A product space is meant to remember two kinds of motion at once. A point of $X \times Y$ has an $X$-coordinate and a $Y$-coordinate, so a path may move in the first coordinate, the second coordinate, or both. The first problem is not how to name the set $X \times Y$, but how to measure distance in it without losing either coordinate. If we measure only the first coordinate, different points with the same $X$-coordinate collapse. If we add the two coordinate distances, we get a usable answer, but it behaves differently from the Euclidean formula on $\mathbb R^2$. Product metrics are the systematic ways to combine coordinate distances into a genuine metric on a Cartesian product. The guiding example is familiar: the plane is a product $\mathbb R \times \mathbb R$, and the usual Euclidean distance is built from the two coordinate distances by the Pythagorean rule. But topology is less rigid than geometry. For convergence, continuity, and compactness, many different formulas on $X \times Y$ give the same open sets. Product metrics separate two questions that are often blurred together: which formula measures distance, and which topology does that formula generate? [example: A Bad Distance on a Product] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, and suppose $Y$ has at least two distinct points. Choose distinct points $y_1,y_2\in Y$, and choose any $x\in X$. Define \begin{align*} \rho((x_1,y_1),(x_2,y_2))=d_X(x_1,x_2). \end{align*} Then the two points $(x,y_1)$ and $(x,y_2)$ in $X\times Y$ are distinct because their second coordinates are distinct. However, substituting these points into the formula for $\rho$ gives \begin{align*} \rho((x,y_1),(x,y_2))=d_X(x,x). \end{align*} Since $d_X$ is a metric on $X$, the distance from a point to itself is $0$, so \begin{align*} d_X(x,x)=0. \end{align*} Therefore \begin{align*} \rho((x,y_1),(x,y_2))=0. \end{align*} Thus $\rho$ assigns distance $0$ to two distinct points of $X\times Y$, so it fails the identity of indiscernibles and is not a metric. The example shows that a product distance must detect motion in every coordinate. [/example] The failure is small, but it is decisive. A metric is not only a numerical gadget; it determines which sequences converge, which functions are continuous, and which subsets are open. A product metric must keep coordinate information while still interacting well with the product structure. ## Definition Before choosing a formula, we need the ambient object: a Cartesian product equipped with coordinate metric data. The construction begins with two metric spaces and asks for a distance between ordered pairs that responds to motion in both coordinates. [definition: Product Metric] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. A product metric on $X\times Y$ is a function \begin{align*} D:(X\times Y)\times(X\times Y)&\to [0,\infty) \end{align*} which is a metric on $X\times Y$ and is defined from the coordinate metrics $d_X$ and $d_Y$ by a specified product metric formula. [/definition] The associated metric space $(X\times Y,D)$ is often called a product metric space. The phrase is intentionally flexible: there are several standard formulas. The most important family is the $p$-family, which includes the Euclidean-style construction when $p=2$ and the taxicab-style construction when $p=1$. ## Standard Product Metric Formulas For finite $p$, the idea is to measure the two coordinate distances and then apply the usual $\ell^p$ norm to the pair of nonnegative numbers. This imports the geometry of $\mathbb R^2$ into the abstract setting of arbitrary metric spaces. [definition: $p$-Product Metric] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, and let $1 \le p < \infty$. The $p$-product metric on $X \times Y$ is the function $d_p:(X\times Y)\times(X\times Y)\to [0,\infty)$ defined by \begin{align*} d_p((x_1,y_1),(x_2,y_2))=\bigl(d_X(x_1,x_2)^p+d_Y(y_1,y_2)^p\bigr)^{1/p}. \end{align*} [/definition] The finite $p$-formula is excellent for estimates that combine coordinate errors by summing powers. It still leaves a useful limiting question: what if we care only about the worst coordinate error, so that a product ball forces both coordinates to stay inside prescribed balls of the same radius? [definition: Supremum Product Metric] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. The