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An ultrafilter is a proper filter maximal under inclusion. Equivalently, for every $A\subseteq \mathbb{N}$, exactly one of $A$ and $\mathbb{N}\setminus A$ belongs to $\mathcal{U}$.

Solved by gpt-5.6-sol high.

The cofinite subsets of $\mathbb{N}$ form a proper filter. By Zorn lemma, it extends to an ultrafilter $\mathcal{U}$. Since $\mathcal{U}$ contains every cofinite set, it cannot contain a finite set, so it is nonprincipal.

Define the Stone-Čech compactification of the natural numbers $\beta \mathbb{N}$ to be the set of ultrafilters on $\mathbb{N}$, with basic setsSince $\beta \mathbb{N}\setminus \bar{A}=\overline{\mathbb{N}\setminus A}$, these sets are clopen. If $\mathcal{U}\ne \mathcal{V}$, choose $A\in \mathcal{U}\setminus \mathcal{V}$; then $\bar{A}$ and $\overline{\mathbb{N}\setminus A}$ are disjoint neighbourhoods, proving Hausdorffness. For compactness, a family of basic closed sets with the finite-intersection property corresponds to a family of subsets of $\mathbb{N}$ with the finite-intersection property. Extend that family to an ultrafilter; the resulting point belongs to every closed set. The Alexander subbase theorem now proves compactness.

The Hindman theorem states that every finite colouring of $\mathbb{N}$ admits an infinite sequence $x_1,x_2,\dots$ for which every nonempty finite sum of distinct terms has one colour. Let $\mathcal{U}\in \beta \mathbb{N}$ be an additive idempotent, and choose a colour class $A\in \mathcal{U}$. PutIdempotence gives $A^{\ast }\in \mathcal{U}$, and $A^{\ast }-x\in \mathcal{U}$ whenever $x\in A^{\ast }$. Having selected $x_1,ldots,x_n$ with all finite sums in $A^{\ast }$, chooseThis finite intersection belongs to $\mathcal{U}$ and is nonempty. Induction keeps every finite sum in $A^{\ast }\subseteq A$, proving Hindman's theorem.

Solved by gpt-5.6-sol high.