Prove that $c_0$ is closed in $\ell^\infty$

Solution 1:

No, $x^{(k)}$ is assumed to be a sequence of points of $c_0$. Here $k$ is the index. So $x^{(1)}$ is a sequence (namely a point in $c_0$), which has itself indices, which are denoted by $n$. So we write $x^{(1)} = (x^{(1)}_0, x^{(1)}_1,x^{(1)}_2,\ldots)$ .So every $x^{(k)}_n$ is indeed some real number: the $n$-th member of the sequence that is the point $x^{(k)}$ of $c_0$.

Note that there is a difference between writing $x^{(k)}$ (a single point, i.e. a single sequence), and $(x^{(k)})$, i.e. a sequence of points (sequences); the latter should really be written $(x^{(k)})_k$, so show the index set (the one that runs), especially when we make the indices of each sequence point explicit: $(x^{(k)}_n)_k$.