the space is the set of all polynomials on $[0,1]$ with metric sup. the space is not complete.

we need to find explicitly a nested family of closed balls with radius (of each closed ball) goes to zero such that the infinite intersection is empty set.


$P_n(t)= 1 -t + ...+ (-1)^n\frac{t^n}{n}$

radius: $R_n=\frac 1{n!}$

$K_n=K(P_n,R_n)$ : closed ball centered $P_n$ with radius $R_n$

We need help to show that the closed balls $K_n$ satisfies the condition (if it does that is)

  • $\begingroup$ What exactly do you mean by $K(p, r)$? Is it the closed ball with center $p$ and radius $r$? $\endgroup$ – Dan Shved Nov 11 '13 at 13:04
  • $\begingroup$ yes. exactly as you wrote $\endgroup$ – 104078 Nov 11 '13 at 13:09
  • $\begingroup$ The distance between $P_n$ and $P_{n+1}$ in your case is $1/n+1$ so $R_n=1/(n!)$ does not work. $\endgroup$ – Beni Bogosel Nov 11 '13 at 13:44

You should consider $$ p_n(t)=\sum\limits_{k=0}^n (-1)^k \frac{t^k}{k!}\quad\mbox{ and }\quad r_n=\frac{2}{n!} $$ For all $n\in\mathbb{N}$ you will have $$ \Vert p_{n+1}-p_n\Vert=\left\Vert \frac{t^{n+1}}{(n+1)!}\right\Vert=\frac{1}{(n+1)!} $$ If $f\in B(p_{n+1},r_{n+1})$, then $$ \Vert f-p_n\Vert\leq\Vert f-p_{n+1}\Vert+\Vert p_{n+1}-p_n\Vert\leq r_{n+1}+\frac{1}{(n+1)!}=\frac{3}{(n+1)!}<\frac{2}{n!} $$ i.e. $f\in B(p_n,r_n)$. Since $f$ is arbitrary, then $B(p_{n+1},r_{n+1})\subset B(p_n,r_n)$. Clearly $r_n\to 0$ when $n\to \infty$. Thus all conditions of Baire category theorem are satisfied for the sequence of balls $\{B(p_n,r_n)\}_{n=1}^\infty$.

Assume $S:=\bigcap_{n=1}^\infty B(_n,r_n)\neq\varnothing$. Since $r_n\to 0$ for $n\to\infty$, then there is at most $1$ point in $S$, say $x$. Clearly $(p_n)_{n=1}^\infty$ converges to $x$ in metric, i.e. in $\sup$ norm. In particular $(p_n)_{n=1}^\infty$ pointwise converges to $x$. Note thar the pointwise limit of $p_n$ is $e^{-t}$. So $x(t)=e^{-t}$. But this is impossible because by construction $x$ is a polynomial, while $e^{-t}$ is not. Contradiction, so $S=\varnothing$


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