$ Now for any $n\in\mathbb{N}$, we have a set in the filter not containing $n$, for instance $A_n=\mathbb{N}\setminus\{n\}\in F$. So we get an empty intersection $\bigcap_{n\in\mathbb{N}}A_n=\bigcap_{n\in\mathbb{N}}(\mathbb{N}\setminus\{n\}) = \emptyset$.

3: Yes, there are filters on every set. The simplest possible example is the family consisting of the single set $F=\{X\}$. This is the only filter on a set $X$ consisting of a single point. For larger sets you have more examples, e.g. the principal filters $F_x = \{A\subset X:\, x\in A \}$. For finite sets the principal filters (and their intersections) are all you have, but for infinite sets you again get more types of examples, such as the above example of the finite-complement filter.

It is worth noting that as the cardinality of $X$ increases, you get more and more filters, since any filter on a subset $Y\subset X$ determines a filter on $X$ via the map $$\mathcal{P}(\mathcal{P}(Y))\to\mathcal{P}(\mathcal{P}(X)),\quad F\mapsto \{A\cup (X\setminus Y):\, A\in F \}.$$

Answer 2

Every principal filter (that is a filter of the form $\{B\subseteq X : A\subseteq B\}$ for some fixed set $A$) has the infinite intersection property, that is for any set $I$ if $A_i\in\mathcal{F}$ then $\bigcap_{i\in I}A_i\in\mathcal{F}$.

On the other hand there are many filters that do not have the infinite intersection property. Consider the following example: $X=\mathbb{N}$, the set of natural numbers, and define a filter $\mathcal{F}$ as follows: $$A\in\mathcal{F}\quad\iff\quad \mathbb{N}\setminus A\text{ is finite}.$$ It is not hard to see that $\mathcal{F}$ is a filter (but I think it's a good exercise to try this yourself). But it is easy to see that this filter does not have the infinite intersection property: Indeed, consider the set $A_n=\mathbb{N}\setminus\{n\}$. Then each of these sets belong in $\mathcal{F}$ but $\bigcap_{n\in\mathbb{N}}A_n=\varnothing$.

For the final question, yes since every principal filter is a filter.

As I said above all principal filters have the infinite intersection property. So it is only non-principal filters that do not have the infinite intersection property.

Do there exist non-principal filters with the infinite intersection property? On some extend yes. Consider the following. Let $X=\mathbb{R}$ and let $\mathcal{F}$ be $$A\in\mathcal{F}\quad\iff\quad \mathbb{R}\setminus A\text{ is at most countable}.$$ This filter has the countable intersection property (that is for every countable set $I$ if for all $i\in I$ $A_i\in\mathcal{F}$ then $\bigcap_{i\in I}A_i\in\mathcal{F}$), since the countable union of countable sets is countable.

On the other hand it's an interesting fact that we cannot prove the existence (nor the consistency of such existence) of non-principal ultrafilters (that is filters that also have the property that for all $A\subseteq X$ either $A\in\mathcal{F}$ or $X\setminus A\in\mathcal{F}$) that have the countable interesection property.