Check the following statements.

Question

Let $X=\{(x_1,x_2,\dots):x_i\in\Bbb{R}$ and finitely many $x_i$'s are non zero} and $d:X\times X\to\Bbb{R}$ be a metric on $X$ defined by $d(x,y)=\text{sup}_{i\in\Bbb{N}}|x_i-x_j|,x=(x_1,x_2,\dots),y=(y_1,y_2,\dots) $in $X$.

Consider the following statements:

• $(X,d)$ is complete metric spaces.
• The set $\{x\in X:d(\overline{0},x)\leq 1\}$ is compact, where $\overline{0}$is the zero elements of $X$.

I have to check that whether the statements are true or false.

consider the sequence $e_n,n\geq 1$, where $e_1=(\delta/2,0,0,...), e_2=(0,\delta/2,0,0,...)$ and so on. clearly this sequence is Cauchy, but how to show that this question is not convergent? I am having difficulty for writing in my notebook. help me with both statements. Thanks|

Answer 1

Consider the sequence $(e_n)_{n\in\Bbb N}$ of elements of $X$ defined by:

• $e_1=(1,0,0,0,0,\ldots)$;
• $e_2=\left(1,\frac12,0,0,0,\ldots\right)$;
• $e_3=\left(1,\frac12,\frac13,0,0,\ldots\right)$;

and so on. Then$$d(e_m,e_n)=\begin{cases}0&\text{ if }m=n\\\frac1{\min\{m,n\}+1}&\text{ otherwise,}\end{cases}$$and therefore it is a Cauchy sequence. But it doesn't converge in $X$. If it did and if $e=(a_1,a_2,a_3,a_4,\ldots)$ was it s limit, then, for each $m\in\Bbb N$, $a_m$ would be the limit of the sequence of the $m$^th terms of the sequence $(e_n)_{n\in\Bbb N}$, which means that $a_m=\frac1m$. But then $e\notin X$.

And $\left\{x\in X\,\middle|\,d\left(\overline0,x\right)\leqslant1\right\}$ is not compact, since it contains a sequence without a convergent subsequence, which is $(e_n)_{n\in\Bbb N}$.