Question

Let X and Y be metric spaces and let f:=X→Y be a mapping. Pick out the true statements:

a. if f is uniformly continuous, then the image of every Cauchy sequence in X is a Cauchy sequence in Y ;

b. if X is complete and if f is continuous, then the image of every Cauchy sequence in X is a Cauchy sequence in Y ;

c. if Y is complete and if f is continuous, then the image of every Cauchy sequence in X is a Cauchy sequence in Y

Answer 1

$$f: X \to Y$$ $$(\forall \varepsilon > 0) (\exists \delta > 0) (\forall x _ {1}, x _ {2}) \left[ d _ {X} \left(x _ {1}, x _ {2}\right) < \delta \Rightarrow d _ {Y} \left(f \left(x _ {1}\right), f \left(x _ {2}\right)\right) < \varepsilon \right]$$ $$\left\{x _ {n} \right\} _ {n = 1} ^ {\infty}: (\forall \varepsilon > 0) (\exists N > 0) (\forall n, m \geq N) \left\{d _ {X} \left(x _ {n}, x _ {m}\right) < \varepsilon \right\}$$ $$\delta := \varepsilon \Rightarrow d _ {X} (x _ {m}, x _ {n}) < \varepsilon \Rightarrow d _ {Y} (f (x _ {m}), f (x _ {m})) < \varepsilon$$ $$\left\{f \left(x _ {n}\right) \right\} _ {n = 1} ^ {\infty}: (\forall \varepsilon > 0) (\exists N > 0) (\forall n, m \geq N) \left\{d _ {X} \left(f \left(x _ {n}\right), f \left(x _ {m}\right)\right) < \varepsilon \right\}$$

So image of Cauchy sequence is again Cauchy sequence, if $f$ is uniformly continuous.

Since every continuous function defined on compact set is uniformly continuous, and complete space is compact, then image of Cauchy sequence is again Cauchy.

But completeness of $\mathbf{Y}$ cannot guarantee the same statement as above.

So, a and b are true, and c is false.