Prove that $E=\{x=(x_n)\in \ell^\infty(\Bbb N): (x_n)_n~~\text{is periodic}\}$ is not complete.

Question

Let $E=\{x=(x_n)\in \ell^\infty(\Bbb N): (x_n)~~\text{is periodic}\}$

Defintion: $x=(x_n)$ is periodic means there exists $p\in \Bbb N$ such that, $x_{n+p} =x_n ~~~\forall ~~n\in\Bbb N.$

we define on $E$ the distance $$d(x,y) =\|x-y\|_\infty$$

Prove that $(E,d)$ is not complete. I don't how to prove this. thanks for help

Answer 1

By contradiction we assume that E is complete

Consider the map $T:E\to E$ such that for $x= (x_n)_n$ we have $$Tx= \left(0,\frac{x_0 +1}{2},0,\frac{x_1 +1}{2},0,\frac{x_2 +1}{2}, \cdots\right) $$

That is $(Tx)_{2n} =0$ and $(Tx)_{2n +1} =\frac{x_n +1}{2}$. Clearly, for $x,y\in E$ we have, $$\|T(x-y)\|_\infty \le\frac{1}{2}\|x-y\|_\infty$$

That is T is contraction since E is complete then any contraction on $E$ should have a fix point. Let $u\in E$ such that $$u =Tu = \left(0,\frac{u_0 +1}{2},0,\frac{u_1 +1}{2},0,\frac{u_2 +1}{2}, \cdots\right) $$

Therefore, $$\begin{cases}u_{2n} = 0\\ u_{2n+1} =\frac{u_n+1}{2}\end{cases}$$

the relation $u_{2n+1} =\frac{u_n+1}{2}$ with $u_1 =\frac{u_0+1}{2}=\frac{1}{2}$ clearly shows that $(u_n)_n$ is not periodic. In fact,

$$ u_3 =\frac{u_2+1}{2} =\frac{1}{2}, u_7=\frac{u_3+1}{2} =\frac{\frac12+1}{2} = \frac34,$$$$u_{15}=\frac{u_7+1}{2} =\frac{\frac34+1}{2} = \frac78, u_{31}=\frac{u_{15}+1}{2} =\frac{\frac78+1}{2} = \frac{15}{16}\cdots$$

With further investigation one glimpses that if we set $a_n =u_{4n-1}$ then $$a_{n+1} =\frac{a_n +1}{2}~~~\text{which is not periodic}$$

Then, $u\not\in E$. Conclusion T has no fix point in $E$, therefore, E is not complete.

Answer 2

Let us define a sequence $x^i=(x_n^i)_{i,n\in\omega}$ which is Cauchy but not convergent (sorry for the cumbersome notation).

We set $x^0$ as the sequence identically $0$, so $x_n^0=0$ for all $n$. Then, let $x^1$ be the sequence such that $x_{2n}^1=0$ and $x_{2n+1}^1=1$(so that we obtain the sequence $(0,1,0,1,\dots)$). As $x^2$, we set the sequence such that $x^2_{2n+1}=1$, $x^2_{4n}=1/2$ and has $0$ in the remaining positions (this is the sequence $(0,1,1/2,1,0,1,\dots))$. We continue this way: the sequence $x^{j+1}$ coincides with $x^j$ in every position except the ones that are divisible by $2^{j+1}$, which are assigned value $1/2^{j}$. This sequence is Cauchy, since $d(x^j,x^k)$, for $j<k$, is $2^{-k}$. But the limit of the sequence is not periodic, since it has $0$ only in its first position.