Convergence in product metric spaces

A sequence in X×Y converges iff each coordinate sequence converges
Convergence in product metric spaces

Let (X,dX)(X,d_X) and (Y,dY)(Y,d_Y) be . On the X×YX\times Y, define the metric d((x,y),(x,y))=max{dX(x,x),dY(y,y)}. d_\infty\bigl((x,y),(x',y')\bigr)=\max\{d_X(x,x'),\,d_Y(y,y')\}. (Any equivalent product metric, such as d1=dX+dYd_1=d_X+d_Y, yields the same notion of .)

Proposition (coordinatewise convergence): A sequence ((xn,yn))((x_n,y_n)) in X×YX\times Y converges to (x,y)(x,y) (with respect to dd_\infty) if and only if xnx in Xandyny in Y. x_n\to x \text{ in } X \quad\text{and}\quad y_n\to y \text{ in } Y. Likewise, ((xn,yn))((x_n,y_n)) is in X×YX\times Y iff (xn)(x_n) is Cauchy in XX and (yn)(y_n) is Cauchy in YY.

This proposition justifies treating product convergence as “simultaneous convergence of components.”

Proof sketch: By definition, d((xn,yn),(x,y))0max{dX(xn,x),dY(yn,y)}0, d_\infty\bigl((x_n,y_n),(x,y)\bigr)\to 0 \quad\Longleftrightarrow\quad \max\{d_X(x_n,x),d_Y(y_n,y)\}\to 0, which holds iff both dX(xn,x)0d_X(x_n,x)\to 0 and dY(yn,y)0d_Y(y_n,y)\to 0. The Cauchy statement is identical with (x,y)(x,y) replaced by (xm,ym)(x_m,y_m).