supremum product metric on $X \times Y$ is the function $d_\infty:(X\times Y)\times(X\times Y)\to [0,\infty)$ defined by \begin{align*} d_\infty((x_1,y_1),(x_2,y_2))=\max\{d_X(x_1,x_2),d_Y(y_1,y_2)\}. \end{align*} [/definition] Before these formulas can support convergence or continuity arguments, they must pass the metric axioms. Positivity and symmetry are inherited from the coordinate metrics, but the main question is whether combining coordinate distances by a $p$-formula still produces a valid triangle inequality on the product. [quotetheorem:8620] This theorem justifies the terminology. From this point onward, the formulas are not candidates or analogies; they are genuine distances on the product. The next issue is how these distances compare with each other, since different choices of $p$ measure the same coordinate errors with different emphasis. [example: Familiar Product Metrics on $\mathbb R^2$] Take $X=Y=\mathbb R$ with the usual metric $d(s,t)=|s-t|$, and let $a=(0,0)$ and $b=(3,4)$. The coordinate distances are \begin{align*} d(0,3)=|0-3|=|-3|=3. \end{align*} and \begin{align*} d(0,4)=|0-4|=|-4|=4. \end{align*} For the taxicab product metric, \begin{align*} d_1(a,b)=d(0,3)+d(0,4)=3+4=7. \end{align*} For the Euclidean product metric, \begin{align*} d_2(a,b)=\bigl(d(0,3)^2+d(0,4)^2\bigr)^{1/2}=(3^2+4^2)^{1/2}. \end{align*} Since $3^2=9$ and $4^2=16$, this becomes \begin{align*} d_2(a,b)=(9+16)^{1/2}=25^{1/2}=5. \end{align*} For the supremum product metric, \begin{align*} d_\infty(a,b)=\max\{d(0,3),d(0,4)\}=\max\{3,4\}=4. \end{align*} Thus the same pair of points has distances $7$, $5$, and $4$ under $d_1$, $d_2$, and $d_\infty$, respectively. Both coordinate distances are nonzero, so all three metrics detect motion in both coordinates; by *[Product Metrics Induce the Product Topology](/theorems/8622)*, all three generate the usual topology on $\mathbb R^2$. [/example] ## Coordinate Control and Convergence ### Comparing Product Metrics A product metric is useful because it translates statements about ordered pairs into statements about coordinates. This is the main reason product metrics appear throughout analysis: convergence in a product should mean convergence in each coordinate, and continuity into or out of a product should be testable coordinate by coordinate. The basic quantitative comparison says that the supremum metric is bounded above and below by any finite $p$-product metric up to constants. These inequalities are the bridge between the different formulas, because they turn estimates for one product metric into estimates for another. [quotetheorem:8621] The constants say that a small $d_p$-ball contains a small $d_\infty$-ball and is contained in a slightly larger $d_\infty$-ball. Thus the choice of $p$ changes numerical geometry but not local nearness. ### Coordinatewise Limits Convergence is the first place where this comparison pays off. A sequence in $X \times Y$ should converge exactly when each coordinate converges; otherwise a product limit would either ignore a coordinate or demand more than coordinate convergence can provide. [quotetheorem:956] This theorem is the operational meaning of the product metric. When a sequence of pairs converges, neither coordinate can hide from the metric; when both coordinates settle down, the pair settles down. [example: A Sequence with One Bad Coordinate] In $\mathbb R\times \mathbb R$ with the Euclidean product metric, define \begin{align*} z_n=\left(\frac{1}{n},(-1)^n\right). \end{align*} The first coordinate tends to $0$ because, for every $n\ge 1$, \begin{align*} \left|\frac{1}{n}-0\right|=\frac{1}{n}, \end{align*} and $\frac{1}{n}\to 0$ in $\mathbb R$. We show that the full sequence $(z_n)$ does not converge in $\mathbb R^2$. Suppose instead that $z_n\to(a,b)$. Then the even subsequence also converges to $(a,b)$, so \begin{align*} d_2(z_{2k},(a,b))=\left(\left(\frac{1}{2k}-a\right)^2+(1-b)^2\right)^{1/2}\to 0. \end{align*} Since $(1-b)^2\le \left(\frac{1}{2k}-a\right)^2+(1-b)^2$, taking square roots gives \begin{align*} |1-b|\le d_2(z_{2k},(a,b)). \end{align*} Therefore $|1-b|=0$, so $b=1$. The odd subsequence must also converge to $(a,b)$, and for it we have \begin{align*} d_2(z_{2k-1},(a,b))=\left(\left(\frac{1}{2k-1}-a\right)^2+(-1-b)^2\right)^{1/2}\to 