Let $F:={\mathbb{Z_2}}$ be the field with two elements and let $F^{\mathbb{N}}:=\{f:\mathbb{N}\to\mathbb{Z_2}:$ with usual pointwise addition and multiplication by scalers $\}$
Then define
$V:=\{f\in F^{\mathbb{N}}:\operatorname{supp}(f)$ is finite $\}$ where $\operatorname{supp}(f)=\{n\in \mathbb{N}:f(n)=1\}$
Now we can think about the elements of V as seqeunces alternating between $0$ and $1$.
We define the dual space $V'$ as
$V'=\{\phi:V\to \mathbb{Z_2}:$ $\phi$ is a linear functional $\}$
So far I have shown that if we define for $S\subset \mathbb{N}$, $\phi_S(f):= \sum_{n\in S}^{}f(n)$ then $V'=\{\phi_S: S\subset \mathbb{N}\}$
Would anyone be able to help me show that $V$ is countable and $V'$ is uncountable. I am hoping what I have shown already can help show $V'$ is uncountable, if I am correct in that there are an uncountable number of subsets of $\mathbb{N}$ then we have already shown $V'$ is uncountable.
EDIT: As the set $\{S:S\subset \mathbb{N}\}$ is uncountable so is $V'$
EDIT#2: How about $\forall f \in V$, $|\operatorname{supp}(f)|=n\in \mathbb{N}$
So let us call $V=\{f:|\operatorname{supp}(f)|=1\}\cup\{f:|\operatorname{supp}(f)|=2\}\cup\cdots\cup\{f:|\operatorname{supp}(f)|=n\}\cup\cdots$ this forms a countable number of subsets of $V$
So by considering $\{f:|\operatorname{supp}(f)|=n\}$, the set of functions with have $n$ such values such that $f(x)=1$, if we can find a way to make the ways $n$ values can by arranged over $\mathbb{N}$ countable then $V$ must also be countable?
EDIT#3: Could we do this inductively? Take n=1 and define,
$g:\mathbb{N} \to \{f:|\operatorname{supp}(f)|=1\}$ where $g(n)=f(x)$ where $f(x)=1$ for $x=n$ and $f(x)=0$ for $x\neq n$
Therefore countable for $n=1$ assume countable for $n$.
Define $S_i:=\{f:|\operatorname{supp}(f)|=i\}$
By assumption $\exists g:\mathbb{N} \to S_n$ then $\forall f \in S_{n+1}$ we can write
$f=f_n+f_1$ for some $f_n \in S_n$ and $f_1 \in S_1$ as $S_1$ is countable and by assumption $S_n$ is countable so is $S_{n+1}$, therefore $S_n$ is countable for $n\in \mathbb{N}$. Therefore as the set $\{S_1,S_2,\dots\}$ is countable so is $V$?