Metric Space & Cauchy Sequence

Question

Question: Consider a metric space $(X,d)$

1) Show that if $x$ is a convergent sequence in $(X,d)$, then it is a Cauchy sequence in $(X,d)$.

Suppose $(X,d)$ is complete. Let $f\colon X \to X$ be a contraction, i.e., there exits a $\beta\in(0,1)$ such that, for all $x,y \in X$, we have $d( f(x) , f(y) )\le \beta d(x,y)$. Let $x_0\in X$. Define the sequence $x$ inductively by the formula $x_n=f(x_{n-1})$ for $n\in\mathbb{N}$.

2) Show that $x$ is a Cauchy sequence, and therefore convergent.

3) Show that a limit point of $x$, say $\alpha$, is a fixed point of $f$, i.e., $\alpha=f(\alpha)$.

4)Show that $f$ has a unique fixed point.

My try for part 1:

If $x$ is a convergent sequence in $(X,d)$ then there exists $x_0 \in X$ and $M \in \mathbb{R}$ such that for all $x \in X$, $d(x,x_0) \le M$

This implies $|x-x_0| \le M$ . This proves it is Cauchy sequence.

Answer 1

Your attempt to prove (1) does not work. Consider the sequence $\langle(-1)^n:n\in\Bbb N\rangle$ in $\Bbb R$. It is certainly true that there are $y,M\in\Bbb R$ such that $|(-1)^n-y|<M$ for all $n\in\Bbb N$: take $y=0$ and $M=2$, for instance. But the sequence is neither Cauchy nor convergent.

You cannot do mathematics without paying attention to definitions. Here you need to know the definitions of Cauchy sequence and convergent sequence, and it’s clear from your attempted proof that you don’t actually understand the definitions of either of these concepts.

The sequence $\langle x_n:n\in\Bbb N\rangle$ in the metric space $\langle X,d\rangle$ is Cauchy if for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $d(x_k,x_n)<\epsilon$ whenever $k,n\ge m_\epsilon$.

This is a moderately complicated definition, and you’ll probably need to spend some time thinking about it in order to understand it. Informally, however, it says that if you take a tail of the sequence starting far enough out, you can make the diameter of the tail as small as you please. Specifically, if you want to make the diameter at most $\epsilon$, the tail $\langle x_n:n\ge m_\epsilon\rangle$ will do the job.

The sequence $\langle x_n:n\in\Bbb N\rangle$ in the metric space $\langle X,d\rangle$ converges to the point $y\in X$ if for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $d(x_n,y)<\epsilon$ whenever $n\ge m_\epsilon$.

This is just the generalization to arbitrary metric spaces of the usual definition of convergence of a sequence in $\Bbb R$ that you learned back in calculus: if you start far enough out, you can all of the terms in a tail of the sequence to be within $\epsilon$ of $y$.

HINTS: For (1) you want to show that if $\langle x_n:n\in\Bbb N\rangle$ is convergent in $\langle X,d\rangle$, then it is a Cauchy sequence. Since it’s convergent, it converges to some $y\in X$. By definition this tells you that for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $d(x_n,y)<\epsilon$ whenever $n\ge m_\epsilon$. What happens if you take two terms of the sequence that are in the tail $\langle x_n:n\ge m_\epsilon\rangle$? If $k,n\ge m_\epsilon$, then by the triangle inequality we must have

$$d(x_k,x_n)\le d(x_k,y)+d(y,x_n)<\epsilon+\epsilon=2\epsilon\;.$$

• I’ll let you do the little that’s left to the proof. What can you say about $d(x_k,x_n)$ if $k,n\ge m_{\epsilon/2}$? Why does this show that $\langle x_n:n\in\Bbb N\rangle$ is Cauchy?

For (2) use the fact that if $\beta\in(0,1)$, $\sum_{n\ge 0}\beta^n$ converges.

For (3) use the continuity of $f$: show (if you’ve not already done so) that if $y=\lim\limits_{n\to\infty}x_n$, then $\lim\limits_{n\to\infty}f(x_n)=f(y)$.

For (4) just observe that if $y,z\in X$ and $y\ne z$, then $d(f(y),f(z))<d(y,z)$. If $f(y)=y$ and $f(z)=z$, however, $d(f(y),f(z))=\ldots$ what?