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Every connected component $C$ of a nonempty compact proper subset $D$ of a connected Hausdorff space meets its boundary. If not, the Šura-Bura lemma identifies $C$ as a quasi-component in the compact Hausdorff subspace $D$. Finitely many clopen neighborhoods separate it from the compact boundary, giving a nonempty relatively clopen $H\subseteq \mathrm{int}D$. Such $H$ is open in the ambient space, and is also closed there because $D$ is closed. It is proper because $D$ is proper. This contradicts connectedness.

A closed combinatorial surface is a finite connected simplicial complex in which every simplex is a face of a triangle and the link of a simplex at every vertex is a cycle graph. Every edge then lies in exactly two triangles, and the triangles at a vertex form a single cyclic fan. The edge-incidence condition alone is insufficient: gluing two triangulated spheres only at one vertex gives two disjoint circles in that vertex's link.

Its geometric realization is a compact surface without boundary. Triangle interiors have disc neighbourhoods, edge interiors have such neighbourhoods from the two incident triangles, and the cyclic fan gives one at each vertex. A finite realization is a Hausdorff space with a countable basis of a topology, being a finite union of geometric simplexes in Euclidean space. If connectedness is omitted by convention, require these properties separately on each connected component.

For a continuous map $f:X\to Y$ with Hausdorff space $Y$, its graph of a function is closed in the product topology. If $y\ne f(x)$, choose disjoint neighborhoods of those two points in $Y$. Continuity supplies a neighborhood of $x$ whose image lies in the neighborhood of $f(x)$; its product with the other neighborhood misses the graph.

The cofinite topology on a set $X$ declares the empty set and all cofinite subsets to be open sets: a nonempty open set may exclude only a finite set of points. There is no assumed metric. On an infinite set, this gives a particularly small collection of open sets while still making every singleton a closed set. On a finite set, every subset has finite complement, so the cofinite topology is the discrete topology.^[ref]

For an original example, use $X={\mathbb{N}}_{>0}$ and chooseBoth are open sets, and their intersection is $U\cap V=X\setminus \{1,2,3,5,7,8\}$. Thus $4,6,9,10,\dots$ are in both. Their finite exclusions cannot separate the infinite remainder. The finite intersection of any finite collection of nonempty open sets is again a nonempty open set, since its complement is a finite union of finite sets. An arbitrary nonempty union is open because its complement is contained in the finite complement of any one nonempty member.

The cofinite topology is useful for separating hypotheses that familiar metric spaces package together. An infinite cofinite topological space is a T1 space, because its singletons are closed, but is not a Hausdorff space, because two nonempty open sets cannot be disjoint sets. It is a hyperconnected space and hence a connected space. Moreover, every subset of a cofinite space is compact, and sequence convergence in the cofinite topology can have many limits. These are compatible properties, rather than contradictions: the usual uniqueness and closedness results use the Hausdorff space hypothesis.^[ref]

The map $x\mapsto (x,f(x))$ is continuous into the product topology. It maps a compact space $X$ onto the graph of a function of a continuous $f$. Hence that graph is compact, whether or not the codomain is a Hausdorff space. Projection to $X$ is its continuous inverse.

A compact group is a topological group whose underlying topological space is a compact Hausdorff space. The circle group is a compact group: the unit circle is a closed bounded subset of the complex plane, so the Heine-Borel theorem gives compactness. The additive group $(\mathbb{R},+)$ is a locally compact group and a Hausdorff space, but is not a compact group: the open cover $\{(-n,n):n\ge 1\}$ has no finite subcover. A finite group with the discrete topology is another compact group; compactness does not require a smooth manifold structure.

The great advantage of a compact group is that one can average over every element without choosing a preferred origin. Precisely, every locally compact group that is a Hausdorff space admits a nonzero left Haar measure $\mu$, a regular Borel measure finite on compact sets, unique up to a positive scalar. Left invariance means $\mu (gA)=\mu (A)$ for every Borel set $A$ and every $g\in G$. There is also a right Haar measure; in general left and right Haar measures can differ. On a compact group, $0<\mu (G)<\infty$, and normalizing to $\mu (G)=1$ gives a unique regular Borel probability measure invariant under both left and right translation. See Tornier, Haar Measures, Theorem 2.2 for existence and uniqueness, and Ram's notes on left and right Haar measure for the distinction.

For a finite group $G$, normalized Haar measure assigns each element mass $1/|G|$, so integration is simply $|G{|}^{-1}{\sum }_{g\in G}f(g)$. For the circle group, the corresponding integral is uniform angle averaging, computed below. On $(\mathbb{R},+)$, Lebesgue measure is a Haar measure, but its total mass is infinite: no positive scalar multiple makes it a probability measure. Thus the existence of Haar measure does not make every topological group a compact group.

