Cauchy Sequence some challenge

Question

i read this sentence in one of math books:

‌Every convergent sequence in metric space is a cauchy sequence.

would you please some one add more detail, why this is true?

thanks.

Answer 1

Suppose $\;x_n\xrightarrow[n\to\infty]{}x\;$ , and let

$$\;\epsilon>0\implies \;\exists\,N_\epsilon\in\Bbb N\;\;\text{such that}\;\;n>N_\epsilon\implies d(x_n,x)<\frac\epsilon2$$

thus, for $\;n,m>N_\epsilon\;$ we get

$$d(x_n,x_m)\le d(x_n,x)+d(x,x_m)<\frac\epsilon2+\frac\epsilon2=\epsilon\implies$$

$\;\{x_n\}\;$ is Cauchy.

Answer 2

Let d be the metric and the sequence $a_1, a_2, ...$ converges to $a$, which means
For given $\epsilon>0$ exist N such that $n\geq N$ implies $d(a_n,a)<\epsilon/2$
For that N, if $n\geq N$ and $m\geq N$, we have $d(a_n,a_m)\leq d(a_n,a)+d(a,a_m)<2\cdot\epsilon/2=\epsilon$ therefore the sequence is Cauchy.

Answer 3

Let $X$ be a metric space. Take a convergent sequence, say $(x_n) \subset X $. For convenience, I will write the metric of $X$ as: $d( x, y ) = |x-y| $. However, you must keep in mind that we are in an arbitrary metric space, and that I am using the standard metric just for convenience. Now, since $(x_n)$ is convergent (Say $x_n \to x$) we know that for a given $\epsilon > 0$, we can choose $N \in \mathbb{N}$ such that for any $n \geq N$, we have

$$ |x_n - x | < \epsilon/2 $$

In particular, this also holds for any $m > N$:

$$ |x_m - x| < \epsilon/2 $$

Now, by using the triangle inequality ($|a+b| \leq |a| + |b| $), we have

that for any $n,m > N$,

$$ |x_m-x_n| = |x_m - x + x - x_n| \leq |x_m - x| + |x-x_n| < \epsilon/2 + \epsilon/2 = \epsilon$$

Therefore, by definition $(x_n)$ must be cauchy.