Cofinite filters and directed set filters

olivierd

May 16, 2023 3 min read

Cofinite filters and directed set filters

We assume that $E$ is an infinite set.

Cofinite filters

The cofinite filter $\mathcal F_{\text {cof}}$ over $E$ is the set of subsets $F$ of $E$ such that $E \backslash F$ is finite.

Since $E$ is infinite, $\emptyset$ is not in $\mathcal F_{\text {cof}}$ .

It is trivial to check that any superset in $E$ of an element of $\mathcal F_{\text {cof}}$ is in $\mathcal F_{\text {cof}}$ , and that the intersection of any two elements of $\mathcal F_{\text {cof}}$ is in $\mathcal F_{\text {cof}}$ . Thus $\mathcal F_{\text {cof}}$ is a filter over $E$ .

The cofinite filter over $E$ is defined independently of any ordering on $E$ .

Property:

If $\mathcal F$ is an ultrafilter over $E$ , then either it is a principal ultrafilter, or finer than the cofinite filter, but not both.

Proof:

Directed set filters

Directed set = ensemble ordonné filtrant, a partially ordered set $(E, \le)$ such that for any $x, y \in E$ , there exists $a \in E$ with $x \le a$ and $y \le a$ .

The directed set filter $\mathcal F_{\text{dsf}}$ is defined by the filter base $\mathcal B_{\text{dsf}} = \{ \ \{x \in E\ |\ a \le x \}\ \}_{a \in E}$ ; that is that the elements of $\mathcal B_{\text{dsf}}$ are the “segments” of all elements of $E$ greater than any $a \in E$ .

Property:

If $(E, \le)$ has no maximal element (that is, for all elements of $E$ , there is another element of $E$ that is strictly greater), then the directed set filter is finer than the cofinite filter.

Proof:

Let $F$ be an element of the cofinite filter over $E$ .

Then $E \backslash F$ is finite. Since $(E, \le)$ is a directed set, there exists an $a_1 \in E$ that is greater than all elements of $E \backslash F$ . There exists an $a \in E$ that is strictly greater than all elements of $E \backslash F$ . All elements in the segment $\{x \in E\ |\ a \le x \}$ are strictly greater, hence different from, all elements of $E \backslash F$ , that is, that they belong to $F$ . Thus $\{x \in E\ |\ a \le x \} \subseteq F$ , and since $\{x \in E\ |\ a \le x \}$ belongs to the filter base $\mathcal B_{\text{dsf}}$ , $F$ belongs to the filter $\mathcal F_{\text{dsf}}$ .

Thus the cofinite filter is a subset of the directed set filter.