Historically, Alfréd Haar introduced the general invariant-measure construction in Der Massbegriff in der Theorie der kontinuierlichen Gruppen, Annals of Mathematics 34 (1933), pp. 147-169. This bibliographic identification is also recorded in André Weil's archived seminar notes, Mesure de Haar, reference Haa33. The conceptual step is to let group symmetry itself determine the weights used in integration, rather than relying on coordinates in an ambient Euclidean space.

Let $K\subseteq X$, where the ambient topological space $X$ is Hausdorff and $K$ is compact in its subspace topology. Then $K$ is a closed set in $X$: the complement $X\setminus K$ is an open set in $X$. The ambient space need not be compact. For example, $[0,1]$ is a compact closed interval in the non-compact Hausdorff real line.^[ref]

The ambient Hausdorff hypothesis matters even when $K$ itself is Hausdorff. In the Sierpiński space $S=\{0,1\}$, the open sets are $\varnothing ,\{1\},S$, so the closed sets are $\varnothing ,\{0\},S$. Thus $K=\{1\}$ is compact and Hausdorff in its one-point subspace topology, but is not closed in $S$: its complement $\{0\}$ is not an open set. Every open neighborhood of $0$ contains $1$, so separation from $K$ fails. In contrast, the discrete topology on the same underlying set is Hausdorff and makes both singletons closed.

The diagram lists every open set and closed set, so the distinction can be checked directly. All subsets in both panels are compact: for any open cover of a finite set, choose one covering open set per point, obtaining a finite subcover. Compactness alone therefore cannot distinguish the two panels.

Here is the finite-selection mechanism in the general proof. For $x\in X\setminus K$ and each $k\in K$, the Hausdorff property gives disjoint open neighborhoods $U_k$ of $k$ and $V_k$ of $x$. The $U_k$ give an ambient open cover of $K$. Select a finite subcover $U_{k_1},\dots ,U_{k_m}$ and setThen $K\subseteq U$, $x\in V$, and $U\cap V=\varnothing$. The finite intersection makes $V$ an open set; an infinite intersection would not guarantee this. Thus every point of $X\setminus K$ has an open neighborhood contained in that complement, proving that $K$ is closed. The empty set is already closed.

This theorem combines with the continuous image of a compact space theorem to make every continuous map from a compact space to a Hausdorff space a closed map: a closed subset of its domain is compact, and its image is compact and hence closed in the target. This is the reason the inverse in the compact-to-Hausdorff continuous bijection theorem is continuous. The converse implication, from closed to compact, needs additional hypotheses: the real line is closed in itself but not compact; in a compact space, every closed subset is compact.

A Hausdorff space is completely regular if a point $x$ outside a closed set $F$ can be separated from $F$ by a continuous function $u:X\to [0,1]$ with $u(x)=1$ and $u{|}_F=0$. These functions determine the original topology: for any neighborhood $O$ of $x$, choose $F=X\setminus O$ to obtain $x\in \{u>1/2\}\subseteq O$. This is exactly the point-separation property needed for the evaluation embedding into the weak-star dual of bounded continuous functions.

Let $X$ be a Hausdorff space, let $K_n$ be a nested sequence of nonempty compact subsets of $X$, and let $f:X\to Y$ be a continuous map into a T1 space. Then the displayed equality holds. The forward inclusion is true for every map. For the reverse, fix $y\in {\bigcap }_{n}f(K_n)$. Each $F_n=K_n\cap f^{-1}(\{y\})$ is nonempty and closed in the compact set $K_1$: the $K_n$ are closed by the Hausdorff assumption, while the fiber is closed because $Y$ is T1 and $f$ is continuous. The nested sets $F_n$ have the finite intersection property, so compactness gives a point in their intersection. Its image is $y$.

The proof only needs the relevant fibers to be closed in $K_1$, rather than all of $Y$ to be T1. A Hausdorff space as codomain is a familiar sufficient special case. Mere continuity into an arbitrary space does not suffice.

A continuous map that is also an injective function. A continuous injection need not be a topological embedding: its inverse on its image need not be continuous. If its source is a compact space and its target is a Hausdorff space, however, the compact-to-Hausdorff continuous bijection theorem makes that inverse continuous.

If $X$ is an infinite set with its cofinite topology and $Y$ is a Hausdorff space, every continuous map $f:X\to Y$ is a constant function. Otherwise two distinct image points would have disjoint open neighbourhoods. Their inverse images would be nonempty disjoint open sets of $X$, contradicting its hyperconnected space property.

For example, $f(n)=n$ from ${\mathbb{N}}_{>0}$ with its cofinite topology into the real line with its usual Euclidean metric is not continuous: the inverse image of the open interval $(1/2,3/2)$ is $\{1\}$, which is not open in the infinite domain. A continuous real-valued readout on that domain can only report one value. Finite cofinite domains are different: their discrete topology allows arbitrary functions into any topological space.

