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A zero-dimensional space has a base of clopen sets. A totally disconnected space has no connected subset with more than one point, equivalently all its connected components are singletons. We use these definitions in the stated Hausdorff setting.

If a Hausdorff space has a clopen base, any two different points can be separated by a clopen set. A connected subset containing both would be split into two nonempty disjoint relatively open sets. Thus zero-dimensional implies totally disconnected here.

The compact converse, including the component lemma. For a point $x$ of a compact Hausdorff $K$, putThis is its quasicomponent, a closed compact set. We prove the Šura-Bura component lemma that $Q_x$ equals the connected component of $x$.

Suppose $Q_x=A\cup B$ were a separation into nonempty disjoint relatively closed sets, with $x\in A$. Both are compact, hence closed in $K$. Normality of compact Hausdorff spaces gives disjoint open neighborhoods $U\supseteq A$, $V\supseteq B$. The compact set $H=K\setminus (U\cup V)$ misses $Q_x$. For each $y\in H$, choose a clopen neighborhood $N_y$ of $x$ missing $y$; these exist by the definition of $Q_x$. The open complements cover $H$, so finitely many suffice. Their intersection $N$ is clopen, contains $Q_x$, and lies in $U\cup V$. If $H$ is empty take $N=K$.

Now $N\cap U$ is clopen in $K$: it is open, and its complement is $(K\setminus N)\cup (N\cap V)$, also open. It contains $x$ and misses $B$, contradicting that $B\subseteq Q_x$ lies in every clopen neighborhood of $x$. Hence $Q_x$ is connected. Conversely every connected subset containing $x$ lies in every clopen neighborhood of $x$, so the connected component is contained in $Q_x$. Equality follows.

If $K$ is totally disconnected, then $Q_x=\{x\}$. Given an open neighborhood $O$ of $x$, for each $y\in K\setminus O$ choose a clopen neighborhood of $x$ missing $y$. Compactness again gives a finite intersection $N$ with $x\in N\subseteq O$. These $N$ form a clopen base. Therefore

The Stone-Čech characterization. The stated clopen separation property is strong zero-dimensionality in this normal Hausdorff setting. A normal T1 space is a Tychonoff space, so its Stone-Čech compactification exists. We use its extension property for maps into compact Hausdorff spaces and the density of $X$.

First assume $\beta X$ is a zero-dimensional space, and let $C,D$ be disjoint closed subsets of $X$. Urysohn's lemma gives $f:X\to [0,1]$ equal to zero on $C$ and one on $D$. Its continuous extension to $\beta X$ puts ${\overline{C}}^{\beta X}$ and ${\overline{D}}^{\beta X}$ in disjoint zero and one level sets. Using a clopen base, cover the first compact set by finitely many clopen sets missing the second; their union $H$ is clopen and separates them. Thus $U=H\cap X$ and $V=X\setminus U$ are disjoint clopen sets containing $C,D$.

Conversely, assume the stated clopen separation property on $X$. Let $A,B$ be disjoint closed subsets of $\beta X$. Normality supplies a continuous function $f:\beta X\to [0,1]$ with $f{|}_A=0$, $f{|}_B=1$. The setsare disjoint and closed in $X$. Select clopen $U\subseteq X$ containing $C$ and missing $D$; the given disjoint pair of clopen neighborhoods supplies such a $U$. Its characteristic function, one on $U$ and zero outside, extends from $X$ to a continuous map $\beta X\to \{0,1\}$. Let $H$ be its one level set, a clopen set.

Every $a\in A$ is in ${\overline{C}}^{\beta X}$: any neighborhood of $a$, intersected with $f^{-1}([0,1/3))$, is an open neighborhood meeting the dense set $X$. Likewise $B\subseteq {\overline{D}}^{\beta X}$. Since $C\subseteq H$ and $D\subseteq \beta X\setminus H$, it follows that $A\subseteq H$ and $B\cap H=\varnothing$.

Apply this to $A=\{z\}$ and $B=\beta X\setminus O$, for any point $z$ and open neighborhood $O$. It yields a clopen $H$ with $z\in H\subseteq O$, proving that $\beta X$ has a clopen base. Hence the strong zero-dimensionality and Stone-Čech compactification criterion is

A topological group is a group equipped with a topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous maps. The product uses the product topology. Translation and inversion are therefore homeomorphisms. Hausdorffness is sometimes included in the definition; the two open-subgroup arguments below do not require it.

For the inverse system, take a directed set $I$, topological spaces $X_i$, and continuous maps $f_{ij}:X_j\to X_i$ for $i\le j$, with $f_{ii}=\mathrm{id}$ and $f_{ik}=f_{ij}{f}_{jk}$. An inverse limit is a space $X$ with compatible continuous maps ${\pi }_i:X\to X_i$ such that every compatible family $q_i:Y\to X_i$ factors uniquely as $q_i={\pi }_{i}q$ through a continuous map $q:Y\to X$. Existence follows from the concrete constructionwith the subspace topology of the product topology. The only possible map $q$ is $q(y)=(q_i(y){)}_i$. Compatibility puts its image in $X$, and coordinatewise continuity makes it continuous. This proves the universal property of an inverse limit, even when $X$ is empty. Given two inverse limits, their universal properties supply maps each way; the composites have the same coordinates as the identity, so uniqueness makes them identities. The inverse limit exists and is unique up to the unique homeomorphism respecting its projections.

A profinite group can be defined as an inverse limit of finite discrete groups, or as a compact Hausdorff space that is a totally disconnected space, equipped with continuous group operations. For the first implication, use the Tychonoff theorem. The compatibility equalities define a closed subgroup of the compact Hausdorff product. Two distinct points are distinguished by a finite discrete coordinate, so no connected subset contains both. Consequently the inverse limit is a compact Hausdorff totally disconnected topological group.

For the converse, use the topological result that a compact Hausdorff totally disconnected space has a basis of clopen sets. Choose a clopen set $U$ containing $1$ inside any specified identity neighbourhood. For each $u\in U$, continuity gives an identity neighbourhood $V_u$ and an open neighbourhood $W_u$ of $u$ with $V_u{W}_u\subseteq U$. Finitely many $W_u$ cover the compact set $U$. Intersect their $V_u$ and its inverse to obtain a symmetric identity neighbourhood $V$ with $VU\subseteq U$. Symmetry also gives $V^{-1}U\subseteq U$, and hence $vU=U$ for every $v\in V$. The setis a subgroup containing $V$, so it is open. It is contained in $U$, since $1\in U$ and $h=h\cdot 1\in hU=U$. By part (ii), $H$ has finite index. Its normal core of a subgroup is a finite intersection of conjugates of $H$, hence an open normal subgroup $N$ contained in $U$. This proves the open normal subgroup basis of a profinite group.

The natural group homomorphismis injective because the intersection of these $N$ is $\{1\}$. For a compatible family of cosets, any finitely many have nonempty intersection: their intersection subgroup is another indexing subgroup, and its prescribed coset lies in each. The finite intersection property and compactness give a point belonging to all the cosets, proving surjectivity. Finally a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. The two definitions of profinite group are equivalent.

This is always true. Every open ball in an ultrametric space is also closed: a point outside a ball has a disjoint ball of the same radius around it. Distinct points can therefore be separated by clopen sets, so every connected subset is a singleton and $K$ is totally disconnected.

A space is zero-dimensional when its topology has a base of clopen sets. In the Hausdorff setting these sets separate distinct points, so the space is totally disconnected. Conversely a compact Hausdorff totally disconnected space has a clopen base, by the Šura-Bura component lemma and finite subcovers of the complement of a neighborhood.