The set $F$ of all functions $f:\Bbb{N}\to \{0,1\}$ that are "eventually zero" is countable Prove or disprove the set $F$ of all functions $f:\Bbb{N}\to \{0,1\}$ that are ''eventually zero'' is countable.
For each $n\in \Bbb{N}$, let $F_n = \{f: \Bbb{N}\to\{0, 1\}:f(i) = 0 \forall i > n\}$. Then it is easy to see that $F_n$ is finite.
How to prove rigorously?
function $f$ is eventually zero means $f(n)=0$ $\forall$ $n\geq N$, $N\in \Bbb{N}$.
Define  a map $\psi: \Bbb{N} \to \{0,1\}^\Bbb{N}$ such that
$\psi(i)=\{f(i):i\in\{0,2,...,n-1\}$
$\psi(i)=\{0:i\notin\{0,2,...,n-1\}$.
Does this map work?
Any help will be appreciated.
 A: Just writing numbers from $\mathbb{N}$ in binary effectively identifies every number in $\mathbb{N}$ with a function of the form you describe. Conversely every such function is basically telling you how to write down a binary number, from right to left.
For example, suppose $$\begin{align}f:{}&1\to0\\&2\to1\\&3\to0\\&4\to0\\&5\to1\end{align}$$ and all else to $0$. This identifies with the binary integer $$\stackrel{5}{1}\stackrel{4}{0}\stackrel{3}{0}\stackrel{2}{1}\stackrel{1}{0}$$ which is $18$ in binary. So this $f\leftrightarrow 18$.
And for example, start with the number $23$, which is $10111$. This defines $$\begin{align}g:{}&1\to1\\&2\to1\\&3\to1\\&4\to0\\&5\to1\end{align}$$ So $g\leftrightarrow23$.
More formally, there is a map $\varphi:\mathbb{N}\to F$, such that if $n$ is $b_kb_{k-1}\cdots b_1$ in binary, then $\left(\varphi(n)\right)(m)=b_m$, where $b_m=0$ for $m>k$.
And there is $\varphi^{-1}:F\to \mathbb{N}$ where $\varphi^{-1}(f)=f(k)\cdots f(2)f(1)$ (read as a binary concatenation) where $k$ is the largest number for which $f$ returns $1$.
So there is a very direct enumeration of these functions, which makes them countable.
A: The goal was to establish a bijection between the set of maps and a countable set, but you proposed a map from $\mathbb N \rightarrow \{0,1\}^\mathbb N$.  The second one is uncountable, so it’s unclear how that’s going to help.
Anyhow, the thing that occurs to me first is to consider this following. Enumerate the prime numbers in $\mathbb N$ as $p_0,p_1,p_2,\ldots$ and so on.  Then consider the map $\theta$ that takes $\theta(f)=\prod_{i=0}^\infty p_i^{f(i)}$. This is well defined since the sequences are eventually $0$.  Show $\theta$ is an injection of that set of functions into the positive integers.
In fact, the larger set of functions $\mathbb N\to\mathbb N$ which are eventually zero is bijective with the positive integers using this map. (This is just the fundamental theorem of arithmetic.)
By only considering the functions into $\{0,1\}$ you are obtaining the restriction of the map and its image is onto the squarefree positive integers.
A: First of all note that If $f:\mathbb{N}\to \{0, 1\}$ is eventually zero, it means that there is some $M$ such that $f(i)=0$ for all $i>M.$ Therefore, $$\{f| f:\mathbb{N}\to \{0, 1\}, f \text{is eventually zero}\}\subseteq \bigcup_{M\in \mathbb{N}} \{f: f(i)=0 \hspace{2mm} \forall i>M\}=:\mathcal{F}_M.$$
It is now clear that it suffices to show that for each $M\in \mathbb{N}$ we have that $\{f: f(i)=0 \hspace{2mm} \forall i>M\}$ is at most countable. But it is easily seen that $|\mathcal{F}_M|\le 2^M.$
A: Let's denote by $\{0,1\}^\infty$ to the set of all functions from $\mathbb N$ to $\{0,1\}$ that are eventually zero.
Then, for $f \in \{0,1\}^\infty$ let $n_f := \min\{N \in \mathbb N : \forall n \geq N (f(n) = 0)\}$ and consider the map
\begin{align}
\{0,1\}^\infty & \to \bigcup_{n \in \mathbb N} \{0,1\}^n \\[1mm]
f & \mapsto (f(1),\dots,f(n_f)) \in \{0,1\}^{n_f}
\end{align}
with the convention that $\{0,1\}^0 = \{0\}$.
