Principal Ultrafilter is All Sets Containing Cluster Point

Theorem

Let $S$ be a set.

Let $\powerset S$ denote the power set of $S$.

Let $\FF \subset \powerset S$ be a principal ultrafilter on $S$.

Let its cluster point be $x$.

Then $\FF$ is the set of all subsets $T$ of $S$ such that $x \in T$.

Proof

We have by hypothesis that the cluster point of $\FF$ is $x$.

Aiming for a contradiction, suppose there is some $A \subseteq S$ such that $x \in A$ and $A \notin \FF$.

Then, by definition of relative complement:

$x \notin \relcomp S A$

By definition of cluster point, $x$ is in every element of $\FF$.

But as $x \notin \relcomp S A$, it follows that:

$\relcomp S A \notin \FF$

This contradicts $\FF$ being an ultrafilter:

for every $A \subseteq S$, either $A \in \FF$ or $\relcomp S A \in \FF$

Therefore, by Proof by Contradiction, every $A \subseteq S$ such that $x \in A$ is in $\FF$.

Thus, $A \subseteq S$ is in $\FF$ precisely when $x \in A$.

This needs considerable tedious hard slog to complete it. In particular: We have demonstrated that $\FF$ contains (as a subset) the set of all subsets of $S$ which contain (as an element) $x$. We still need to show that the set of all subsets of $S$ which contain (as an element) $x$ contains (as a subset) $\FF$ To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction: Filters