Subsets implies subspace in free vector space Let $X$ and $Y$ be sets such that $Y \subseteq X$. Let $\mathbb{F}$ to be a field. Denote $\mathbb{F} \langle S \rangle$ be a subspace of $Fun(S, \mathbb{F})$ (all function $f: S \to \mathbb{F}$ such that $f^{-1}[\mathbb{F} - \{0\}]$ is finite) spanned by the characteristic function $\chi_{s}$, where
$$
\begin{equation*}
\chi_s(x) := \begin{cases}
    1, \, x = s \\
    0, \, x \ne s
  \end{cases}
\end{equation*}
$$
This means that $k \in \mathbb{F} \langle S \rangle \iff k = \sum^{n}_{i = 1}a_i \chi_{s_i}$ and $s \mapsto \sum^{n}_{i = 1}a_i \chi_{s_i}(s)$ for all $s \in S$, with $a_i \in \mathbb{F}$ and $1 \le i \le n$.
Now, for simplicity, I will denote $\mathbb{F}_S := \mathbb{F} \langle S \rangle$ from now on. This $\mathbb{F}_S$ is a free vector space over $\mathbb{F}$ generated by $S$.
My question is, if $Y \subseteq X \Longrightarrow \mathbb{F}_Y$ is a subspace of $\mathbb{F}_X$?
 A: Morally, it's the other way around! The set of finitely supported functionals on $X$ is bigger and naturally maps onto the set of finitely supported functionals on $Y$:
$$
Y \subseteq X 
\implies 
\mathbb{F}_X \twoheadrightarrow \mathbb{F}_Y
$$
The map is given by restriction: $f \mapsto f\big|_Y$ for any $f \in \mathbb{F}_X$. You have to check that it is well-defined. In particular, you have to verify that $f \mapsto f\big|_Y$ also has finite support, but that's straightforward.
Why is this restriction map surjective? Take any $g \in \mathbb{F}_Y$. In order to find an extension (a preimage $\tilde{g} \in \mathbb{F}_X$ such that $\tilde{g}|_Y = g$), we need to define $\tilde{g}$ on $X \setminus Y$. Here's where the fact that vector spaces are free objects come in handy: we can define the set map however we like. One way is to define $\tilde{g}:X \to \mathbb{F}$ by
$$
\tilde{g}(x) = \begin{cases}
g(x), & x \in Y, \\
0,    & x \in X \setminus Y.
\end{cases}
$$
Note: even though, given $g \in \mathbb{F}_Y$ we've constructed a corresponding $\tilde{g} \in \mathbb{F}_X$ such that $\tilde{g}|_Y = g$, it involved choices where there were many possibilities for $\tilde{g}$, so it's not natural to identify $g \mapsto \tilde{g}$ in the other direction. Moreover, the map is not injective unless $X = Y$, so it's certainly not reasonable to think of of the functionals on $Y$ as a subset of functionals on $X$.
A: Formally not, since elements of $\mathbb F_Y$ are functions $f\colon Y \to \mathbb{F}$ and elements of $\mathbb F_X$ are functions $f\colon X \to \mathbb{F}$. Therefore if $X\neq Y$ then $\mathbb F_X\cap \mathbb F_Y=\emptyset$.
On the other hand we can identify each $f\colon Y \to \mathbb{F}$ with the function $f'\colon X \to \mathbb{F}$ defined by $$f'(x)=\begin{cases}f(x),&x\in Y\\ 0,&x\in X\setminus Y.\end{cases}$$ This identification is a monomorphism $\mathbb F_Y\to \mathbb F_X$ and therefore it's justifiable to say $\mathbb F_Y\subset \mathbb F_X$ as long as we know this identification.