0. \end{align*} Again, \begin{align*} |-1-b|\le d_2(z_{2k-1},(a,b)), \end{align*} so $|-1-b|=0$, hence $b=-1$. This contradicts $b=1$. Thus $(z_n)$ does not converge in the product metric, because the second coordinate keeps alternating between two separated values. [/example] ## Product Topology from Balls ### Basic Product Neighbourhoods Metric spaces carry topologies through open balls. If a product metric is the right metric on $X \times Y$, then its open sets should match the topology generated by products of open sets in $X$ and $Y$. The supremum metric makes this relationship visible because its balls are coordinate boxes. To state the comparison cleanly, we first isolate the topological structure that product metrics are meant to recover. This definition answers the question: what should an open neighbourhood of a pair be allowed to control independently in each coordinate? [definition: Product Topology] Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces. The [product topology](/page/Product%20Topology) on $X \times Y$ is the topology generated by all sets of the form $U \times V$, where $U \in \tau_X$ and $V \in \tau_Y$. [/definition] The definition says that basic neighbourhoods in a product prescribe independent neighbourhoods in each coordinate. In metric spaces, these coordinate neighbourhoods may be chosen as balls, so the central compatibility question is whether the metric balls generate exactly this topology. [quotetheorem:8622] This is the key structural result. It explains why analysts often choose whichever product metric is easiest for the task: $d_2$ for Euclidean geometry, $d_1$ for additive estimates, and $d_\infty$ for open-set arguments. ### Shape of Balls The next example shows how the supremum metric turns topological neighbourhoods into rectangles. This is the geometric picture behind the compatibility theorem, and it is the reason $d_\infty$ is often the simplest metric for product topology arguments. [example: Supremum Balls are Rectangles] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, let $(x,y)\in X\times Y$, and let $r>0$. We show that the open ball for the supremum product metric is exactly the product of the coordinate balls: \begin{align*} B_{d_\infty}((x,y),r)=B_X(x,r)\times B_Y(y,r). \end{align*} First, take $(u,v)\in B_{d_\infty}((x,y),r)$. By the definition of an open ball, this means \begin{align*} d_\infty((x,y),(u,v))<r. \end{align*} Using the definition of $d_\infty$, this is \begin{align*} \max\{d_X(x,u),d_Y(y,v)\}<r. \end{align*} Since each entry of a maximum is at most the maximum, we have \begin{align*} d_X(x,u)\le \max\{d_X(x,u),d_Y(y,v)\}<r. \end{align*} Similarly, \begin{align*} d_Y(y,v)\le \max\{d_X(x,u),d_Y(y,v)\}<r. \end{align*} Thus $u\in B_X(x,r)$ and $v\in B_Y(y,r)$, so $(u,v)\in B_X(x,r)\times B_Y(y,r)$. Conversely, take $(u,v)\in B_X(x,r)\times B_Y(y,r)$. Then $u\in B_X(x,r)$ and $v\in B_Y(y,r)$, so \begin{align*} d_X(x,u)<r. \end{align*} and \begin{align*} d_Y(y,v)<r. \end{align*} The maximum of two [real numbers](/page/Real%20Numbers) that are both less than $r$ is also less than $r$, hence \begin{align*} \max\{d_X(x,u),d_Y(y,v)\}<r. \end{align*} By the definition of $d_\infty$, this says \begin{align*} d_\infty((x,y),(u,v))<r. \end{align*} Therefore $(u,v)\in B_{d_\infty}((x,y),r)$. The two inclusions prove the equality, so supremum balls are precisely rectangular products of coordinate balls with the same radius. [/example] For finite $p$, the balls are rounded rather than rectangular, but the comparison theorem places them between rectangular balls. This is why they generate the same open sets. ## Continuity and Uniform Estimates ### Projection Maps Product metrics are not only about convergence of points. They organize maps whose inputs or outputs have several components. A map into a product is continuous precisely when its coordinate maps are continuous. A map out of a product is continuous when it responds continuously to simultaneous changes in both coordinates. We begin by naming the coordinate maps that extract information from the product. They are simple, but they are the mechanism through which products interact with the rest of topology and analysis. [definition: Coordinate Projection] Let $X$ and $Y$ be sets. The coordinate projections from $X \times Y$ are the maps $\pi_X:X\times Y\to X$ and $\pi_Y:X\times Y\to Y$ defined by \begin{align*} \pi_X(x,y)=x \end{align*} and \begin{align*} \pi_Y(x,y)=y. \end{align*} [/definition] Metric product spaces are meant to let us read off coordinate information without amplifying errors. A map $f:(M,d_M)\to(N,d_N)$ is called $1$-Lipschitz if $d_N(f(a),f(b))\le d_M(a,b)$ for all $a,b\in M$. The next issue is whether the product metric actually gives the coordinate projections this stability. If extracting one coordinate could enlarge distances, then product convergence would not reliably imply coordinatewise convergence. The projection estimate rules this out by showing that each coordinate map is $1$-Lipschitz for the product metric. [quotetheorem:8623] This estimate is stronger than continuity. It says that a small error in the product controls the error in each coordinate with no loss of constants. ### Maps Into Products Coordinate projections let us disassemble maps out of a product. For maps into a product, the complementary question is how to assemble coordinate maps into one product-valued map without creating a new continuity problem. [quotetheorem:962] The theorem is often used without naming it. Whenever a curve $t\mapsto (x(t),y(t))$ is declared continuous, the actual check usually happens coordinate by coordinate. [example: A Continuous Curve into a Product] Let $f:[0,1]\to\mathbb R$ and $g:[0,1]\to\mathbb R$ be the coordinate maps \begin{align*} f(t)=t^2 \end{align*} and \begin{align*} g(t)=\sin(2\pi t). \end{align*} Then, for every $t\in[0,1]$, \begin{align*} F(t)=(f(t),g(t)). \end{align*} The map $f$ is continuous because $t\mapsto t^2$ is a polynomial function. The map $g$ is continuous because $t\mapsto 2\pi t$ is continuous, $\sin$ is continuous on $\mathbb R$, and compositions of continuous functions are continuous. By *Continuity into a Product*, the map $F:[0,1]\to\mathbb R\times\mathbb R$ is continuous for every product metric $d_p$ with $1\le p\le\infty$. In particular, $F$ is continuous for $d_1$, $d_2$, and $d_\infty$. For two parameters $s,t\in[0,1]$, the three product metrics measure the same displacement by different formulas. For $d_1$, \begin{align*} d_1(F(s),F(t))=|s^2-t^2|+|\sin(2\pi s)-\sin(2\pi t)|. \end{align*} For $d_2$, \begin{align*} d_2(F(s),F(t))=\bigl(|s^2-t^2|^2+|\sin(2\pi s)-\sin(2\pi t)|^2\bigr)^{1/2}. \end{align*} For $d_\infty$, \begin{align*} d_\infty(F(s),F(t))=\max\{|s^2-t^2|,|\sin(2\pi s)-\sin(2\pi t)|\}. \end{align*} Thus the numerical length of a displacement depends on the chosen product metric, while the continuity of the curve is determined coordinate by coordinate. [/example] ### Uniform Estimates [Uniform convergence](/page/Uniform%20Convergence) and Lipschitz estimates are often proved coordinate by coordinate. Product metrics provide the bookkeeping rule that combines those coordinate estimates into a single bound for the product-valued map. [example: Combining Error Bounds] Suppose $f,g:Z\to \mathbb R$ and $f_n,g_n:Z\to \mathbb R$ satisfy the uniform coordinate bounds \begin{align*} \sup_{z\in Z}|f_n(z)-f(z)|\le a_n \end{align*} and \begin{align*} \sup_{z\in Z}|g_n(z)-g(z)|\le b_n. \end{align*} For each $z\in Z$, define $F_n(z)=(f_n(z),g_n(z))$ and $F(z)=(f(z),g(z))$. By the definition of the Euclidean product metric on $\mathbb R\times\mathbb R$, \begin{align*} d_2(F_n(z),F(z))=\bigl(|f_n(z)-f(z)|^2+|g_n(z)-g(z)|^2\bigr)^{1/2}. \end{align*} Since $|f_n(z)-f(z)|\le \sup_{w\in Z}|f_n(w)-f(w)|\le a_n$, we have \begin{align*} |f_n(z)-f(z)|^2\le a_n^2. \end{align*} Similarly, \begin{align*} |g_n(z)-g(z)|^2\le b_n^2. \end{align*} Adding these two inequalities gives \begin{align*} |f_n(z)-f(z)|^2+|g_n(z)-g(z)|^2\le a_n^2+b_n^2. \end{align*} The square-root function is increasing on $[0,\infty)$, so \begin{align*} d_2(F_n(z),F(z))\le (a_n^2+b_n^2)^{1/2}. \end{align*} This bound holds for every $z\in Z$, hence \begin{align*} \sup_{z\in Z}d_2(F_n(z),F(z))\le (a_n^2+b_n^2)^{1/2}. \end{align*} Therefore, if $a_n\to 0$ and $b_n\to 0$, then $(a_n^2+b_n^2)^{1/2}\to 0$, so the product-valued maps $F_n$ converge uniformly to $F$ in the $d_2$ product metric. [/example] ## Completeness, Compactness, and Separability ### Complete Products Many analytic arguments rely on structural properties of the ambient space: Cauchy sequences should