$c_0$ is a closed subspace of $l^{\infty}$

Question

Put

$$ l^{\infty} = \{ (x_n) \subseteq \mathbb{C} : \forall j \; \;\ \;|x_j| \leq C(x)\} $$

I want to show that $c_0$, the space of all sequences of scalars that converges to $0$ is closed subspace of $l^{\infty}$.

MY atttemt: Take arbitrary sequence $(x_n) \in c_0$. We know $x_n \to 0$. but $0 \in c_0$. Therefore, every sequence in $c_0$ converges to an element in $c_0$. Hence, $c_0$ must be closed

Is this correct? thanks

Answer 1

No, it's not. You're confusing the notion of convergence of a sequence of numbers with convergence of a sequence of elements of $\ell^\infty$, that is to say, a sequence of sequences.

Of course, for any $(x_n)_n\in c_0$ we have $x_n\to 0$, but this is not the same zero as $0\in c_0$, as the former is a number, and the latter is a sequence, comparison between those two doesn't make sense, you might as well say that 3 meters is the same as 3 degrees celsius.

Instead, take a sequence of elements of $c_0$, indexed by say $x_{n,m}$ (so that each $x_{n,m}$ is a number and each $(x_{n,m})_n$ is in $c_0$), convergent in $\ell^\infty$. That means that for each $m$ we have $\lim_nx_{n,m}=0$ and for fixed $n$ the sequence $(x_{n,m})_m$ converges to some $y_n$, and the convergence is uniform with respect to $n$. Then you need to show that $y_n\to 0$.