Continuity, uniform continuity and preservation of Cauchy sequences in metric spaces.

Question

Let $ X $ and $ Y $ be metric spaces, and let $ f: X \to Y $ be a mapping. Determine which of the following statements is/are true.

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

(a) is true. Just apply the definition of uniform continuity of a function

Answer 2

Let me give you some hints that will allow you to complete a solution on your own.

---

(a) If $ f: (X,d_{X}) \to (Y,d_{Y}) $ is a uniformly continuous function, then by definition, for every $ \epsilon > 0 $, there exists a $ \delta > 0 $ such that $$ (*) \quad \forall x,y \in X: \quad {d_{X}}(x,y) < \delta ~ \Longrightarrow ~ {d_{Y}}(f(x),f(y)) < \epsilon. $$ Let $ (x_{n})_{n \in \mathbb{N}} $ be a Cauchy sequence in $ (X,d_{X}) $. Fix an $ \epsilon > 0 $, and find a $ \delta > 0 $ so that $ (*) $ is satisfied. There exists an $ N \in \mathbb{N} $ sufficiently large such that for all $ m,n \in \mathbb{N}_{\geq N} $, we have $ {d_{X}}(x_{m},x_{n}) < \delta $. What can you say now about $ {d_{Y}}(f(x_{m}),f(x_{n})) $ for all $ m,n \in \mathbb{N}_{\geq N} $?

---

(b) If $ (X,d_{X}) $ is a complete metric space, then by definition, every Cauchy sequence $ (x_{n})_{n \in \mathbb{N}} $ in $ (X,d_{X}) $ has a limit. As $ f: (X,d_{X}) \to (Y,d_{Y}) $ is a continuous function, what can you say about the sequence $ (f(x_{n}))_{n \in \mathbb{N}} $ in $ (Y,d_{Y}) $? Does it converge? Does the convergence of a sequence in an arbitrary metric space imply that the sequence is Cauchy?

---

(c) Let $ f: \left\{ \dfrac{1}{n} \right\}_{n \in \mathbb{N}} \to \{ \pm 1 \} $ be defined by $$ \forall n \in \mathbb{N}: \quad f \left( \frac{1}{n} \right) \stackrel{\text{def}}{=} (-1)^{n}. $$ Equip the sets $ \left\{ \dfrac{1}{n} \right\}_{n \in \mathbb{N}} $ and $ \{ \pm 1 \} $ with the metric inherited from $ \mathbb{R} $. Then

• $ \left\{ \dfrac{1}{n} \right\}_{n \in \mathbb{N}} $ becomes a non-complete discrete metric space,

• $ \{ \pm 1 \} $ becomes a complete discrete metric space (to prove completeness, think about what the Cauchy sequences in $ \{ \pm 1 \} $ are) and

• $ f $ becomes a continuous function (any function from a discrete topological space to another topological space is automatically continuous).

Note: Discrete metric spaces are not necessarily complete, but they are always completely metrizable.

Observe that $ \left( \dfrac{1}{n} \right)_{n \in \mathbb{N}} $ is a Cauchy sequence in $ \left\{ \dfrac{1}{n} \right\}_{n \in \mathbb{N}} $. Its image under $ f $ is $ ((-1)^{n})_{n \in \mathbb{N}} $. Is this a Cauchy sequence in $ \{ \pm 1 \} $?