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If distinct points $x,y$ of a dynamical system on a compact metric space satisfy $d(T^{n}x,T^{n}y)\to 0$ in one compatible metric, no compatible metric can make $T$ an isometry. On a compact space, all compatible metrics give the same asymptotic-pair property by uniform continuity, whereas an isometry preserves the strictly positive distance between distinct points. In a full shift, a constant sequence and a sequence differing at just one coordinate converge to each other under forward shifts, proving this obstruction directly in the product topology.

A basis of a topology is a family of open sets such that every open set is a union of members of that family. Equivalently, every point of every open set lies in a basis member contained in it. The rectangles $U\times V$ form a basis for the product topology of two spaces. This topological meaning differs from a linear-algebra basis.

The box topology on ${\prod }_{i\in I}{X}_i$ has a basis of a topology consisting of all Cartesian products ${\prod }_i{U}_i$ with every $U_i$ an open set in $X_i$; infinitely many coordinates may be restricted. It contains the product topology and agrees with it for finite $I$.^[ref]

For copies of the usual real line, the box topology is strictly finer when $I=\mathbb{N}$. The box $W={\prod }_{j\ge 1}(-1/2,1/2)$ is not an open set in the product topology: a basic cylinder set containing zero leaves some coordinate $j$ unrestricted, so it contains $e_j\notin W$. This is a construction for every finite coordinate set, not evidence from testing a finite prefix.

Cantor space is the binary sequence space with the product topology of discrete two-point factors. Here the Bernoulli-space synonym refers to this topology, without selecting a particular probability measure. The metric $d(x,y)={\sum }_{j\ge 1}{2}^{-j}|x_j-y_j|$ induces the topology. A diagonal subsequence argument gives compactness. Fixing a finite prefix gives a clopen cylinder set. The map $x\mapsto {\sum }_{j}2x_j{3}^{-j}$ is a homeomorphism to the usual Cantor set.

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 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.

Suppose $X,Y$ are nonempty connected spaces, and $A⊊X$, $B⊊Y$ are proper subsets. Then $C=(X\times Y)\setminus (A\times B)$, with its subspace topology from the product topology, is a connected space. No closed set or open set hypothesis on $A,B$ is needed.

Choose $x_0\in X\setminus A$, $y_0\in Y\setminus B$. The two topological subspaces $\{x_0\}\times Y$ and $X\times \{y_0\}$ are homeomorphic to the factors, intersect at $(x_0,y_0)$, and lie in $C$. Their union is a connected subset $H$. Every $(x,y)\in C$ has $x\notin A$ or $y\notin B$. In the first case its entire vertical copy $\{x\}\times Y$ lies in $C$ and meets $H$; in the second case the horizontal copy $X\times \{y\}$ does so. Thus $C$ is a connected union with a connected hub.

Both properness assumptions are needed for this general result. Taking $X=Y=\mathbb{R}$, $A=X$ and $B=\{0\}$ gives the two disjoint open half-planes $y<0$ and $y>0$, a separation of a topological space of the complement.

For a topological space $Y$ and topological spaces $X_i$, a function $f:Y\to {\prod }_i{X}_i$ is continuous into the product topology exactly when every coordinate function $f_i={\pi }_i\circ f$ is continuous.^[ref]

Proof outline: one direction uses function composition with the continuous projection maps. For the other, the preimage of a basic cylinder set isa finite set intersection of open sets in $Y$. This gives continuity on every basic test. It explains, for example, why $t\mapsto (t,t^2,t^3,\dots )$ from the real line to ${\mathbb{R}}^{\mathbb{N}}$ is continuous in the product topology: each coordinate is a polynomial function. The general coordinate criterion fails for the box topology, where infinitely many preimages can be required at once. For example, $g(t)=(t,t,\dots )$ has continuous coordinate functions, but the preimage of the box ${\prod }_{j\ge 1}(-1/j,1/j)$ is $\{0\}$, which is not an open set in the real line.

For arbitrary topological spaces $X_i$, an index set $I$, a sequence $x^{(n)}\in {\prod }_i{X}_i$, and a point $x$ in that Cartesian product, convergence in a product space to $x$ in the product topology is equivalent to coordinatewise convergence in every factor. No countability or Hausdorff assumption on the factors is needed. For a product space of copies of the real line, regarding each point as a function $I\to \mathbb{R}$ identifies this with pointwise convergence.

Proof outline: the forward implication follows because each projection map is continuous. Conversely, given a basic neighborhood $B(F,U)$ of $x$, each of its finitely many coordinate tests holds for all $n\ge N_i$. Taking $N={\max }_{i\in F}{N}_i$ makes all tests hold simultaneously; if $F$ is empty, any threshold works. This proves eventual membership in every neighborhood, rather than only in a sampled collection of cylinders. The same equivalence holds for nets: a directed set supplies a common upper bound for finitely many thresholds. Sequences alone need not characterize the whole topology of an arbitrary product space.

The displayed metric on a countable Cartesian product of metric spaces combines the bounded metric transform with summable positive weights. It is finite even when individual factor metrics are unbounded. The triangle inequality follows from subadditivity of $t/(1+t)$ and summation. Zero distance forces equality in each coordinate.

