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## Homework Statement

I need a example of a continuous function f:(X, d) -> Y(Y, p) does NOT map a Cauchy sequence [xn in X] to a Cauchy sequence of its images [f(xn) in Y] in the complex plane between metric spaces.

## Homework Equations

If a function f is continuous in metric space (X, d), then it continuous at every point in X.

Definition of a sequence converges to a point in metric space:

If {xn} -> x, then for all e>0, there exists an N such that d(x, xn)<e, for all n<= N

Definition of Cauchy sequence in metric space:

- {xn} is a Cauchy sequence if for all e>0, there exists an N such that d(xn, xm)<e, for all n, m <= N

- Every convergent sequence in metric space is a Cauchy sequence.

## The Attempt at a Solution

I have counterexample in mind in R but not complex plane in metric space though.

Let f:R-{0} -> R-{0} defined by f(x)= 1/x, so f(x) is continuous on all domain.

Let {xn} = 1/n, then {xn} converges to 0 and therefore is a Cauchy sequence in R–{0}, but its image f(xn) = 1/xn = 1/(1/n) = n diverges, therefore is NOT a Cauchy sequence.

But I need an example of a continuous function f:(X, d) -> Y(Y, p) does NOT map a Cauchy sequence [xn in X] to the Cauchy sequence of its images [f(xn) in Y] between complex planes in the metric spaces?

Any suggestions?