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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.

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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?

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  • $\begingroup$ What do you mean by "finitely supported real sequences"? Also, $a_n$ and $b_n$ cannot be a real number. It has to be a sequence. $\endgroup$ – Aniruddha Deshmukh Apr 7 at 3:31
  • $\begingroup$ Sorry I will edit this and clarify and add in some bits $\endgroup$ – Alexander Quinn Apr 7 at 3:34
  • $\begingroup$ So you mean to say that $a_n = \left( 1, 1, 1, \cdots, 1, 0, 0, \cdots \right)$ ($1$ occurs $n$ times) and $b_n = \left( 1, \dfrac{1}{2}, \cdots, \dfrac{1}{n}, 0, 0, \cdots \right)$? $\endgroup$ – Aniruddha Deshmukh Apr 7 at 3:36
  • $\begingroup$ @AniruddhaDeshmukh yes exactly $\endgroup$ – Alexander Quinn Apr 7 at 3:42
  • $\begingroup$ So, can you see that the limit of $b_n$ is $\left( 1, \dfrac{1}{2}, \dfrac{1}{3}, \cdots \right)$ which is not finitely supported and hence not in the space. Therefore, $b_n$ does not converge although it is Cauchy $\endgroup$ – Aniruddha Deshmukh Apr 7 at 4:08
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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 $c_{00}$.

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  • $\begingroup$ I have clarified the question $\endgroup$ – Alexander Quinn Apr 7 at 6:11
  • $\begingroup$ There are still problems with your question. You have not defined any norm explicitly and there is confusion between $c_0$ and $c_{00}$. Please define $c_0$ and clarify whether the question is about convergence in $c_0$ or in $c_{00}$. $\endgroup$ – Kavi Rama Murthy Apr 7 at 6:25
  • $\begingroup$ I have clarified the notation and the norm is explicitly similar to that of $d_∞ (v,b_k)$ $\endgroup$ – Alexander Quinn Apr 7 at 7:03

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