My textbook says that Proposition 2 can be considered as a generalization of Proposition 1 to an infinite case. I don't understand the above statement, although I understand the proofs of both propositions. Could anybody help me understand the relation between the two propositions?
Proposition 1:
Let $V$ and $W$ be $K$-linear spaces, and $x_{1}, \ldots x_{n}\in V$ be basis of $V$. For any $y = (y_1, \ldots, y_n) \in W^n$, there exists a linear map $f: V \to W$ which satisfies $f(x_i) = y_i$ for all $i\in \{1, \ldots, n\}$.
Proposition 2:
Define $K^{(X)} = \{f: X \to K| \#\{x \in X|f(x)\neq0 \}\neq \infty\}$, which is a linear space. Define $e_x \in K^{(X)}$ such that for $x, y\in X$, $e_{x}(y) = 1$ if $x=y$ and else, $e_{x}(y) = 0$. Let $V$ be a $K$-linear space. Define a map $F:\{\mathrm{Linear\,\,map\,}K^{(X)}\to V \}\ni f \to g\in\{map\, X \to V\}$ where $g(x) = f(e_{x})$. Then, $F$ is a bijective map.
(I must say that the statement and propositions are translated to English from my native language, and I hope my English is not that bad.)
Proposition $1$ states that, if $V$ is vector space with dimension $n$, i.e. if $V\simeq K^n$, we have a surjective map \begin{align} \mathscr L(V,W)&\mathrel{—}\twoheadrightarrow W^n\simeq W^{\mathcal B},\\ f &\longmapsto (f(x_1),\dots,f(x_n). \end{align} where $\mathcal B=\{x_1 \dots,x_n\}$ denotes a (finite) basis for $V.
Sticking to the same naming conventions as in Proposition $1$, Proposition $2$ asserts that for any vector space $V$ with (not necessarily finite) basis $\mathcal B$ and any vector space $W$, we have a bijective map \begin{align} \mathscr L(V,W)&\overset{\sim}{\;\mathrel{—}\longrightarrow} W^{\mathcal B},\\ f &\longmapsto \bigl(f(x)\bigr)_{x\in \mathcal B} \end{align}
In other words, the second proposition says a little more than the first: a linear map from a vector space $V$ to a vector space $W$ is entirely determined by the images in $W$ of the vectors of a basis of $V$.