converge, compactness should pass to limits, and countable dense sets should allow approximation. Product metrics preserve these properties in the expected finite-product setting. Completeness is the most direct. A [Cauchy sequence](/page/Cauchy%20Sequence) of pairs is exactly a pair of Cauchy coordinate sequences, so we first name the property whose preservation will matter for approximation and existence arguments. [definition: Complete Metric Space] A metric space $(X,d)$ is complete if every Cauchy sequence in $X$ converges to a point of $X$. [/definition] Completeness matters because it turns approximation into existence. If a construction produces a Cauchy sequence of pairs, the theorem below allows us to finish the argument by checking the two coordinates separately. [quotetheorem:8624] For completeness, the product has the structural property exactly when the factors have it. Compactness asks for a different kind of control, because it concerns all open covers or, in metric spaces, all sequences. ### Compact Products Compactness is stronger than boundedness and closedness in general metric spaces, and it is the property that makes subsequence arguments possible. To discuss product compactness without relying on a special Euclidean criterion, we use the open-cover definition. [definition: Compact Metric Space] A metric space $(X,d)$ is compact if every [open cover](/page/Open%20Cover) of $X$ has a finite subcover. [/definition] A finite product of compact spaces should still have enough finiteness to control all coordinates simultaneously. The product theorem below is the compactness version of the coordinate philosophy: finite coordinate compactness combines into compactness of the pair. [quotetheorem:1071] The finite-product hypothesis is important for the metric formula on this page. Infinite products require new choices: a naive supremum metric may be infinite, while weighted sums introduce their own scale choices. [example: Compact Square from Compact Intervals] Let $X=Y=[0,1]$ with the usual metric, and equip $[0,1]\times[0,1]$ with the Euclidean product metric $d_2$. Since $[0,1]$ is compact, *Compactness of Finite Metric Products* gives that $[0,1]\times[0,1]$ is compact under $d_2$. The subsequence argument makes the compactness concrete. Let $((x_n,y_n))_{n=1}^\infty$ be a sequence in $[0,1]\times[0,1]$. The first-coordinate sequence $(x_n)$ lies in the compact interval $[0,1]$, so by *[Sequential Compactness](/page/Sequential%20Compactness) of Compact Metric Spaces* there are indices $n_1<n_2<\cdots$ and a point $x\in[0,1]$ such that \begin{align*} x_{n_k}\to x. \end{align*} Now consider the second coordinates along this subsequence, namely $(y_{n_k})_{k=1}^\infty$. Again this is a sequence in $[0,1]$, so there are indices $k_1<k_2<\cdots$ and a point $y\in[0,1]$ such that \begin{align*} y_{n_{k_j}}\to y. \end{align*} Since a subsequence of a convergent sequence has the same limit, we also have \begin{align*} x_{n_{k_j}}\to x. \end{align*} We now verify convergence in the product metric. For each $j$, \begin{align*} d_2((x_{n_{k_j}},y_{n_{k_j}}),(x,y))=\bigl(|x_{n_{k_j}}-x|^2+|y_{n_{k_j}}-y|^2\bigr)^{1/2}. \end{align*} Let $\varepsilon>0$. Since $x_{n_{k_j}}\to x$, there is $J_X$ such that $j\ge J_X$ implies \begin{align*} |x_{n_{k_j}}-x|<\frac{\varepsilon}{\sqrt{2}}. \end{align*} Since $y_{n_{k_j}}\to y$, there is $J_Y$ such that $j\ge J_Y$ implies \begin{align*} |y_{n_{k_j}}-y|<\frac{\varepsilon}{\sqrt{2}}. \end{align*} For $j\ge \max\{J_X,J_Y\}$, squaring these two inequalities gives \begin{align*} |x_{n_{k_j}}-x|^2<\frac{\varepsilon^2}{2}. \end{align*} and \begin{align*} |y_{n_{k_j}}-y|^2<\frac{\varepsilon^2}{2}. \end{align*} Adding them gives \begin{align*} |x_{n_{k_j}}-x|^2+|y_{n_{k_j}}-y|^2<\varepsilon^2. \end{align*} Because the square-root function is increasing on $[0,\infty)$, \begin{align*} d_2((x_{n_{k_j}},y_{n_{k_j}}),(x,y))<\varepsilon. \end{align*} Thus the subsequence $((x_{n_{k_j}},y_{n_{k_j}}))$ converges to $(x,y)$ in the product metric, so compactness of the square is exactly the ability to extract coordinatewise convergent subsequences and then combine their limits. [/example] ### Separable Products Approximation arguments often need a countable supply of test points. In a product, the natural candidates are grid