A metric space of equivalent Cauchy sequence classes (Z, rho) is defined using a metric of the sequence elements in the space (X,d), where d is from XX to R (real numbers). The metric of the sequence classes is rho = lim d(S, T), where S and T are the elements of the respective sequences. To show that the limit exists, the authors* apply the triangle inequality in the metric space of the sequence elements and state:(adsbygoogle = window.adsbygoogle || []).push({});

(1) /d(S, T) - d(S, T)/ < d(S, S) + d(T, T)

The argument continues from (1) that if the sequences are Cauchy sequences (rt side -> 0), then the left side goes to zero too and lim d(S, T) exists in the complete space which is the range of d.

I am having difficulty obtaining (1) from the triangle inequality, which I take to be:

(2) d(S, T) < d(S, S) + d(T, T)

In (1) and (2), I have omitted the subscripts, but you may surmise that on the left sides of the two inequalities they are mixed; whereas on the right sides they are not.

In my naive effort to see through this derivation, I considered the space RxR and a quadrilateral with corners, S, S, T, T and subscripts m, n. In this simple representation, one can see that (1) is says that the sum of two opposite sides of the quad is greater than the difference of the other two sides; or that the sum of the diagonals is greater than the difference of two opposite sides (depending on how the vertices are ordered.

But this is not a general proof. That is what I should like some help to understand.

Not an exciting problem, I know, but I would appreciate the help greatly.

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* "Into. to Topology", by Gamelin and Greene (Dover, 1999), p 13, ex 7 (b) with hints on p197.

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# A metric space of equivalent sequence classes

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