Norm of difference of repeated limit is zero implies existence of limit point? Let $(x_n)_{n=0}^\infty$ be a sequence in a normed vector space $X$ (not necessarily complete). Denote the norm by $\|\cdot \|$. Assume that
$$
\lim_{j\to\infty} ( \lim_{k\to\infty} \|x_j -x_k\|) = 0
$$
and
$$
\lim_{k\to\infty} ( \lim_{j\to\infty} \|x_j -x_k\|) = 0
$$
Does this imply that there is a limit point $x\in X$ such that $x= \lim_{n\to \infty} x_n$?
 A: Notice that your condition implies that $(x_n)_n$ is a Cauchy sequence. Namely, for any $\varepsilon > 0$ by definition of $\lim_{j\to\infty}$ there exists $j_0 \in \Bbb{N}$ such that
$$j \ge j_0 \implies \lim_{k\to\infty} \|x_j-x_k\| \le \frac\varepsilon2.$$
Now for any $j \ge j_0$ by definition of $\lim_{k\to\infty}$ there exists $k_0 \in \Bbb{N}$ such that $$k \ge k_0 \implies \|x_j-x_k\| \le \varepsilon$$
so for all $j,k \ge \max\{j_0,k_0\}$ we have $\|x_j-x_k\| \le \varepsilon$.
So we have to find a counterexample in an incomplete space. Consider $c_{00}$, the space of finitely supported sequences, equipped with the norm $\|\cdot\|_1$. Define
$$x_n = \left(\frac12, \frac1{2^2}, \ldots, \frac1{2^n}, 0, 0, \ldots\right).$$
Then $(x_n)_n$ satisfies your condition. Indeed, assuming $k \ge j$ since we first let $k\to\infty$, we have
\begin{align}\|x_j-x_k\|_1 &= \left\|\left(0, \ldots, 0, -\frac1{2^{j+1}}, -\frac1{2^{j+2}}, \ldots, -\frac1{2^k}, 0, 0, \ldots\right)\right\|_1\\
&= \sum_{i=j+1}^k \frac1{2^i} \xrightarrow{k\to\infty} \sum_{i=j+1}^\infty \frac1{2^i} = \frac1{2^j} \xrightarrow{j\to\infty} 0.
\end{align}
However, $(x_n)_n$ doesn't converge in $c_{00}$ because it converges to $\left(\frac1{2^n}\right)_n \in \ell^1\setminus c_{00}$.
