Let $l^{1}$ be the space of real sequences with the usual metric. Consider the sequence $(x_{n})_{i=1}^{\infty}$ where we define for $x_{i}$, $x_{i}^{n}=1+\frac{1}{i}$ if $n=i$ and $x_{i}^{n}=0$ otherwise. I want to know if the set $(x_{n})_{i=1}^{\infty}$ is closed or not.
My argument was thinking of showing that any sequence in $(x_{n})_{i=1}^{\infty}$ that converges is ultimately constant (else, we can show that the sequence is not Cauchy, and hence will not be convergent since $l^{1}$ is complete). Then, clearly any convergent sequence from the set will have that its limit belongs to the set (since its constant).
Would my argument be correct? Maybe I'm missing something important.