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The quotient maps define a continuous homomorphismIts kernel is ${\bigcap }_{U\in \mathcal{U}}U=\{1\}$, because $\mathcal{U}$ is a neighborhood basis and $G$ is Hausdorff. Thus $\Phi$ is injective. The standard compactness argument for inverse limits makes it surjective: a compatible family of cosets has the finite intersection property, and the corresponding closed cosets in compact $G$ have nonempty intersection. Finally, a continuous bijection from compact $G$ to the Hausdorff inverse limit is a homeomorphism. Henceas topological groups.

Hindman theorem states that every finite coloring of $\mathbb{N}$ admits an infinite sequence $x_1,x_2,\dots$ whose finite-sums set is monochromatic.

Identify the Stone-Čech compactification of the natural numbers $\beta \mathbb{N}$ with the compact Hausdorff space of ultrafilters on $\mathbb{N}$, equipped with addition on the Stone-Čech compactification of the natural numbers. This makes $\beta \mathbb{N}$ a compact Hausdorff left-topological semigroup. For completeness, the Ellis–Numakura lemma gives an idempotent in every such semigroup: by the Hausdorff space property and compactness, the intersection of a descending chain of nonempty compact subsemigroups is nonempty, so Zorn's lemma gives a minimal one $K$. For $a\in K$, the compact subsemigroup $K+a$ equals $K$. Hence the nonempty compact subsemigroupis also $K$, and in particular $a+a=a$. Choose the resulting idempotent ultrafilter on the natural numbers $\mathcal{U}$.

One color class $A$ belongs to $\mathcal{U}$. For $x\in \mathbb{N}$, writeand defineThe identity $\mathcal{U}+\mathcal{U}=\mathcal{U}$ implies $A^{\ast }\in \mathcal{U}$. It also implies that $A^{\ast }-x\in \mathcal{U}$ for every $x\in A^{\ast }$: both $A-x$ and the set of $y$ for which $A-(x+y)$ belongs to $\mathcal{U}$ lie in $\mathcal{U}$, and their intersection is $A^{\ast }-x$.

Choose $x_1\in A^{\ast }$. Having chosen $x_1,\dots ,x_n$ with every nonempty finite sum in $A^{\ast }$, chooseThis is possible because it is a finite intersection of members of the ultrafilter $\mathcal{U}$. Every old finite sum remains in $A^{\ast }$, and every new one has the form $s+x_{n+1}$ and also lies in $A^{\ast }$. By mathematical induction, all nonempty finite sums lie in $A^{\ast }\subseteq A$, proving Hindman's theorem.

Now putfor each $A\in \mathcal{U}$. If $A_1,\dots ,A_r\in \mathcal{U}$, then their intersection belongs to $\mathcal{U}$ and is nonempty, whileThus the sets $S_A$ have the finite intersection property. Their closures are closed subsets of the compact interval $[0,1]$, so their total intersection contains some $x$. Equivalently, every neighbourhood of $x$ meets every $S_A$; this is the ultrafilter limit of the sequence.

The point is unique. If distinct points $x,y$ both had this property, choose disjoint neighborhoods $U,V$. The index set $I_U=\{n:x_n\in U\}$ must belong to $\mathcal{U}$, because otherwise its complement would belong to $\mathcal{U}$ and the associated $S_A$ would miss $U$. Similarly $I_V\in \mathcal{U}$. Their intersection is empty, contradicting the definition of an ultrafilter.

A topological space is compact if and only if every ultrafilter on it converges to at least one point, assuming the ultrafilter lemma. No Hausdorff assumption is needed. In a compact space, the closures of the filter members have the finite intersection property, so choose a point in their intersection. Every open neighbourhood of that point belongs to the ultrafilter, since otherwise its closed complement would be a filter member. Conversely, extend the complements of a cover with no finite subcover to an ultrafilter; a limit point lies in a covering open set, contradicting the simultaneous membership of that set and its complement.

Let $S$ be a nonempty set and $\mathcal{P}\subseteq \mathcal{P}(S)$ be upward closed. An ultrafilter all of whose members lie in $\mathcal{P}$ exists exactly when every finite set partition of $S$ has a cell in $\mathcal{P}$. For sufficiency, the complements of sets outside $\mathcal{P}$ have the finite intersection property: a finite cover by such sets could be disjointified into a partition with no cell in $\mathcal{P}$. The ultrafilter lemma extends the generated proper filter. Necessity follows because an ultrafilter selects exactly one cell of every finite partition. This criterion requires partition richness of the whole ground set; it does not assert that the complement family is closed under unions.