Question: How do I show that $c_0$, which is shorthand for $C_0(\mathbb{N})$, is closed in $l_\infty$?
I've tried to work with the fact that if $(f_n)$ is a sequence in $c_0$ converging to $f \in l_\infty$, $|f(k)|\leq|f(k) - f_n(k)| + |f_n(k)|$ and that I can choose n so that $|f(k) - f_n(k)|$ is small independent of k. However, I don't get very far..
Thanks in advance!
Suppose that $x=(x_n)$ is a sequence in $\ell_\infty$ that is not in $c_0$, which means it's not convergent to $0$. Negating the definition of convergence shows that this means there exists some $\varepsilon >0$ such that:
$$\forall N \in \mathbb{N}: \exists n \in \mathbb{N}: (n > N \land |x_n| \ge \varepsilon )$$
or equivalently, there is an infinite subset $M$ of indices such that $|x_m| \ge \varepsilon$ for all $m \in M$.
Now if $y=(y_n)$ is a sequence in $\ell_\infty$ such that $d(x,y) < \frac{\varepsilon}{2}$ ,this means that $|y_n -x_n| \le d(x,y) < \frac{\varepsilon}{2}$ for all $n$ and in particular that
$$\forall m \in M: |y_m| \ge \frac{\varepsilon}{2}$$
showing that $y \notin c_0$ as well.
Hence, $B(x, {\varepsilon \over 2}) \cap c_0 = \emptyset$, showing that no point outside of $c_0$ is a limit point of $c_0$, i.e. $c_0$ is closed.