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I came across this fact in my notes that in given without proof, and I am having trouble proving it.

$c_0 = \overline{span\{e_n:n\ge1\}}$? where $e_n$ is the sequence with a $1$ in the n-th component and $0$ elsewhere and $c_0$ is the set of infinitesimal sequences with the sup norm

My try:

I need to prove that $\forall \varepsilon> 0 \forall x \in c_0\exists y \in span\{e_n:n\ge1\}$ such that $sup_{k \ge 1}|x_k-y_k|<\epsilon$

So $y$ must be of the form $y=\sum_{n=1}^{N} y_ne_n$ for some $N \in \mathbb{N}$ and $y_n \in \mathbb{C}$ .

Furthermore by definition of limit, since $x \in c_0$, we have that $\forall \varepsilon> 0 \exists K(\varepsilon) \in \mathbb{N}$ such that $\forall k \ge K(\varepsilon)$, $|x_k|\le \varepsilon$

This is as far as I got, how do I finish it?

some_math_guy
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1 Answers1

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Let $E$ the span of the $e_n$. It is clear that $E \subset c_0$ and $c_0$ is closed hence $\overline{E} \subset c_0$.

Suppose $x \in c_0$, then $x(n) \to 0$, let $x_k$ be equal to $x$ for the first $k$ components and zero afterwards. Then we see that $\|x-x_k\| \to 0$. Since $x_k \in E$, we see that $x \in \overline{E}$.

copper.hat
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  • How do we know $c_0$ is closed? How do I prove rigorously that $sup_{n \ge k+1}|x(n)| \to 0$, for$ k\to \infty$? – some_math_guy Feb 13 '22 at 19:19
  • For the first, look at https://math.stackexchange.com/a/2462750/27978. – copper.hat Feb 13 '22 at 19:22
  • For the second, you can't, it is not true and I do not understand the relevance. – copper.hat Feb 13 '22 at 19:23
  • Do we have to consider $c_0 $ closed with respect to $l^\infty$? Can't we consider it closed as a subspace of itself? – some_math_guy Feb 13 '22 at 19:23
  • To prove this $|x-x_k| \to 0$ I get to the expresion in my first comment, so it mush be correct, right? Intuitively I am taking the sup over a smaller set when n grows, so I wanted to formalize it – some_math_guy Feb 13 '22 at 19:26
  • If you consider $c_0$ as the space within which you are taking the closure, then you do not need to show that it is closed as the whole space. (That is not the same as saying it is closed in $l_\infty$, of course.) To prove $|x-x_k| \to 0$ you just need to note that $x(n) \to 0$ and since $x_k(n) = 0$ for $n >k$, the result follows. – copper.hat Feb 13 '22 at 19:31
  • Note, for example, that the set $A=\mathbb{Q} \cap [0,1]$ is closed within $\mathbb{Q}$, but generally unless specified otherwise, one would not refer to $A$ as being closed without further elaboration. Since $c_0$ is complete this is not an issue here. – copper.hat Feb 13 '22 at 19:34
  • does the same sequence $x_k$ works to prove $ l^1=\overline{span{e_n:n\ge1}}$ ? – some_math_guy Feb 13 '22 at 21:48
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    If you mean the closure with the $1$-norm, then yes. – copper.hat Feb 13 '22 at 22:06