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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]

A filter base is a nonempty family of nonempty subsets directed under reverse inclusion. Its upward closure is a proper filter. Every finite intersection contains a base member, so it has the finite intersection property.

A finite intersection is an intersection of finitely many sets. The indexing set is finite, regardless of the size of the intersected sets. For an empty indexed family of subsets of a specified ambient set $X$, the convention is that the intersection equals $X$.

Use these standard ultrafilter facts: a space is compact if and only if every ultrafilter on it converges; the image ultrafilter under a map is an ultrafilter; and a convergent ultrafilter contains every neighborhood of its limit.

Let $X={\prod }_{\lambda \in \Lambda }{X}_{\lambda }$ with all factors compact. An empty factor makes $X$ empty and hence compact. Otherwise, given an ultrafilter $\mathcal{U}$ on $X$, push it forward by each product projection to obtain an ultrafilter ${\mathcal{U}}_{\lambda }$ on $X_{\lambda }$. Choose a limit $x_{\lambda }$ of each, and put $x=(x_{\lambda }{)}_{\lambda }$. A basic product neighborhood of $x$ is an intersection of finitely many inverse images of coordinate neighborhoods. Each inverse image belongs to $\mathcal{U}$, so their finite intersection does too. Thus $\mathcal{U}$ converges to $x$, proving compactness of the product. Uniqueness of coordinate limits is not needed, so the factors need not be Hausdorff.

For local compactness, suppose only the factors indexed by a finite set $F$ may be noncompact. At $x\in X$ choose, for each $\lambda \in F$, a compact neighborhood $K_{\lambda }$ of $x_{\lambda }$, with an open neighborhood $U_{\lambda }\subseteq K_{\lambda }$. For other indices put $K_{\lambda }=X_{\lambda }$. The compact product ${\prod }_{\lambda }{K}_{\lambda }$ contains the open product neighborhood restricting just the finitely many coordinates in $F$ to $U_{\lambda }$. Hence every point has a compact neighborhood and the product is locally compact.

The unrestricted assertion is false. Each factor of ${\mathbb{R}}^{\mathbb{N}}$ is locally compact, but its product is not. Any basic neighborhood restricts only finitely many coordinates. Choose a free coordinate $j$; its projection of that neighborhood is all of $\mathbb{R}$. A compact neighborhood containing it would consequently have compact image equal to $\mathbb{R}$, impossible. This shows directly that no point has a compact neighborhood.

For $x$ in a compact Hausdorff space $K$, its quasi-component $Q$ is the intersection of all clopen sets containing $x$. It is closed and therefore compact. Suppose $Q=A\bigsqcup B$ is a separation of a topological space into two nonempty relatively closed sets, with $x\in A$. The sets $A,B$ are compact and hence closed in $K$. Normality of a compact Hausdorff space gives disjoint open neighborhoods $U,V$ of $A,B$.

The compact set $K\setminus (U\cup V)$ is disjoint from $Q$. For each point $y$ of that set choose a clopen $C_y$ containing $x$ but not $y$. Finitely many complements cover $K\setminus (U\cup V)$, so their corresponding intersection $C$ is clopen, contains $Q$, and is contained in $U\cup V$. Now $C\cap U$ is clopen in $C$, hence clopen in $K$, and contains $x$ while missing $B$. This contradicts the definition of $Q$. Thus $Q$ is connected. Any connected set containing $x$ lies in every clopen neighborhood of $x$, so the connected component of $x$ equals $Q$.

For the second assertion, $D$ is closed in the ambient Hausdorff space $X$, and is itself compact Hausdorff. Let $C$ be a connected component of $D$ disjoint from ${\partial }_{X}D$. Choose $p\in C$. By the first assertion $C$ is the quasi-component of $p$ in $D$. For each $y\in {\partial }_{X}D$ choose a clopen subset of $D$ containing $p$ but not $y$. Compactness of ${\partial }_{X}D\subseteq D$ gives a finite intersection $H$ of these sets which contains $C$ and avoids the whole boundary. Thus $H\subseteq {\mathrm{int}}_{X}D$. It is closed in $X$, because it is closed in the closed set $D$, and is open in $X$, because it is relatively open in $D$ and contained in its interior. It is nonempty and proper, contradicting connectedness of $X$. Hence every connected component of $D$ meets its boundary.