A continuous map from a nonempty finite product space of infinite sets with their cofinite topologies to a Hausdorff space is a constant function. Fixing all but one coordinate gives a continuous map from an infinite cofinite topology, so changing that coordinate does not change the value. Any two points in a finite Cartesian product can be joined by finitely many one-coordinate changes. This proves the assertion. In particular every real-valued continuous function on $Z\times Z$ is constant when $Z$ is the real line equipped with the cofinite topology.

Give $\mathbb{N}$ the topology whose nonempty open sets are $\mathbb{N}$ and the finite initial segments. Every nonempty open set contains $1$, so any two nonempty open sets intersect. A continuous map from this topological space into a Hausdorff space is constant: distinct image points would have disjoint open neighbourhoods, whose inverse images would be disjoint nonempty open sets. This includes continuous real-valued maps.

A filter on a set equipped with a topology converges to $x$ if it contains every neighbourhood of $x$. A Hausdorff space gives every convergent filter at most one limit: disjoint neighbourhoods of distinct purported limits would have empty intersection in the filter.

A sequence $(x_n)$ in a topological space converges to $x$ if every open neighbourhood of $x$ contains all sufficiently late terms:For a metric space, this is the usual limit of a sequence condition expressed through open balls. For a Hausdorff space, distinct proposed limits have disjoint open neighbourhoods, so a sequence cannot eventually stay in both and the limit is unique. The sequence convergence in the cofinite topology demonstrates what can change when that separation property is absent.^[ref]

Continuous functions $X\to [0,1]$ on a normal Hausdorff space separate points and closed sets by Urysohn's lemma. The evaluation map into the product of their codomains is injective and continuous. For each point in an open set, a coordinate equal to one there and zero outside the open set detects a smaller coordinate neighborhood, proving the inverse topology.

Every subset $A$ of a cofinite topological space is a compact subset. To see the finite reduction directly, take an open cover of a nonempty $A$ by open sets of $X$ and choose one member $U$ containing a point of $A$. Since $U$ has finite complement, the set $A\setminus U$ is finite. Choose a covering member for each of those remaining points. Those members together with $U$ are a finite subcover. The empty set is already compact.

Equivalently, the subspace topology on $A$ is itself cofinite: for finite $F\subseteq X$, the trace of $X\setminus F$ is $A\setminus (F\cap A)$. The argument applies again inside $A$. A concrete open cover $\{X\setminus \{k\}:k\in X\}$ of $X={\mathbb{N}}_{>0}$ needs only the two members $X\setminus \{1\}$ and $X\setminus \{2\}$ when these labels are distinct: each fills the other's single gap.

On $X={\mathbb{N}}_{>0}$, the subset of even integers is compact but not closed, because it is infinite and proper. A compact subset of a Hausdorff space is guaranteed to be closed, whereas compactness without the Hausdorff space assumption does not suffice. Changing the topology changes compactness: the same infinite set with its discrete topology has the singleton open cover, which has no finite subcover. The finite-exception argument is special to finite complements; in the cocountable topology, compact subsets of a cocountable space must instead be finite.

Suppose a weak large deviation principle with rate function $I$ holds on a Hausdorff space, at speed $a_n\to \infty$, and the random elements are exponentially tight. For a finite $r$, choose a compact set $K$ whose complement has upper exponential rate strictly below $-M$, with $M>r$. The open-set lower bound on $K^c$ implies ${\inf }_{K^c}I>M$, so $\{I\le r\}\subset K$. Lower semicontinuity makes this sublevel set closed, hence compact. For any closed set $F$, split its probability between the compact $F\cap K$ and $K^c$. The upper exponential rate is at most $\max \{-{\inf }_{F}I,-M\}$. Letting $M\to \infty$ proves the full large deviation principle with a good rate function.

A topological space is extremally disconnected when the closure of every open set is open. This concerns open sets, not arbitrary subsets. In particular, a nonisolated singleton in a Hausdorff space is closed but not open, so its closure gives a counterexample to that stronger condition.

A left Haar measure on a locally compact group that is a Hausdorff space is a nonzero regular Borel measure, finite on compact sets, satisfying $\mu (gA)=\mu (A)$ for every Borel set $A$. Such a measure exists and is unique up to a positive scalar. Right Haar measure instead satisfies $\mu (Ag)=\mu (A)$; the two need not coincide. On a compact group, normalization gives total mass one and invariance on both sides. See Tornier, Haar Measures, Definition 2.1 and Theorem 2.2.

For a finite group, normalized Haar measure is uniform counting; for the circle group, it is normalized angle integration on the circle. These examples turn symmetry into a rule for averaging.