Separability of $l^2(\mathbb{Z}^2)$ Let  $l^2(\mathbb{Z}^2)$ be the space of all functions(sequences) $x: \mathbb{Z}^2 \longrightarrow \mathbb{C}$  that have the property $\sum_{(m,n) \in \mathbb{Z}^2}|x(m,n)|^2 < \infty$.
The norm of this space is defined as the square root of the above sum.
Prove that this space is separable (it contains a countable dense subset)
I  have not worked with sequences of double index.
Can someone help me plase?
Thank you in advance!
 A: Let $U_d$ be the set of sequences in $l^2$ where, for any point $p\in\mathbb Z^2$, if $|p|>d$ then a sequence $a\in U_d$ has $a(p) = 0$.  Essentially, this is the set of all sequences that are constantly zero further from the origin than $d$.
There are finitely many points within $d$ from the origin.  Now, let $U_d'$ be the set of all sequences $f:l^2(\mathbb Z^2)\to\mathbb Q^2$ (where I'm viewing $\mathbb Q^2\subseteq\mathbb C$ to be a dense countable subset).  As for each of the finitely many points within the circle of radius $d$ has countable many places in the codomain it can be sent, it follows that $U_d'$ is countable.
Now, we just need to show that it's dense.  For any $X(m,n)$ in our original space, we have that:
$$\sum_{m= -\infty}^\infty\sum_{n = -\infty}^\infty|x(m,n)|^2<\infty$$
Now, it follows that given $\epsilon>0$ there exits $M,N$ such that:
$$\sum_{m \not \in [-M,M]}\sum_{n\not\in[-N,N]}|x(m,n)|^2<\epsilon$$
This is just saying the tails of these sequences die down enough for it to converge.
Let $K = \max(M,N)$  We thus have that:
$$\sum_{(m,n)\not\in[-K,K]^2} |x(m,n)|^2<\epsilon$$
Now, for each pair $(m',n')\in [-K,K]^2$, we can approximate the complex number $x(m',n')$ arbitrairly closely by a pair of rational numbers (using the density of $\mathbb Q^2\subseteq\mathbb C$).
For each of these points, let $|r(m',n')-x(m',n')|<\frac{\epsilon}{4K}$.
Let $r(m,n) = 0$ for $(m,n)\not\in[-K,K]^2$.  Then, we have that:
\begin{align*}
|r-x|_2^2 & = \sum_{(m,n)\in\mathbb Z^2} |r(m,n)-x(m,n)|^2 \\
& = \sum_{(m,n)\in[-K,K]^2} |r(m,n)-x(m,n)|^2 +\sum_{(m,n)\not\in[-K,K]^2} |0-x(m,n)|^2 \\
& \leq \sum_{(m,n)\in [-K,K]^2} \frac{\epsilon}{4K}+\sum_{(m,n)\not\in[-K,K]^2} |x(m,n)|^2 \\
& < \epsilon+\epsilon = 2\epsilon
\end{align*}
So, we've approximated $x$ arbitrairly closely by $r$, which is a member of a countable subset $U' = \cup_{d\in\mathbb N} U_d'$.
This all might seem a little contrived, but the basic idea is simple:
1. Use that $|x|_2 < \infty$ to say that outside some things near the origin, everything has to be less than $\epsilon$ (so we can ignore it).
2. For the finitely many things that are close enough to the origin to matter, approximate each of them very closely (using that $\mathbb Q^2$ is dense in $\mathbb C$).
3. Break $||x-r||_2^2$ into the two parts - the "approximate closely" and the "is far enough out so is less than $\epsilon$" part), and apply the approximation to get that $||x-r||_2^2 <\epsilon$.
This is generically how you show a $l^2$ space is separable.
A: Let $d_{m,n}$ be defined as $1$ at $(m, n)$ and zero elsewhere.  These elements form a countable (closed) spanning set.  If a normed linear space has a countable spanning set, it is separable (The proof of this is an fairly straightforward exercise).
