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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.