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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.

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What have you tried so far? Any ideas for any of these questions? –  alexwlchan Jun 22 '13 at 21:39
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Most of this is pretty straightforward; have you made any progress on any part of it? Note that (2), (3), and (4) are independent of (1), that it’s entirely possible to do (3) without having done (2), and that (4) is easily done without having done any of the rest. –  Brian M. Scott Jun 22 '13 at 21:40
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I don't understand the point of questions like this. The OP's first question is something that every textbook that defines metric space and Cauchy sequence should prove. The rest of the questions are giving a proof of the Banach Fixed Point Theorem, which is also included in such texts (and the proof is probably too hard for an unassisted homework exercise). In case the OP didn't know that this was the name of what s/he has been asked to show, I included a wikipedia link about it, which of course gives a proof. No response from the OP. –  Pete L. Clark Jun 23 '13 at 0:03
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As an experiment I literally cut and paste "Show that if x is a convergent sequence in (X,d), then it is a Cauchy sequence in (X,d)." into google: the first page contains several links which give the proof. Then I cut and paste "Suppose (X,d) is complete. Let f:X→X be a contraction" The first hit is to beautifully written lecture notes of Keith Conrad, which put the result in context, give a proof, and proceed to go on to deeper things. Given that all this already exists on the internet, what is the point of typing out an answer here? –  Pete L. Clark Jun 23 '13 at 0:06
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Finally, the end of the post shows that the OP does not understand what a Cauchy sequence is, without which the proof of Banach's theorem will certainly not be understood. It all seems futile. If this is from a course, the OP needs to talk to the instructor. If there isn't a course there probably should be: most people cannot easily learn this material on their own. –  Pete L. Clark Jun 23 '13 at 0:12

1 Answer 1

up vote 1 down vote accepted

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?

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Thank you so much. Actually after Mr Clark pointed out I revised Banach fixed point theorem and parts 2,3,4 became clear. Thanx for your valuable inputs which helped me finish part 1 as well. –  Mathy Jun 23 '13 at 10:00
    
@Mathy: You’re very welcome. –  Brian M. Scott Jun 23 '13 at 10:01

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