Concept of Cauchy sequence

Question

Definition: A sequence $\{x_n\}_{n\geqslant 1}$ in a metric space $X$ is said to be a Cauchy sequence if $\forall \varepsilon>0$ $\exists N\in \mathbb{N}$ such that $d(x_n,x_m)<\varepsilon$ if $n\geqslant N$ and $m\geqslant N$

Wee see that this definition does not depends on metric space. Here appears only metric $d$ on $X$.

I have two questions:

1) If $E$ some nonempty subset of $X$ and $\{x_n\}$ is a Cauchy sequence in $E$. Will $\{x_n\}$ be a Cauchy sequence in $X$?

2) If $E$ some nonempty subset of $X$ and $\{x_n\}$ is a Cauchy sequence in $X$. Will $\{x_n\}$ be a Cauchy sequence in $E$?

I hope that you'll help me disassemble these questions.

If you are not changing the definition of a metric then it is line you have a collection of points and you are checking whether they get close to each other. And the answer is yes since the distance is not really changing! — Makarand Sarnobat, Aug 14 '15 at 19:06
For 2) if some $x_n$ does not belonge to $E$. What about this case? Maybe there is some counterexample?! — RFZ, Aug 14 '15 at 19:21

score 2 · Answer 1 · answered Aug 14 '15 at 19:37

When you say that $E$ is a subset of $X$, and that $\{x_n\}$ is a cauchy sequence in $X$, I assume you mean that $X$ is also a metric space and that the metric of $E$ coincides with that of $X$ in their intersection. In that case the answer to 1) is yes because the value of $d(x_n,x_m)$ will not change when $x_n$ and $x_m$ are considered to be elements of $X$.

In the second case the answer is no, because a sequence in $X$ need not even be contained in $E$

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

1

By definition, $1)$ is true.

$2)$ is not true. Consider $X = [0, 1]$, $E = \mathbb{Q} \cap X$. Let $$a_n = \{n\pi\} = n\pi - \lfloor n\pi\rfloor, \quad n = 1, 2, \ldots.$$

It can be shown that $\{a_n\}$ is dense in $X$. Hence there exists a subsequence $\{a_{n_k}\}$ such that $$\lim_{k \to \infty} a_{n_k} = 0.$$ Thus $\{a_{n_k}\}$ is Cauchy in $X$, but clearly each $a_{n_k} \notin E$ (otherwise $n_k\pi$ would be a rational), therefore it is not Cauchy in $E$.

edited Apr 13 '17 at 12:20

Community

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answered Aug 14 '15 at 19:28

Zhanxiong

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Easier example is surely with $[0,1]$, taking the Cauchy sequence $x_n = \frac{1}{2}$ and $E = [0, 1] \setminus {\frac{1}{2} }$? – Patrick Stevens Aug 14 '15 at 19:35
@PatrickStevens Excellent! – Zhanxiong Aug 14 '15 at 19:36
Dear Zhanxiong! 1)You wrote "Thus ${a_{n_k}}$ is Cauchy in $X$". It's Cauchy because it's converges in $X$? Am I right? 2) When we say that ${x_n}$ is Cauchy in $E$ we meant that all $x_n\in E$? – RFZ Aug 14 '15 at 19:48
@Pacman Absolutely right. – Zhanxiong Aug 14 '15 at 19:51
Dear Zhanxiong thank you very much! Your post helped me. – RFZ Aug 14 '15 at 19:52

Concept of Cauchy sequence

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