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The product topology on $X\times Y$ consists of arbitrary unions of rectangles $U\times V$ with $U\subseteq X$, $V\subseteq Y$ open. Such rectangles form a basis of a topology: every point lies in one, and intersections of two rectangles are rectangles with open factors.

If $d_X,d_Y$ induce the factor topologies, setDefiniteness and symmetry are immediate. Each factor triangle inequality bounds its distance by the sum of the two relevant product distances; taking the maximum proves the product triangle inequality. A ball of radius $r$ is exactly $B_X(x,r)\times B_Y(y,r)$, so every metric-open set is product-open. Conversely an open rectangle containing $(x,y)$ contains a product ball with radius the minimum of two suitable factor radii. Hence the maximum product metric induces precisely the product topology.

A compact space is one for which every open cover has a finite subcover. Let $\mathcal{U}$ cover $X\times Y$, with both factors compact; if either is empty the result is immediate. For each fixed $x$ and each $y$, choose a rectangle $U_{x,y}\times V_{x,y}$ containing $(x,y)$ and lying in a member of $\mathcal{U}$. The sets $V_{x,y}$ cover $Y$, so finitely many, indexed by $y_1,\dots ,y_{m_x}$, suffice. Their $X$-neighbourhoods have open intersectioncontaining $x$. The finitely many selected members of $\mathcal{U}$ cover all of $U_x\times Y$. Now the sets $U_x$ cover $X$, so select finitely many of these by compactness of $X$. Taking the union of their finite selections from $\mathcal{U}$ gives a finite cover of the whole product. The product of two compact spaces is compact, without requiring the factors to be Hausdorff.

Nonnegativity and symmetry of the maximum product metric follow from those of the two factor metrics. Its value is zero precisely when both coordinates agree. For three points $u=(x,x^{\prime })$, $v=(y,y^{\prime })$, $w=(z,z^{\prime })$, the two triangle inequalities giveAll metric axioms therefore hold.

When the two factors are the same metric space, consider an off-diagonal point $(x,y)$ and set $\delta =d(x,y)>0$. A product-metric ball of radius $\delta /3$ cannot contain any $(z,z)$: such membership would imply $d(x,z),d(y,z)<\delta /3$ and then $d(x,y)<2\delta /3$, a contradiction. The complement of the diagonal is thus open, soThis proves the closed diagonal of a metric space. Equality of the factors here means equality as metric spaces, as intended; merely sharing an underlying set while allowing unrelated metrics would be a different assertion.

For metric spaces $(X_1,d_1)$ and $(X_2,d_2)$, the sum metric on $X_1\times X_2$ is the displayed formula. Nonnegativity, symmetry and the triangle inequality follow by adding the respective inequalities in the two factors, and zero distance forces equality in both coordinates. It induces the finite product topology, as does the maximum product metric, since their distances satisfy $d_{\max }\le d_{\mathrm{sum}}\le 2d_{\max }$. A sequence converges for this metric exactly when both coordinate sequences converge. Two successive extractions of a convergent subsequence therefore show that the product of two sequentially compact spaces is sequentially compact.