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?
$c_{00}$ 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.