Let xn and yn be Cauchy sequences in a metric space Xd Prove

Let (x_n) and (y_n) be Cauchy sequences in a metric space (X,d). Prove that the sequence (d(x_n,y_n)) converges.

Solution

Let ( x_n ) and ( y_n ) be Cauchy sequences in a metric space ( X , d ).

Because X is not necessarily complete, we cannot rely on the convergence of xn and yn. The fact that the sequences are Cauchy means that for all > 0,there exists an N_x( ) such that m; n N_x( ) =>d(xm; xn) < and there exists an N_y( ) such that m; n N_y() => d(y_m; y_n) < 0.

We will use this to show that the sequence d(x_n; y_n) is Cauchy.Then because d(x_n; y_n) is in R and R is complete, it must converge.

First let us make note of two facts which come from repeated application of the triangle inequality:

d(x_n; y_n) d(x_n; x_m) + d(x_m; y_m) + d(y_m; y_n)

d(x_m; y_m) d(x_m; x_n) + d(x_n; y_n) + d(y_n; y_m)

Rearranging these (by isolating the expression d(x_m; y_m) - d(x_n; y_n)) yields

- (d(x_m; x_n) + d(y_m; y_n)) d(x_m; y_m) - d(x_n; y_n) d(x_m; x_n) + d(y_m; y_n)

| d(x_m; y_m) - d(x_n; y_n) | d(x_m; x_n) + d(y_m; y_n)

Now given > 0, choose N( ) > max { N_x( / 2 ); N_y( / 2 ) } .

Then n N( ) => | d(x_m; y_m) - d(x_n; y_n) | d(x_m; x_n) + d(y_m; y_n) < / 2 + / 2 = :

So d(xn; yn) is Cauchy and consequently converges.