How to put this into set notation?

Question

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!

Answer 1

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 $\ddot\smile$

Answer 2

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.

Answer 3

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}