Finally suppose a closed subset $A$ of the square has the point-separating clopen property specified in the paper. Every nonempty relatively open subset of the square contains a nontrivial line segment: it contains an interior point of the square and a small ball about that point. If such a segment lay in $A$, a clopen subset of $A$ separating its endpoints would induce a separation of a topological space of the connected segment. This is impossible. Therefore $A$ contains no nonempty relatively open subset of the square. Being closed, it is nowhere dense in the square, and also nowhere dense in ${\mathbb{R}}^2$.

For a connected intersection of a closed filter base, the filter-base property implies the finite intersection property. Compactness, in its closed-set formulation, therefore gives $K=\bigcap \mathcal{F}\ne \varnothing$. It is closed in $X$, hence compact.

Suppose additionally that each base member is connected but $K$ is disconnected. Write $K=A\cup B$ with $A,B$ disjoint nonempty closed subsets of $K$. They are compact, hence closed in the Hausdorff space $X$. Normality of a compact Hausdorff space gives disjoint open sets $U,V$ containing $A,B$, respectively. The compact set $W=X\setminus (U\cup V)$ misses $K$. For every $w\in W$ some $F_w\in \mathcal{F}$ misses $w$, so the open sets $X\setminus F_w$ cover $W$. A finite subcover gives $F_1\cap \cdots \cap F_m\subseteq U\cup V$. Choose $F_0\in \mathcal{F}$ inside this finite intersection. If $W$ is empty, choose any base member instead.

Since $F_0$ contains $K$, it meets both $U$ and $V$; but it lies in their disjoint union. These two relatively open nonempty pieces disconnect $F_0$, a contradiction. Thus the intersection is nonempty and connected. Compactness is used to pass from an infinite exclusion condition to a single base member.

A filter base on a set $X$ is a nonempty family $\mathcal{F}$ of nonempty subsets such that for every $F,G\in \mathcal{F}$ there is $H\in \mathcal{F}$ with $H\subseteq F\cap G$. Repeated application gives a member contained in any specified finite intersection. In particular every finite intersection is nonempty, so the family has the finite intersection property. Its upward closure $\{A\subseteq X:\exists F\in \mathcal{F},F\subseteq A\}$ is the corresponding proper filter.

A metric space is a set $X$ with a function $d:X\times X\to [0,\infty )$ satisfying $d(x,y)=0⟺x=y$, symmetry and the triangle inequality. A subset $U$ is open if every $x\in U$ has some open open ball $B(x,r)$ contained in $U$.

For finitely many open sets $U_1,\dots ,U_k$ and $x$ in their intersection, choose radii $r_j>0$ with $B(x,r_j)\subseteq U_j$. The positive minimum radius gives a ball contained in the intersection. Thus every finite intersection is open; the empty intersection is $X$, also open.

For distinct points put $\delta =d(x_1,x_2)>0$. The balls $B(x_1,\delta /3)$ and $B(x_2,\delta /3)$ are disjoint: a common point would make $d(x_1,x_2)<2\delta /3$ by the triangle inequality. This proves the Hausdorff condition.

A singleton $\{x\}$ is closed because each $y\ne x$ has a ball of radius $d(x,y)/2$ avoiding it, making its complement open. Finite unions of closed sets are closed, as their complements are finite intersections of open sets. Hence every finite subset is closed, including the empty subset.

First assume both factors are nonempty. If the product is compact, its continuous coordinate projections are onto, so each factor is compact by the continuous image of a compact space theorem.

For the converse, let $X,Y$ be compact and let $(W_{\lambda })$ be an open cover of their product. Fix $x\in X$. For every $y\in Y$, choose a basis rectangle $U_{x,y}\times V_{x,y}$ containing $(x,y)$ and lying in a cover member $W_{\lambda (x,y)}$. Compactness of $Y$ gives finitely many $V_{x,y_j}$ covering $Y$. Thenis an open neighbourhood of $x$, and $U_x\times Y$ is covered by the corresponding finitely many $W_{\lambda (x,y_j)}$.

The sets $U_x$ cover $X$. Compactness gives finitely many of them covering $X$, and the finitely many cover members associated to these choices now cover $X\times Y$. Hence the product is compact. This is the finite-product Tychonoff theorem proof, with the finite intersection creating a uniform neighbourhood along each fibre.

As printed, the assertion omits the nonempty hypothesis. If $X=\varnothing$ and $Y=\mathbb{R}$, the product is compact while $Y$ is not. The fully correct version without any nonemptiness assumption is