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

Define $c_0$ to be the space of real sequences in $\ell^\infty$ that converge to $0$ and $\ell^\infty$ be the set of all bounded sequences. Show that $c_0$ is closed in $\ell^\infty$.

I am reading in the solution in here and I don't understand how we define $$x^{k}_n - \omega_n$$

Because $(x_n^k) = ( \{x^1_1,x_2^1,\dots \}, \{x_1^2, x_2^2,\dots \}, \dots)$ and $(\omega_n) = (\omega_1, \omega_2, \dots)$

The objects in $(x_n^k)$ are sequences (sets) and the objects in $\omega_n$ are real numbers, how can you subtract a number from a set?

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: $(x^{(k)}_n)_k$.
• As I said, this is true. So $x^{(k)}_n$ is a number. – Henno Brandsma Sep 20 '14 at 21:41
• Yes, I agree $x^{k}$ is a number. But we are dealing with the collection $(x^{k}_n)_{k \geq 1}$ which is a set of sequences. – Hawk Sep 20 '14 at 21:44
• $x^{(k)}$ is not a number, but a sequence (=point). The proof writes $|x^{(k)}_n - \omega_n|$, which is a simple absolute value of a difference between reals. Note that $k$ and $n$ are bound variables there! – Henno Brandsma Sep 20 '14 at 21:47
• I meant to say "$x_n^k$", I have a cold right now so I am making a bunch of typos. – Hawk Sep 20 '14 at 21:52