Let $X = \prod_{k=1}^\infty \{0,1\}$
Given that for each $k$ = $1, 2, 3$, . . . the projection onto the $k^{th}$ coordinate $π^k$ : X → {0, 1}, given by $π^k$($\{x_n\}^∞_{n=1}$) is Lipschitz
If $\{a_n\}^∞_{n=1}$ is a sequence in $X$
$a_1 = ( a_1,_1, a_1,_2, a_1,_3, a_1,_4, . . .)$
$a_2 = ( a_2,_1, a_2,_2, a_2,_3, a_2,_4, . . .)$
$a_3 = ( a_3,_1, a_3,_2, a_3,_3, a_3,_4, . . .)$
that is Cauchy, then for each coordinate $k$ $\{a_n,_k\}^∞_{n=1}$ is eventually constant.
We also know X is complete since there is a bijection $f : X → C_3$ onto the “middle third” Cantor set that has $f$, $f^−1$ uniformly continuous, and that $C_3$ is itself complete. Thus $X$ is complete.
$Now$
Let $c_{00}$ be the vector space of finitely supported real sequences (as seen before) with the norm $ \|\cdot\|_∞$ and the metric $d_∞$ from this norm
The sequence $\{a_n\}^∞_{n=1}$ ⊆ $c_{00}$ is defined below
$a_1 = ( 1, 0, 0, 0, 0, . . .)$
$a_2 = ( 1, 1, 0, 0, 0, . . .)$
$a_3 = ( 1, 1, 1, 0, 0, . . .)$
$a_4 = ( 1, 1, 1, 1, 0, . . .)$
And sequence $\{b_n\}^∞_{n=1}$ ⊆ $c_{00}$ defined below is
$b_1 = ( 1, 0, 0, 0, 0, . . .)$
$b_2 = ( 1, 1/2, 0, 0, 0, . . .)$
$b_3 = ( 1, 1/2, 1/3, 0, 0, . . .)$
$b_4 = ( 1, 1/2, 1/3, 1/4, 0, . . .)$
$\{a_n\}^∞_{n=1}$ ⊆ $c_{00}$ where $a_n$ = $1^n$ and $\{b_n\}^∞_{n=1}$ ⊆ $c_{00}$ where $b_n$ = $1/n$ given that both sequences are divergent and that $\{a_n\}^∞_{n=1}$ is not cauchy but $\{b_n\}^∞_{n=1}$ is, how does one show that $\{b_n\}^∞_{n=1}$ does not not converge in $c_{00}$
To prove this I Suppose that $b_n \to V $ $\in c_{00}$ If we can find a $\varepsilon>0$ such that for all large enough k, $d_∞ (v,b_k) \ge \varepsilon$
This is where I am having trouble do we have to do this by triangle inequality?
Convergence in $c_{00}$ implies convergnce of each coordinate. If $(b_n)$ converges then, taking limits of its coordinates we see that the limit has to be $(1,1/2,1/3,...)$. This is a contradiction to the definition of $c_{00}$, so $(b_n)$ does not converge in $c_{00}$.