points formed by choosing one dense coordinate from each factor. [definition: Separable Metric Space] A metric space $(X,d)$ is separable if there exists a countable subset $A\subset X$ whose closure is $X$. [/definition] The product version explains why rational grids approximate Euclidean spaces and many function-space constructions built from product spaces. For finite products, choosing countable dense sets in each coordinate gives a countable dense grid. For countably many metric factors, one uses the same idea with an extra restriction: fix base points in all coordinates, and allow only finitely many coordinates to vary through their countable dense sets. This produces only countably many test points, and it meets every basic product neighborhood because such a neighborhood restricts only finitely many coordinates. The metric hypothesis matters here. Separability alone is not preserved by arbitrary countable products of topological spaces without additional assumptions such as metrizability or second countability. In the metric setting used on this page, separable factors have countable local data, so finite-coordinate grids give the intended approximation principle. ## Geometry of Different Product Metrics ### Additive Geometry Although the topology does not depend on $p$, the geometry does. Balls have different shapes, Lipschitz constants change, and path lengths respond differently to coordinate motion. This distinction is important in analysis, where topology may control convergence while geometry controls estimates. The $d_1$ metric is additive in coordinate distances. It deserves its own name because many estimates in analysis are proved by splitting an expression into two pieces and adding the resulting bounds. [definition: Taxicab Product Metric] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. The taxicab product metric on $X\times Y$ is the function $d_1:(X\times Y)\times(X\times Y)\to [0,\infty)$ defined by \begin{align*} d_1((x_1,y_1),(x_2,y_2))=d_X(x_1,x_2)+d_Y(y_1,y_2). \end{align*} [/definition] Additive estimates appear constantly: split a difference into two pieces, bound each piece, and add the bounds. The taxicab metric records exactly that structure. [example: Splitting an Estimate] Let $f:X\to \mathbb R$ and $g:Y\to \mathbb R$ be $L_f$- and $L_g$-Lipschitz functions, and define $H:X\times Y\to\mathbb R$ by \begin{align*} H(x,y)=f(x)+g(y). \end{align*} For two points $(x_1,y_1),(x_2,y_2)\in X\times Y$, we compute \begin{align*} H(x_1,y_1)-H(x_2,y_2)=f(x_1)+g(y_1)-f(x_2)-g(y_2). \end{align*} Regrouping the terms gives \begin{align*} H(x_1,y_1)-H(x_2,y_2)=\bigl(f(x_1)-f(x_2)\bigr)+\bigl(g(y_1)-g(y_2)\bigr). \end{align*} By the triangle inequality in $\mathbb R$, \begin{align*} |H(x_1,y_1)-H(x_2,y_2)|\le |f(x_1)-f(x_2)|+|g(y_1)-g(y_2)|. \end{align*} Since $f$ is $L_f$-Lipschitz and $g$ is $L_g$-Lipschitz, \begin{align*} |f(x_1)-f(x_2)|\le L_f d_X(x_1,x_2). \end{align*} and \begin{align*} |g(y_1)-g(y_2)|\le L_g d_Y(y_1,y_2). \end{align*} Substituting these two bounds yields \begin{align*} |H(x_1,y_1)-H(x_2,y_2)|\le L_f d_X(x_1,x_2)+L_g d_Y(y_1,y_2). \end{align*} Let $L=\max\{L_f,L_g\}$. Then $L_f\le L$ and $L_g\le L$, so \begin{align*} L_f d_X(x_1,x_2)\le L d_X(x_1,x_2). \end{align*} and \begin{align*} L_g d_Y(y_1,y_2)\le L d_Y(y_1,y_2). \end{align*} Adding the inequalities gives \begin{align*} L_f d_X(x_1,x_2)+L_g d_Y(y_1,y_2)\le L\bigl(d_X(x_1,x_2)+d_Y(y_1,y_2)\bigr). \end{align*} By the definition of the taxicab product metric, \begin{align*} d_1((x_1,y_1),(x_2,y_2))=d_X(x_1,x_2)+d_Y(y_1,y_2). \end{align*} Therefore \begin{align*} |H(x_1,y_1)-H(x_2,y_2)|\le Ld_1((x_1,y_1),(x_2,y_2)). \end{align*} Thus $H$ is $L$-Lipschitz from $(X\times Y,d_1)$ to $\mathbb R$: the taxicab metric records exactly the additive split of the two coordinate estimates. [/example] ### Euclidean Geometry Additive control is not the only natural geometry. When the factor distances come from Euclidean norms or inner-product geometry, squared coordinate distances combine through the Pythagorean rule, and the product metric should preserve that structure. [definition: Euclidean Product Metric] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. The Euclidean product metric on $X\times Y$ is the function $d_2:(X\times Y)\times(X\times Y)\to [0,\infty)$ defined by \begin{align*} d_2((x_1,y_1),(x_2,y_2))=\bigl(d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2\bigr)^{1/2}. \end{align*} [/definition] The definition is abstract, but it is designed to agree with the standard Euclidean distance when the factors are Euclidean spaces. This agreement is the anchor that connects the metric-space construction with multivariable calculus. [quotetheorem:8625] The supremum metric, by contrast, is geometrically square in finite-dimensional Euclidean pictures. Its advantage is not rotational symmetry but coordinate control. [example: Different Unit Balls] In $\mathbb R^2=\mathbb R\times\mathbb R$, centered at $(0,0)$, the coordinate distances from $(0,0)$ to $(x,y)$ are $|x-0|=|x|$ and $|y-0|=|y|$. Therefore the $d_1$-unit ball is \begin{align*} B_{d_1}((0,0),1)=\{(x,y): |x|+|y|<1\}. \end{align*} This is the open diamond whose boundary is described by $|x|+|y|=1$. For the Euclidean product metric, \begin{align*} d_2((0,0),(x,y))=\bigl(|x|^2+|y|^2\bigr)^{1/2}. \end{align*} Since $|x|^2=x^2$ and $|y|^2=y^2$, this becomes \begin{align*} d_2((0,0),(x,y))=\bigl(x^2+y^2\bigr)^{1/2}. \end{align*} Thus $d_2((0,0),(x,y))<1$ holds exactly when \begin{align*} \bigl(x^2+y^2\bigr)^{1/2}<1. \end{align*} Both sides are nonnegative, so squaring preserves the inequality and gives \begin{align*} x^2+y^2<1. \end{align*} Hence the $d_2$-unit ball is \begin{align*} B_{d_2}((0,0),1)=\{(x,y): x^2+y^2<1\}, \end{align*} the usual open disk. For the supremum product metric, \begin{align*} d_\infty((0,0),(x,y))=\max\{|x|,|y|\}. \end{align*} So the $d_\infty$-unit ball is \begin{align*} B_{d_\infty}((0,0),1)=\{(x,y): \max\{|x|,|y|\}<1\}. \end{align*} The inequality $\max\{|x|,|y|\}<1$ holds exactly when $|x|<1$ and $|y|<1$, which is exactly the condition $x\in(-1,1)$ and $y\in(-1,1)$. Therefore \begin{align*} B_{d_\infty}((0,0),1)=(-1,1)\times(-1,1). \end{align*} The containment relations can also be seen numerically. If $|x|+|y|<1$, then $|x|\le |x|+|y|<1$ and $|y|\le |x|+|y|<1$, so \begin{align*} \max\{|x|,|y|\}<1. \end{align*} Thus \begin{align*} B_{d_1}((0,0),1)\subseteq B_{d_\infty}((0,0),1). \end{align*} Conversely, if $\max\{|x|,|y|\}<r$, then $|x|<r$ and $|y|<r$, so \begin{align*} |x|+|y|<2r. \end{align*} Choosing $r=\frac12$ gives \begin{align*} B_{d_\infty}\left((0,0),\frac12\right)\subseteq B_{d_1}((0,0),1). \end{align*} Similarly, if $\max\{|x|,|y|\}<r$, then $x^2<r^2$ and $y^2<r^2$, hence \begin{align*} x^2+y^2<2r^2. \end{align*} Choosing $r=\frac{1}{\sqrt2}$ gives \begin{align*} B_{d_\infty}\left((0,0),\frac{1}{\sqrt2}\right)\subseteq B_{d_2}((0,0),1). \end{align*} Also, if $x^2+y^2<1$, then $x^2\le x^2+y^2<1$ and $y^2\le x^2+y^2<1$, so $|x|<1$ and $|y|<1$, giving \begin{align*} B_{d_2}((0,0),1)\subseteq B_{d_\infty}((0,0),1). \end{align*} Thus the unit balls are shaped like a diamond, a disk, and a square, but each metric ball around the origin contains a smaller ball for the others, so the three metrics determine the same local neighbourhoods of the origin. [/example] ## Finite and Infinite Products ### Finite Products The definition graph for product metrics usually starts with two factors, but the same construction works for any finite list. This is the setting used in multivariable analysis, finite-dimensional normed spaces, and systems with finitely many components. To avoid hiding the finite-dimensional pattern, we state the general finite version explicitly. The finite formula answers the same question as before: how should a single distance combine finitely many coordinate distances? [definition: Finite $p$-Product Metric] Let $(X_i,d_i)$ be metric spaces for $i\in\{1,\dots,n\}$, where $n\in\mathbb N$. For $1\le p<\infty$, the finite $p$-product metric on $X_1\times\cdots\times X_n$ is \begin{align*} d_p:(X_1\times\cdots\times X_n)^2\to [0,\infty) \end{align*} defined by \begin{align*} d_p((x_1,\dots,x_n),(y_1,\dots,y_n))=\left(\sum_{i=1}^n d_i(x_i,y_i)^p\right)^{1/p}. \end{align*} The finite supremum product metric is \begin{align*} d_\infty:(X_1\times\cdots\times X_n)^2\to [0,\infty) \end{align*} defined by \begin{align*} d_\infty((x_1,\dots,x_n),(y_1,\dots,y_n))=\max_{1\le i\le n} d_i(x_i,y_i). \end{align*} [/definition] Finite products behave like the two-factor product in all the basic ways: coordinatewise