0
$\begingroup$

The space $(F,d_\infty)$ is not complete. Find explicitly a nested family of closed balls $(K_n)_{n=1}^\infty$ with $r_n \mapsto 0 $ such that $\cap_{n=1}^\infty K_n = \emptyset$ Here, $F$ is the set of all finite sequences and $d_\infty(x,y) = sup_n |x_n - y_n|$

For the solution I know that for complete spaces $\cap_{n=1}^\infty K_n \ne \emptyset$, and it implies that for an incomplete space $\cap_{n=1}^\infty K_n = \emptyset$ but I do not know how to find closed balls to prove this.

$\endgroup$
  • 1
    $\begingroup$ Hint: use a Cauchy sequence that does not converge (there are standard examples) and for each point $x_n$ in the sequence choose an appropriate radius $r_n$ such that $r_n\to0$, then take $K_n=B(x_n, r_n) $. $\endgroup$ – Jason Nov 8 '17 at 19:40
  • $\begingroup$ @Jason how do I take closed ball as equal to open ball ? $\endgroup$ – Pumpkin Nov 8 '17 at 19:42
  • $\begingroup$ I am referring to the closed ball. $\endgroup$ – Jason Nov 8 '17 at 19:44
3
$\begingroup$

I assume you meant by finite sequence a sequence in $\mathbb R$ or $\mathbb C$ with finite support.

For $n\in \mathbb N$ let $a^n \in F$ be definite by $$ a^n_m = \begin{cases} 2^{-m}, & m \le n \\ 0, & m > n. \end{cases} $$ Further let $r_n = 2^{-n}$ and $$K_n = \{ x\in F \mid d(x, a^n) \le r_n \}. $$

Then, $K_n$ is decreasing and the intersection is empty.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.