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Is $c_{00} $ open or closed in $\ell^\infty$

My professor gave the following definition for $c_{00}$

$$c_{00} = \{(x_n)_{n\geq 1} | x_i \neq 0 \;\text{for only finitely many values of i}\}$$

from which I made my own inference that

$$c_{00}= \{(x_n) | \; (x_n) \; \text{eventually constant to 0} \}$$

is this a reasonable implication ??

I will use this to show $c_{00}$ is closed in $\ell\infty$

Let $(x_n) \in c_{00}$ such that $(x_n)\to x_0$ where $x_0 \in \ell\infty$

WTS: $x_0 \in c_{00}$

We have $||x_n - x_0||_\infty \to 0$ in $R$

$\Rightarrow sup_{k \in N}{|x_n^k -x_0^k|} \to 0$ in $R$

$\Rightarrow |x_n^k -x_0^k| \to 0$ in $R \; \forall k\in N$

as $(x_n) \in c_{00} \exists t\in N $ such that $x_n^k =0\;\forall k\geq t$

$\Rightarrow x_0^k =0 \; \forall k\geq t$ Hence $x_0 \in c_{00}$

For openness :

$0\in c_{00}$

I don't think it can be open in $\ell_\infty$ since otherwise we can expand $c_{00}$ to the whole of $\ell\infty$ because we are dealing with normed linear spaces.

But how do I prove it?

Robert Shore
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chesslad
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1 Answers1

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$c_{00}$ is neither open nor closed in $l_\infty$.

Your proof that it is closed is incorrect at one point: yes, each $x^k$ is eventually $0$. But perhaps they are eventually $0$ later and later. For instance, the following sequences are each in $c_{00}$: $$ x^k_i = \frac{1}{i} \text{ if $i < k$, } 0 \text{ otherwise} $$ This converges in $l_\infty$ to the sequence $x$, where $x_i = \frac{1}{i}$. Of course, this is not in $c_{00}$ because it never turns into the zero sequence.

To show that it is not open, show that each $x \in c_{00}$ is arbitrarily close to some element of $l_\infty$ which is not eventually $0$ - for instance, you can turn the zero tail of $x$ into all $\varepsilon$s, where $\varepsilon$ is as small as you like.

Andrew Tindall
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  • Very nice example. heartbreaking because I was convinced with my proof, that's why I post here for proof verification. Do you think $c_00$ be closed in $\ell 1$ then ?or $\ell 2$ for that matter? – chesslad Sep 19 '19 at 18:34
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    Nope, it is not closed in any of them. In fact, the regular way to define an $l_n$ sequence involves taking partial sums, which is essentially estimating your $x \in l_n$ by a sequence $x^k \in c_{00}$. For instance, $\frac{1}{n^2} \in l_1$ can be approximated with the $c_{00}$ sequence of elements $x^k_i = \frac{1}{i^2}$ if $i < k$, 0 otherwise. You can do the same thing with your favorite elements of $l_2$, $l_3$, etc. – Andrew Tindall Sep 19 '19 at 18:37
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    The important think is that for any $1\leq p<\infty,$ $c_{00}$ is dense in $\ell^p.$ This is not true for $\ell^{\infty}$ because $x=(0,1/2,2/3,3/4,\dots)$ has $|x|{\infty}=1$ but no sequence in $c{00}$ converges to this $x$. – Thomas Andrews Sep 19 '19 at 18:43
  • So by the same example we can also show that $c_{00}$ is not closed in $(c_0,||.||\infty)$ where $c_0$ is the set of sequence tending to zero? Is $c{00}$ closed anywhere in a well known space? – chesslad Sep 19 '19 at 20:06
  • It's closed in itself :)

    But no, it's not closed in any other sequence spaces that I know of - not any $\ell$ spaces, not $c_0$... I don't really know any other examples?

     – Andrew Tindall Sep 19 '19 at 20:13