Let $(x_n)$ be a Cauchy sequence in $(X,d)$ and $a \in X$. Show that there exist $M>0$ such that $x_n \in B(a,M)$ for all $n \in \mathbb{N}$.

Question

Let $(x_n)$ be a Cauchy sequence in the metric space $(X,d)$ and $a \in X$. Show that there exist $M>0$ such that $x_n \in B(a,M)$ for all $n \in \mathbb{N}$.

If $(x_n)$ is Cauchy we have that $\forall \varepsilon >0$ there exists $k \in \mathbb{N}$ such that $d(x_n,x_m) < \varepsilon$, when $n,m \geqslant k$.

Let $\varepsilon =1$ I have $d(x_n,x_m)<1$ So $$d(a,x_n) \leqslant d(a,x_m) + d(x_n,x_m) = d(a,x_m)+1$$

I'm a bit stuck here. It seems that I want to get to $d(a,x_n)<M$, but I'm not sure how to achieve this?

Answer 1

$M$. is for you to choose: as you observe, you can find an $n$ such that $d(x_n, x_m) < 1$ whenever $m \ge n$. Then for $m \ge n$:

$$ d(a, x_m) \le d(a, x_n) + d(x_n, x_m) < d(a, x_n) + 1 $$ So the ball of radius $d(a, x_n) + 1$ contains all but the first $n-1$ of the $x_i$. So you can take $M$ to be any number larger than $$\max\{d(a_, x_1), d(a, x_2), \ldots, d(a, x_{n-1}), d(a, x_n) + 1\} $$

Answer 2

Let $\{x_n\}$ be a sequence in $(X,d)$ with $a \in X.$

Since $\{x_n\}$ is Cauchy,

For all $\varepsilon >0,$ there is $N \in X$ such that for all $x_n, x_m \in X$, when $n,m>N$ it follows that $d(x_n,x_m) < \varepsilon$

Let $\varepsilon = 1$

Since $\{x_n\}$ is Cauchy, there is an $N \in X$ such that for all $m,n >N, d(x_n,x_m) < 1$

Since the $x_i$'s with $i>N$ are less than 1 away from each other, we can establish that $|x_n|<1+|x_{N+1}|$. This is true for all $i > N$. This bounds all the terms beyond the $N^{th}$.

To bound the terms before the $N^{th},$ consider the statement:

$|x_n| \leq$ max$\{|x_1|,|x_2|,...,|x_N|\}$. This is true for all $n \leq N$.

Set $M = $ max$\{1+|x_{N+1}|,$ max$\{|x_1|,...,|x_N|\}$ $\}$

Then $x_n \in B(a,M)$ for all $n \in \mathbb{N}$