How does one show that $\{b_n\}^∞_{n=1}$ does not not converge in $c_o$

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?

• 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. – Aniruddha Deshmukh Apr 7 at 3:31
• Sorry I will edit this and clarify and add in some bits – Alexander Quinn Apr 7 at 3:34
• So you mean to say that $a_n = \left( 1, 1, 1, \cdots, 1, 0, 0, \cdots \right)$ ($1$ occurs $n$ times) and $b_n = \left( 1, \dfrac{1}{2}, \cdots, \dfrac{1}{n}, 0, 0, \cdots \right)$? – Aniruddha Deshmukh Apr 7 at 3:36
• @AniruddhaDeshmukh yes exactly – Alexander Quinn Apr 7 at 3:42
• 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 – Aniruddha Deshmukh Apr 7 at 4:08

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 $$c_{00}$$.
• 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 $c_{00}$. – Kavi Rama Murthy Apr 7 at 6:25
• I have clarified the notation and the norm is explicitly similar to that of $d_∞ (v,b_k)$ – Alexander Quinn Apr 7 at 7:03