Show $\{e^{(k)}:k \geq 1\}$ is a closed set in $(\ell_1, \| \cdot \|_1)$ I have seen the following problem floating around on Math.SE and, coincidentally in my textbook and decided to take a shot at it.

Let $e^{(k)}=(0, \ldots, 0, 1, 0, \ldots)$ where the $k$th entry is $1$ and the rest are $0$s. Show that $\{e^{(k)}: k\ge 1\}$ is closed as a subset of $\ell_1$.

Show $\{e^{k}:k \geq 1\}$ is a closed set in $(\ell_1, \| \cdot \|_1)$. Note when they say in the problem that $e^k = (0,0,0,...1,0,0,0...)$ where the $k^{th}$ entry is $1$ and the rest are $0$s, they mean: $e^1 = (1,0,0,0,...), e^2 = (0,1,0,0,...), e^3 = (0,0,1,0,...)$ and so on. Now, observe the following computations using the $1$-norm: $\| e^2 - e^1 \|_1 = 2$, obviously $\|e^1 - e^1\|_1 = 0$, and lastly where $0_{\ell_{1}} = \{0,0,0,0,..\}$ (the zero sequence in $\ell_1$) $\|e^k - 0_{\ell_{1}}\|_1 = 1$. We know that any subset $L$ of a (arbitrary) metric space $M$ is closed if it contains all its limit points. Meaning, if a sequence in $L$ converges to a point in $M$ then that point is also in $L$.
So, let $\{x_m\} \subset \{e^{k}:k \geq 1\}$ and $\{x_m\} \subset \ell_1$. We want to show that if $x_m \to x_n$, $x_n \in \{e^{k} :k \geq 1\}$. Notice that $$\{x_m\} := \{e^1 = (1,0,0,0,...), e^2 = (0,1,0,0,...), e^3 = (0,0,1,0,...),...\}$$ So if $x_m \to x_n$ Then $d(x_m,x_n) =0$ which we know means that $x_m = x_n$ so $x_n$ has to be in $\{e^{k}:k \geq 1\}$. So these sequences converge coordinate-wise and thus $\{e^{k}:k \geq 1\}$ is closed.
Is my solution correct? Any criticism is welcome.
 A: You say let $\{x_m\} \subset \{e^{k} : k \geq 1\}$, but later you say $\{x_m\} := \{e^1, e^2, e^3, \dots\}$ (so I guess $x_1 = e^{1}$ and $x_{2} = e^{2}$ and so on?). Assuming the rest of your logic is correct (which is not clear to me), you are only showing that one particular convergent sequence in $\{e^k\}$ has a limit in $\{e^k\}$, where you need to show that every convergent sequence in $\{e^k\}$ has a limit in $\{e^k\}$. Also, if I'm right in saying that $x_1 = e^1$ etc., then $\{x_m\}$ doesn't converge in $l_1$ to begin with.
Instead, let $\{x_m\}$ be a sequence in $\{e^k\}$. It's important to remember that each $x_m$ is itself a sequence of real numbers. In this particular case, each $x_m$ is a sequence of all 0's with exactly one 1 somewhere in there, but that's about all we can say.
First, suppose that $\{x_m\}$ is eventually constant, say $x_m = e^{j}$ for all large enough $m$; note that the $j$ here is fixed. Then $\lim_{m \to \infty} x_m = e^{j}$ (try to work these details out for yourself if it's not immediately clear), so the limit of $\{x_m\}$ is in $\{e^k\}$, as desired.
We may now assume that $\{x_m\}$ is not eventually constant. Suppose that $\lim_{m \to \infty} x_m = y$ for some $y = (y_1, y_2, y_3, \dots) \in l_{1}$. This means that for all $\epsilon > 0$, there exists some $N \in \mathbb{N}$ such that $||x_m - y||_{l_1} < \epsilon$ for all $m \geq N$. In particular, $||x_N - y||_{l_{1}} < \epsilon$. Letting $n$ be the index where $x_{N}$ has a $1$,
$$
||x_{N} - y||_{l_{1}} = |y_n - 1| + \sum_{i \neq n} |y_i| < \epsilon
$$
Since $\{x_m\}$ is not eventually constant, there is some $N' > N$ such that $x_{N'} \neq x_{N}$. Let $n'$ the index where $x_{N'}$ has a $1$ (and since $x_{N'} \neq x_{N}$, we know that $n' \neq n$), so
$$
||x_{N'} - y||_{l_{1}} = |y_{n'} - 1| + \sum_{i \neq n'} |y_{i}| < \epsilon
$$
However, for small $\epsilon$, these two inequalities are incompatible. Namely, for $\epsilon < \frac{1}{2}$, the first inequality implies that $|y_{n'}| < \frac{1}{2}$. Equivalently,
$$
\begin{aligned}
&-\frac{1}{2} < y_{n'} < \frac{1}{2} \\
\implies & -\frac{3}{2} < y_{n'} - 1 < -\frac{1}{2} \\
\implies & |y_{n'} - 1| > \frac{1}{2} > \epsilon
\end{aligned}
$$
This contradicts the second inequality.
In summary, a sequence $\{x_m\} \subset \{e^k\}$ converges if, and only if, it is eventually constant, in which case the limit is in $\{e^k\}$. Hence, $\{e^k\}$ contains all of its limit points and is therefore closed.