convergence, product topology, preservation of completeness, and preservation of compactness. The proofs repeat the two-factor arguments with sums or maxima over finitely many coordinates. ### Why Infinite Products Need Care Infinite products are subtler because infinitely many coordinate distances must be combined into a finite number. This creates a new problem: there is no single unweighted formula that works in all bounded and unbounded situations. [example: A Naive Infinite Supremum Can Be Infinite] For each $i\in\mathbb N$, let $X_i=\mathbb R$ with the usual metric. On the infinite product $\prod_{i=1}^\infty X_i=\mathbb R^{\mathbb N}$, consider the formula \begin{align*} \rho((x_i),(y_i))=\sup_{i\in\mathbb N}|x_i-y_i|. \end{align*} This formula does not always return a real number. Define two points of $\mathbb R^{\mathbb N}$ by \begin{align*} x_i=0 \end{align*} for every $i\in\mathbb N$, and \begin{align*} y_i=i \end{align*} for every $i\in\mathbb N$. Then, for each $i\in\mathbb N$, \begin{align*} |x_i-y_i|=|0-i|=|-i|=i. \end{align*} Therefore the set of coordinate distances is \begin{align*} \{|x_i-y_i|:i\in\mathbb N\}=\{i:i\in\mathbb N\}. \end{align*} For every real number $M$, choose an integer $i>M$; then $i\in\{j:j\in\mathbb N\}$ and $i>M$. Hence the set $\{i:i\in\mathbb N\}$ is unbounded above in $\mathbb R$, so \begin{align*} \sup_{i\in\mathbb N}|x_i-y_i|=\sup_{i\in\mathbb N}i=\infty. \end{align*} Thus $\rho((x_i),(y_i))=\infty$ for this pair of points, so $\rho$ is not a metric on the whole infinite product $\mathbb R^{\mathbb N}$, since a metric must take finite values in $[0,\infty)$. [/example] This example explains why infinite products usually require bounded coordinate metrics, weights, or a separate topological construction. The finite product metric is the clean core; infinite products belong to a broader theory of product topologies and metrization. ## Beyond and Connected Topics Product metrics are one of the first places where metric and topological thinking meet. The metric formula gives numerical estimates, while the induced topology gives the coordinatewise behaviour needed for continuity and compactness. A natural continuation is [Metric Space](/page/Metric%20Space), where product metrics become part of the larger study of convergence, completeness, and compactness. In topology, the product metric motivates the general product topology, which no longer depends on a metric formula. This is the setting where finite products, arbitrary products, projections, and universal properties are treated uniformly. The Androma notes [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology) are the natural course-level continuation for this viewpoint. In analysis, product metrics underlie multivariable continuity and spaces of vector-valued functions. For instance, a function into $\mathbb R^m\times\mathbb R^n$ is continuous exactly when its coordinate components are continuous, and uniform estimates often combine coordinate errors with $d_1$, $d_2$, or $d_\infty$ depending on the estimate. In differential geometry, products of manifolds inherit product topologies and often product Riemannian metrics. There the distance induced by a Riemannian metric is related to, but not identical in spirit with, the elementary product metrics of metric-space theory. The notes [Cambridge III Differential Geometry](/page/Cambridge%20III%20Differential%20Geometry) and [Cambridge III Riemannian Geometry](/page/Cambridge%20III%20Riemannian%20Geometry) develop the geometric side. The finite product construction also prepares the reader for norm equivalence in finite-dimensional vector spaces. The comparison between $d_p$ and $d_\infty$ is the metric-space shadow of the fact that all norms on a finite-dimensional real [vector space](/page/Vector%20Space) induce the same topology. ## References Androma, [Metric Space](/page/Metric%20Space). Androma, [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes). Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge III Differential Geometry](/page/Cambridge%20III%20Differential%20Geometry). Androma, [Cambridge III Riemannian Geometry](/page/Cambridge%20III%20Riemannian%20Geometry). Munkres, *Topology* (2000). Burago, Burago, and Ivanov, *A Course in Metric Geometry* (2001). Willard, *General Topology* (1970).