A problem on Cauchy sequences in metric spaces.

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 $Y$.

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What are your thoughts on the questions? –  amWhy Jan 10 '13 at 2:01
they are true in R but not sure in general case –  user55686 Jan 10 '13 at 2:13

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 $\{ \pm 1 \}$?

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