Convergence in this metric is exactly coordinatewise convergence. One term is bounded by the sum, so $D(x^{(n)},x)\to 0$ implies each $d_i(x_i^{(n)},x_i)\to 0$. Conversely, choose a finite head whose omitted weight sum is small; convergence in those finitely many coordinates controls the head uniformly, while boundedness of the transformed distances controls the entire tail. Thus it induces the countable product topology. Arbitrary positive summable weights work by the same proof.

A function on a Cartesian product is a cylinder function if it depends on only finitely many coordinates. With finite discrete coordinate spaces it is a continuous function in the product topology. These functions determine finite-dimensional laws and provide a natural domain for local generators of interacting particle systems.

A cylinder set in a product of spaces restricts a finite collection of coordinates $F$ and leaves all other coordinates free. It has the form ${\pi }_F^{-1}(A)$, where ${\pi }_F$ is the coordinate projection onto the finite subproduct. Taking $A$ to be a product of open sets gives the basic open cylinders for the product topology. In a finite-alphabet full shift, specifying exact letters at finitely many coordinates gives a clopen set.

A family of maps with a common domain determines the displayed map into the Cartesian product of their codomains. It is injective exactly when the family separates points. For continuous coordinate maps it is continuous in the product topology. If inverse images of open sets under the coordinates generate the original topology, the inverse from the image is also continuous; thus the map is a topological embedding. A basis of such inverse images, as opposed to a subbasis, is a sufficient stronger hypothesis.

The epigraph is $\{(u,t):E(u)\le t\}$. It is a convex set exactly when $E$ is a convex function, and is a sequentially closed set in the product topology exactly when $E$ is sequentially lower semicontinuous.

The Euclidean topology on ${\mathbb{R}}^n$ is the metric topology induced by the Euclidean norm. A subset is open exactly when each of its points has an open ball contained in it. It agrees with the finite product topology on $n$ copies of the real line: the inequalities $\Vert x{\Vert }_{\infty }\le \Vert x{\Vert }_2\le \sqrt{n}\Vert x{\Vert }_{\infty }$ put a coordinate box inside every sufficiently small ball and a ball inside every sufficiently small coordinate box. The case $n=0$ is the one-point topological space.

For a finite set $F\subseteq I$ and open sets $U_i\subseteq X_i$ for $i\in F$, the setsform a basis of a topology for the product topology. The empty set intersection is $X$. Each basic open set is a cylinder set; all coordinates outside $F$ are free. Arbitrary open sets are unions of these basic tests and need not themselves depend on one fixed finite collection of coordinates.^[ref]

Why finitely many? Any topology making the projection maps continuous must contain their preimages of open sets, and hence finite set intersections of those preimages. Taking arbitrary unions supplies exactly the product topology. Intersecting two basic cylinder sets still restricts only the finite union of their coordinate sets.

If the initial one-spin set $F$ is finite on the square lattice, voter-model duality gives $P({\xi }_t(x)=1)={\sum }_{z\in F}{p}_t(x,z)\to 0$. The transition probabilities tend to zero on this infinite lattice; their Fourier transform also proves the limit of a sequence by dominated convergence. A union bound on any finite set of observation sites then proves weak convergence of probability measures to the all-zero point mass in the product topology.

The two-sided full shift on a finite alphabet $[k]$ consists of all functions $z:\mathbb{Z}\to [k]$, with the product topology and the left shift. It is a compact metric space. A compatible metric isAgreement on increasingly large finite coordinate sets is equivalent to convergence in this topology. The metric above is compatible but is not invariant under the left shift.

Let $Z$ be a nonempty infinite set with the cofinite topology, let $T$ be any topological space, and let $Y$ be Hausdorff. For a continuous map $F:Z\times T\to Y$, every slice $z\mapsto F(z,t)$ is constant. Thus $F(z,t)=g(t)$, where $g(t)=F(z_0,t)$ for any fixed $z_0$. The map $g$ is continuous, since $t\mapsto (z_0,t)$ is continuous into the product topology. It is unique because the coordinate projection ${\pi }_T$ is a surjection. If $F$ is a surjection, so is $g$.

For $T=[0,1]$ and $Y$ homeomorphic to the circle, a surjective $g$ cannot be injective: otherwise the compact-to-Hausdorff continuous bijection theorem would contradict the fact that the circle and interval are not homeomorphic. Hence distinct $a,b\in [0,1]$ satisfy $F(z,a)=F(z,b)$ simultaneously for all $z\in Z$.

For nonempty topological spaces $X,Y$, their product topology is Hausdorff exactly when both factors are Hausdorff spaces. Distinct pairs differ in some coordinate, and disjoint open neighbourhoods in that factor give disjoint open rectangles in the Cartesian product. Conversely, fixing $y_0\in Y$ gives a homeomorphism from $X$ onto the topological subspace $X\times \{y_0\}$; similarly for $Y$. The Hausdorff condition passes to topological subspaces. Nonemptiness matters: an empty set product is Hausdorff regardless of the other factor.