How to put this into set notation? if you could please help me put this sentence into set notation, examine my (flawed?)attempts, and tell me where I went wrong with them?
So we call a function $f\colon \mathbb{N} \to \mathbb{N}$ zero almost everywhere iff $f(n)=0$ for all except a finite number of arguments. 
My question is "how do we represent the the set of functions that are zero almost everywhere?" 
Here is my attempt:
$E=\{f\colon \mathbb{N} \to \mathbb{N}\mid f(n)=0\}$
Thank you!
 A: If I had to write it symbolically I would express "$f$ is zero cofinitely often" (or "$f$ has finite support") as
$$ |f^{-1}(\mathbb N\setminus\{0\})| < \infty $$
(or perhaps $<\aleph_0$ if I felt a need to look more sophisticated). The set of such functions would therefore be
$$ \big\{f:\mathbb N\to\mathbb N\mathrel{\big|} |f^{-1}(\mathbb N\setminus\{0\})| < \infty \big\}$$
However, I would seriously consider if I could get away with describing the set in words instead;; that would be rather more readable.
A: The basic approach would be:
$$E = \{f : \mathbb{N} \to \mathbb{N} \mid f \text{ is zero almost everywhere}\}.$$
To make it shorter (although not necessairly more readable), we can recall that the set of inputs for which the function is non-zero is often called support. Using this we get:
\begin{align}
E &= \{f : \mathbb{N} \to \mathbb{N} \mid f \text{ is of finite support}\}\\
E &= \{f : \mathbb{N} \to \mathbb{N} \mid \operatorname{card}\big(\operatorname{supp}(f)\big) < \infty \}
\end{align}
We could complicate this even fruther by observing that a finite sum of finite elements is finite, hence
$$E = \Big\{f : \mathbb{N}\to\mathbb{N} \ \Big|\  \sum_{k \in \mathbb{N}}f(k) < \infty \Big\},$$
but then, we are in the domain of natural numbers, so
$$E = \Big\{f : \mathbb{N}\to\mathbb{N} \ \Big|\  \lim_{k \to \infty}f(k) = 0 \Big\}$$
or inverting this relation
\begin{align}
E &= \Big\{f : \mathbb{N}\to\mathbb{N} \ \Big|\ \sup\big(\mathbb{N}\setminus f^{-1}(0)\big) < \infty \Big\},\\
E &= \Big\{f : \mathbb{N}\to\mathbb{N} \ \Big|\  \operatorname{card}\big(\mathbb{N}\setminus f^{-1}(0)\big) < \infty \Big\}, \\
E &= \Big\{f : \mathbb{N}\to\mathbb{N} \ \Big|\ \exists k \in \mathbb{N}.\ f(\mathbb{N} + k) = \{0\} \Big\}.
\end{align}
And then you could use the notion of domination, that is, let $g_k(n) = \max(k-n,0)$, then
\begin{align}
E &= \Big\{f : \mathbb{N}\to\mathbb{N} \ \Big|\ \exists k \in \mathbb{N}.\ f \leq g_k \Big\}, \\
E &= \Big\{f : \mathbb{N}\to\mathbb{N} \ \Big|\ g_0 \text{ eventually dominates } f \Big\}.
\end{align}
I hope this helps $\ddot\smile$
A: Your $E$ doesn't ever consider the "for all except a finite number of arguments". I think the most accessible answer (although perhaps not the one you are fishing for) is simply: 
$E=\{f:\mathbb{N} \to \mathbb{N} | \exists S \subset \mathbb{N}$ where $|S|<\infty$ with $f(n) \neq 0$ for $n \in S$ and $f(n)=0$ otherwise}
