Metric Spaces: Proving d is a Metric & Defining \widetilde{} on Cauchy Sequences

[Bachelier]

Question A: Let (xn) and (yn) are two Cauchy sequences in a metric space (X, d), define
d((xn), (yn)) = lim d(xn, yn). It is easy to prove that "d" is a metric on the set of all Cauchy Sequences.

Now let's define $$\widetilde{}$$ on the set of all sequences in a metric space (X, d) by
(xn) $$\widetilde{}$$ (yn) IFF d(xn, yn) tends to 0.
Show that Question A defines a metric on the set $$\hat{X}$$of equivalence classes of Cauchy sequences.


I don't know where to start.

 

[lippyka]

Could you help me?Answer: To show that the definition of d given in Question A defines a metric on the set \hat{X} of equivalence classes of Cauchy sequences, we need to show that it satisfies the three conditions of a metric. 1) Non-negativity: For any two sequences (xn) and (yn), d((xn), (yn)) ≥ 0 because d(xn, yn) ≥ 0 by definition. 2) Symmetry: For any two sequences (xn) and (yn), d((xn), (yn)) = d((yn), (xn)). This follows from the fact that d is a metric on the set of all Cauchy Sequences, which is symmetric by definition. 3) Triangle Inequality: For any three sequences (xn), (yn) and (zn), d((xn), (yn)) + d((yn), (zn)) ≥ d((xn), (zn)). This follows from the fact that d is a metric on the set of all Cauchy Sequences, which satisfies the triangle inequality by definition. Therefore, the definition of d given in Question A defines a metric on the set \hat{X} of equivalence classes of Cauchy sequences, as it satisfies the three conditions of a metric.