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I'm looking over some previous analysis exams and I've come across this question:

Consider the set $\mathbb{R}$ of real numbers and the metric function defined as:

$$d(x, y) = \begin{cases} 1 ~ \text{if} ~ x \ne y \\ 0 ~ \text{if} ~ x = y \end{cases}$$

  1. Is the sequence $(x_n)_{n = 1}^\infty$ defined by $x_n = \frac{1}{n}$ convergent in this metric space?

  2. Is this metric space complete?


For part 1:

Assume that $(x_n)_{n = 1}^\infty$ is convergent to a real $a$. Then, for every $\epsilon > 0$, there exists a $N \in \mathbb{N}$ such that:

$$d(x_n, a) < \epsilon ~ ~ \text{for all} ~ ~ n \geq N$$

Let $\epsilon = \frac{1}{2}$. Then $d(x_n, a) < \frac{1}{2} ~ ~ \implies ~ ~ x_n = a ~ ~ \implies ~ ~ a = \frac{1}{n} ~ ~ \text{for all} ~ ~ n \geq N$. This is a contradiction, so $(x_n)_{n = 1}^\infty$ does not converge in this metric space.

For part 2:

A metric space is complete iff every Cauchy sequence is convergent. A Cauchy sequence is a sequence $(x_n)_{n = 1}^\infty$ such that for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that:

$$d(x_n, x_m) < \epsilon ~ ~ \text{for all} ~ ~ n, m \geq N$$

Reusing part 1's approach, we let $\epsilon = \frac{1}{2}$. Then if $(x_n)_{n = 1}^\infty$ is a Cauchy sequence, it must follow that $x_n = x_m$ for all $n, m \geq N$. So the sequence is constant for $n \geq N$, and as such trivially converges.

We have proved that if $(x_n)_{n = 1}^\infty$ is a Cauchy sequence in this metric space, then it must be convergent. Therefore this metric space is complete.


Are these proofs valid? In particular, is the use of contradiction sound in part 1 and does the reasoning follow in part 2? I've learned to be careful with analysis arguments, so could someone check over my work?

I am confident about part 1 but there's something about part 2 that confuses me. The completeness property only applies to Cauchy sequences, so if my results are correct then the sequence in part 1 is not Cauchy. So did I make a mistake or was $d$ carefully chosen to defy intuition?

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It looks very good to me. The first sequence is not Cauchy with respect to the given distance, and it does not converge. Everything is fine. –  Giuseppe Negro Jun 20 '13 at 12:38
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2 Answers

up vote 4 down vote accepted

You are all right, the metric $d$ makes your metric space being discrete topological space. In the latter, convergence of $(x_n)$ means that there exists $a$ such that $x_n = a$ for all $n$ big enough. You perfectly showed this using $\epsilon = \frac12<1$. As a result, the sequence $\frac1n$ does not converge.

Any Cauchy sequence thus has to have only constant terms after some $n$ because of the Cauchy condition $|x_n -x_m|<\epsilon$ for all $m,n$ big enough. As a result, you obtain a desired sequence with only constants in the tail, so it clearly converges thanks to the first argument. And surely, the sequence $\frac1n$ is not Cauchy w.r.t. the metric $d$.

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Good work. It is simple and clear. +10) –  Paul Jun 20 '13 at 12:21
    
Thank you Ilya :) –  Thomas Jun 20 '13 at 12:37
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YOU are right. Indeed, this space is a discrete space, i.e., $\Bbb R$ with a discrete topology. Every sequence taking and infinite number of values in a discrete space is not convergent and every discrete space is a complete metric space.

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Every sequence in a discrete space is not convergent What do you mean? –  Ilya Jun 20 '13 at 12:29
    
@Ilya: I want to express that any infinite countable subset of discrete space has not an accumulation point. Can it be okey? Please feel free to chang me. –  Paul Jun 20 '13 at 12:32
    
Constant sequences are convergent... –  David Mitra Jun 20 '13 at 12:38
    
@DavidMitra: Thanks for your kind reminding. Something in special case I always forget. I will reedit it. –  Paul Jun 20 '13 at 12:41
    
@Paul: c'mon, converging sequences are exactly those that are eventually constant, but not necessarily entirely. What you shall say that any sequence with an infinite range is not converging. –  Ilya Jun 20 '13 at 12:46
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