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Let $x=(x(1),x(2),…,x(n),…) \in c_0$.

So for any $\varepsilon>0$ there exists a $n_0 \in \mathbb{N}$ such that $|x(n)| \to 0$ as $n \to \infty$ for all $n \geq n_0$.

Now for all $n \geq n_0$ , $||x_n − x||_\infty = \sup \{|x(m)|: m \geq n_0 \} \to 0$ as $m \to \infty$.

I have a doubt that regarding the last line.

That is, how can we say that
$$\sup \{ |x(n+1)|, |x(n+2)|, ......... \} = \sup \{ |x(m)|:m > n \} \to 0$$ as $m \to \infty$ ?
(because we dont know the supremum of this set (i.e. $\sup \{ |x(m)|:m>n \}$ ), it can be infinity also).

So how can we say that $\sup \{ |x(m)|:m>n \} \to 0$ ?
What makes it possible?

Kolmin
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sarani
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  • What are $c_0$ and $c_{00}$ $?$ – user118494 Sep 27 '15 at 14:03
  • C_{0} is the sequences converging to 0 and in c_{00} the elements are sequences of scalars having only a finite number of non zero terms – sarani Sep 27 '15 at 14:07
  • What if we take this sequence in $c_{00}$ : ${x_n}$ s.t $x_n=(1,1/2,1/3,....,1/n,0,0,0,....0,0...)$ then the sequence converges to $x=(1,1/2,1/3,1/4,......,1/n,1/(n+1),......)$ and not to $0=(0,0,0,0,....,0...)$ so this sequence of $c_{00}$ is not in $c_0$ . But closure of $c_{00}$ is supposed to contain $c_{00}$ . – user118494 Sep 27 '15 at 14:30
  • Also the sequence in $c_{00}$ s.t $x_n$ has $1$'s in first $n$ co-ordinates and $0$'s elsewhere then this sequence does not converge at all so this ${x_n }$ is also not in $c_0$ . – user118494 Sep 27 '15 at 14:35

1 Answers1

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Let $ε>0$, because $x(n)$ converges to $0$ we have that there exists a $ξ \in \Bbb N$ so that for every $n>ξ$, $|x(n)|<ε$ if $m>n>ξ$ ($m$ goes to infinity) then $|x(m)|<ε$ and so $sup|x(m)| \le ε$ for $m>n$. Then we send $ε$ to zero and we have it!

Amontillado
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