Q:

Understanding $c_0$ is closed subset in $\ell^{\infty}$

Prove the set of sequences $c_0$ which converge to zero in $l_{\infty}$ is closed. I came across this proof and have a question: Since all $x_n(k)$ is in $c_0$, it's limit point $x(k)$ should be 0, now we only need to show that the sequence of all zeros is in $c_0$ to show that $c_0$ is closed. Is my understanding correct? Any help would be helpful.

A:

To show a subset $Y$ of a normed linear space $X$ (or metric space, more generally) is closed, we need to show the following:

If $y_n$ is a sequence of points in $Y$, such that $y_n$ converges to $x$ in $X$, then $x \in Y$.

What's confusing here is that, in this case $X$ and $Y$ consist of sequences themselves. In particular, they consist of sequences of real numbers: functions from $\Bbb{N}$ to $\Bbb{R}$ (or $\Bbb{C}$, but I'm just going to consistently use $\Bbb{R}$). Each point is a real sequence, and we will be considering sequences of sequences.
In our case we have
\begin{align*}
X &= \ell^\infty = \{\text{Real sequences that are bounded}\} \\
Y &= c_0 = \{\text{Real sequences that converge to } 0\}.
\end{align*}
So, in order to prove $c_0$ is closed in $\ell^\infty$, we assume that we have a sequence of points $(x_n)_{n=1}^\infty$ in $c_0$ that converge to $x$. To clarify, $x$ and $x_1$ and $x_2$ and $x_3$, etc, are each real sequences. We are assuming that $x_1$ and $x_2$ and $x_3$, etc, all converge to the real number $0$. We are not assuming this of $x$; that $x$ limits to $0 \in \Bbb{R}$ is precisely what we want to prove.
The notation used in the linked question is important to clarify too. The $k$th entry of a point (read: real sequence) $y$ (in $\ell^\infty$ and/or $c_0$) is denoted $y(k)$. As we are considering a sequence of points $(x_n)_{n=1}^\infty$, be prepared to see notation like $x_n(k)$, referring to the $k$th entry (a real number) of the $n$th point in the sequence $(x_n)$.
Now, let's quote the linked question:

Let $\varepsilon>0$.  Then there exists an $N>0$ such that 
  $$\parallel x_N-x\parallel_{\infty}:=\sup_{k\in\mathbb{N}}|x_N(k)-x(k)|\leq\varepsilon.$$
Then we have,
  \begin{align}
|x(k)| &= |x(k) - x_N(k) + x_N(k)| \\\\
&\leq |x_N(k) - x(k)| + |x_N(k)| \\\\
&\leq \varepsilon + |x_N(k)|\rightarrow \varepsilon\;\; \text{as}\;\; k\rightarrow\infty. 
\end{align}
Therefore since $\varepsilon$ was chosen arbitrarily we can conclude that $x(k)\rightarrow0$ and thus that $x(k)\in c_0$.

In the above argument, the point was to show that $x$, the limit (in $\ell^\infty$) of the sequence of points $x_n$ in $c_0$, was also in $c_0$. In particular, the membership requirement of $c_0$ is that $x$, a sequence of real numbers, must converge to $0$. That is, $x(k) \to 0$ as $k \to \infty$. Hopefully you can see that the above argument proves this is the case.
To answer your question specifically, no, the limit $x$ does not need to be the $0$ sequence. It just has to be a sequence (of real numbers) that converges to the real number $0$.
To illustrate this, consider an example. Let
$$x_n(k) = \frac{n}{nk+1}.$$
So, for example, the first point $x_1$ is
$$x_1 = \left(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots \right).$$
Note this converges to $0$. The second point $x_2$ is
$$x_2 = \left(\frac{2}{3}, \frac{2}{5}, \frac{2}{7}, \frac{2}{9}, \ldots \right),$$
which also converges to $0$. Similarly
\begin{align*}
x_3 &= \left(\frac{3}{4}, \frac{3}{7}, \frac{3}{10}, \frac{3}{13}, \ldots \right) \\
x_4 &= \left(\frac{4}{5}, \frac{4}{9}, \frac{4}{13}, \frac{4}{17}, \ldots \right),
\end{align*}
etc. Note that each of these points is a sequence of real numbers, approaching the real number $0$.
Now, let $x(k) = \frac{1}{k}$, that is
$$x = \left(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots \right).$$
I claim that $x$ is the limit (in $\ell^\infty$) of $(x_n)_{n=1}^\infty$. That is,
$$\|x_n - x\|_\infty \to 0$$
as $n \to \infty$. Note that,
\begin{align*}
\|x_n - x\|_\infty &= \sup_{k \in \Bbb{N}} |x_n(k) - x(k)| \\
&= \sup_{k \in \Bbb{N}} \left|\frac{n}{nk + 1} - \frac{1}{k}\right| \\
&= \sup_{k \in \Bbb{N}} \left|\frac{nk - (nk - 1)}{k(nk + 1)}\right| \\
&= \sup_{k \in \Bbb{N}} \frac{1}{k(nk + 1)}.
\end{align*}
The supremum is attained at $k = 1$, since this expression is decreasing with respect to $k$, hence
$$\|x_n - x\|_\infty = \frac{1}{n+1} \to 0$$
as $n \to \infty$. Thus, $x_n \to x$ in $\ell^\infty$.
The quoted argument proves now that $x(k) = \frac{1}{k} \to 0$ as $k \to \infty$. Note that $x$ is not the constant $0$ sequence, just a real sequence that (by necessity) limits to the real number $0$.

