Proof that a sequence subspace is perfect (or not) Let $S = \prod_{k=1}^\infty \left\{0, \frac{1}{2^k} \right\} \subset (\ell_1, \|\cdot\|_1)$. I want to find out whether or not $S$ is perfect. I think that $S$ is perfect, and I seem to know how to prove that $S$ contains no isolated points, but I'm having trouble proving that $S$ is also closed.

Part I. $S$ has no isolated points.

Suppose there is a sequence $x=(x_k)_k \in S$ such that $x$ is isolated in $\ell^1$. Then $\exists \varepsilon>0$ such that $B_\varepsilon\setminus \{x\}\cap S=\emptyset$. Let $\varepsilon>0$ and consider $y=(y_k)_k = \begin{cases} 
      0 & x_k= 0 \\
      \frac{1}{2^K} & \mbox{otherwise} 
   \end{cases},$
where $K\in\mathbb{N}$ is chosen such that $k\ge K\implies \frac{1}{2^k}<\frac{\varepsilon}{2(K-1)}$. Also, since $x, y\in \ell^1$, $\forall \varepsilon>0, \exists K' \in\mathbb{N}$ such that $k\ge K'\implies \sum_k|x_k-y_k|<\varepsilon/2$. Let $M:=\max\{K, K'\}$, then
$$\|x-y\|_1=\sum_k |x_k-y_k|=\sum_{k=1}^{M-1}|x_k-y_k|+\sum_{k=M}^\infty |x_k-y_k|<(K-1)\frac{\varepsilon}{2(K-1)}+\frac{\varepsilon}{2}=\varepsilon.$$
Thus $S$ has no isolated points (by contradiction).

Now, for the part with the closedness of $S$.

Suppose $S$ is not closed, then $\ell^1\setminus S$ is not open. Then there exists $p=(p_k)_k\in \ell^1\setminus S$ such that for every $\varepsilon >0$, $B_\varepsilon(p)\cap S\ne \emptyset\ne B_\varepsilon(p)\cap \ell^1\setminus S$...
That's where I'm stuck. I was entertaining the idea that there are sequences in both $S$ and $\ell^1\setminus S$ that converge to the constant sequence $\{0,\dots,0,\dots\}$. Is this right? Then it should be true that neither $S$ nor $\ell^1\setminus S$ is open. Then $S$ is not perfect.
Please correct me if I'm wrong.
 A: Just to make sure, $S$ is defined as:
$$S = \left\{(x_n)_{n=1}^\infty \in \ell^1 : x_n \in \left\{0, \frac{1}{2^n}\right\}, \forall n \in \mathbb{N}\right\}$$
In your proof that $S$ has no isolated points, it seems that $y = x$, since you defined $y_n = 0$ if $x_n = 0$ and $y_n = \frac{1}{2^n}$ if $x_n \ne 0$, that is $x_n = \frac{1}{2^n}$.
To prove that $S$ has no isolated points take $x = (x_n)_{n=1}^{\infty} \in S$ and $\varepsilon > 0$. Let $k \in \mathbb{N}$ be such that $\frac{1}{2^k} < \varepsilon$. Define $y = (y_n)_{n=1}^\infty S$ as:
$$y_n = \begin{cases}
x_n,  & \text{if $n \ne k$} \\
\frac{1}{2^n} - x_n, & \text{if $n = k$}
\end{cases}$$
We have:
$$\|y - x\|_1 = |y_k - x_k| = \left|\frac{1}{2^k} - 2x_k\right| \le \frac{1}{2_k} < \varepsilon$$
So, $y \in \big(B(x, \varepsilon)\setminus \{x\}\big) \cap S$. Thus, $S$ has no isolated points so $S \subseteq S'$.
To prove that $S$ is closed notice that convergence in $\|\cdot\|_1$ implies pointwise convergence. Assume $x^{(n)} \xrightarrow{\|\cdot\|_1} x$, where $\left(x^{(n)}_k\right)_{k=1}^\infty$ for $n \in \mathbb{N}$ and $(x_k)_{k=1}^\infty$ are sequences in $\ell^1$:
$$\left|x_k - x^{(n)}_k\right| \le \left\|x - x^{(n)}\right\|_1 \xrightarrow{n\to\infty} 0, \quad \forall k \in \mathbb{N}$$
Now let $\left(x^{(n)}\right)_{n=1}^\infty$ be a sequence in $S$ which converges to $x \in \ell^1$. By the above observation, $\left(x^{(n)}\right)_{n=1}^\infty$ must also converge pointwise to $x$. Since $x^{(n)}_k \in \left\{0, \frac{1}{2^k}\right\}$ every coordinate sequence $\left(x^{(n)}_k\right)_{n=1}^\infty$ must be eventually constant. Thus, $x_n \in \left\{0, \frac{1}{2^k}\right\}$ so $x \in S$.
Thus, $S$ is closed so $S' \subseteq S$. This gives $S' = S$ so $S$ is perfect.

$\left(x^{(n)}\right)_{n=1}^\infty$ is a sequence of vectors in $\ell^1$, where each vector $x^{(n)}$ is itself a sequence of complex numbers. I used the superscript rather than a subscript to avoid confusion with $x_n$ which usually represents the $n$-th coordinate of a sequence $x$. Anyway, for each of the vectors $x^{(n)}$ let's denote its $k$-th coordinate as $x^{(n)}_k$:
$$x^{(n)} = \left(x^{(n)}_1, x^{(n)}_2, x^{(n)}_3, \ldots\right) = \left(x^{(n)}_k\right)_{k=1}^\infty$$
Now, for a fixed $k \in \mathbb{N}$ you can consider the sequence $\left(x^{(1)}_k, x^{(2)}_k, x^{(3)}_k, \ldots\right) = \left(x^{(n)}_k\right)_{n=1}^\infty$, which is the sequence of $k$-th coordinates of sequences $x^{(n)}$, as $n$ varies. This is usually called a coordinate sequence.
Suppose that $x^{(n)} \xrightarrow{\|\cdot\|_1} x$, i.e. the sequence of vectors $\left(x^{(n)}\right)_{n=1}^\infty$ converges to a vector $x \in \ell^1$ in $\|\cdot\|_1$ norm. I showed that for a fixed $k \in \mathbb{N}$ we then also have that the coordinate sequence $\left(x^{(n)}_k\right)_{n=1}^\infty$ converges (as a sequence in $\mathbb{C}$) to the $k$-th coordinate of $x$, i.e. $x^{(n)}_k \xrightarrow{n\to\infty} x_k$.
In our example, since $x^{(n)}_k \in \left\{0, \frac{1}{2^k}\right\}$, $\forall n \in \mathbb{N}$, we must also have $x_k \in \left\{0, \frac{1}{2^k}\right\}$. This means that $x \in S$, so each limit of a convergent sequence in $S$ must also necessarily be in $S$.
