Either show $c_0=\{x=(x_1,x_2,x_3,...) | x_i\in \mathbb{C}, x_i \rightarrow 0\}$ is complete or closed under supremum norm.

Question

Let

$l^\infty =\{x=(x_1,x_2,x_3,...)|x_i \in \mathbb{C}, \|x\|_\infty=\sup_{i\in \mathbb{N}}{|x_i|}<\infty\}$.

and

$c_0=\{x=(x_1,x_2,x_3,...) | x_i\in \mathbb{C}, x_i \rightarrow 0\}$.

So $l^\infty$ is the set of all infinite sequences of complex numbers such that the supremum norm is finite. And $c_0$ is the set of all infinite sequences of complex numbers which converges to 0.

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Consider $l^\infty$ with the metric $p(x,y)=\|x-y\|_\infty=\sup_{i\in \mathbb{N}}|x_i-y_i|$.

We know that $l^\infty$ is a complete, so that $l^\infty$ is a Banach Space, i.e. every Cauchy sequence in $l^\infty$ has a limit in $l^\infty$.

$(l^\infty, p)$ is a metric space and $c_0 \subseteq l^\infty$. We know that if $l^\infty$ is complete (which it is), then

$c_0$ is complete if and only if $c_0$ is closed.

Question: How could we either show $c_0$ is complete or closed?

Answer 1

Let $x$ in the closure of $c_0$ for $\|\ \|_\infty$. For every $u\gt0$, there exists $y$ in $c_0$ such that $\|x-y\|_\infty\leqslant u$. Since $y$ belongs to $c_0$ there exists $n_u$ such that $|y_n|\leqslant u$ for every $n\geqslant n_u$. Since $|x_n|\leqslant|y_n|+|x_n-y_n|$ this implies that $|x_n|\leqslant|y_n|+\|x-y\|_\infty\leqslant2u$ for every $n\geqslant n_u$. This holds for every $u\gt0$ hence $x_n\to0$ when $n\to\infty$, that is, $x$ belongs to $c_0$.