Let $M\subset l^{\infty}$ be the subspace of $\; l^{\infty}$ consisting of all sequences $(x_{i})$ with at

Question

Please let me know if the solution to the following problem is true. Just a hint can be enough. Problem:

Let $M\subset l^{\infty}$ be the subspace of $\; l^{\infty}$ consisting of all sequences $(x_{i})$ with at most finitely many nonzero terms. Find a Cauchy sequence in $M$ which does not converge in $M$, so that $M$ is not complete.

What I have done:

Define $x_{n}=\{1,\frac{1}{2},\cdots , \frac{1}{n},0,0,\cdots\}$ So we have for any m\leq n, $$ d(x_{n},x_{m})=\vert x_{n}-x_{m}\vert= \{ 0,\cdots,0,\frac{1}{m+1},\dots, \frac{1}{n},0,\cdots\}, $$ so we see that $d(x_{n},x_{m})\to 0\; $ as $\;m,n\to\infty.$ Therefore, ${x_{n}}$ is a Cauchy sequence. On the contrary, $x_{n}\to x$ as $n\to \infty,$ where $x=\{1,\frac{1}{2},\cdots , \frac{1}{n},\frac{1}{n+1},\cdots\}$ which does not belong to $M.$

Answer 1

What you did is just fine. You proved that $M$ is not complete by providing an example of a Cauchy sequence of elements of $M$ which does not converge in $M$.

Answer 2

Your metric outputs a set and not a non-negative real number. The metric I suspect you are using is $d(\{y_j\}_{j\in\mathbb N},\{\tilde y_j\}_{j\in\mathbb N}):=\sup_j |y_j-\tilde y_j|$. Now to prove your statement you should show that your sequence $x_n$ is Cauchy for the metric $d$. Then take any element $m\in M$ and show that there exists $\varepsilon>0$ such that $d(x_n,m)\ge\varepsilon$ for all $n$ large.

Do you see the difference from your approach?