Prove $\{l_x:x\in X\}$ is a basis for $V^*$ Let $V=\mathbb{k}^X$ the vector space over $\mathbb{k}$ of functions of $X\rightarrow \mathbb{k}$. For all $x\in X$ consider the function $l_x:V\rightarrow \mathbb{k}$ defined by $l_x(f) := f(x)$. Prove that $\{l_x:x\in X\}$ is a basis for $V^*$, where $V^*$ is the dual space of $V$,
$$
V^*= End(V,\mathbb{k})$$
I don't have a clear idea of how to solve this, can someone give me a hint for this exercise?
 A: Let $I$ be an index set, then it follows from the universal property of the direct sum that
$$\newcommand{\Hom}{\operatorname{Hom}}\Hom\left(\bigoplus_{i\in I} M_i,k\right)\cong\prod_{i\in I}\Hom(M_i,k).$$
So when all the $M_i=k$, $$\left(k^{\oplus I}\right)^*\cong (k^*)^{I}=k^{I},$$
where $k^{\oplus I}$ is the direct sum of $k$ with itself indexed by $I$, and $k^{I}$ is the product of $k$ with itself $I$ times. 
Now in the question you've asked, $V=k^I$, so $V=\left(k^{\oplus I}\right)^*$, and $l_x$ can basically be regarded as a basis for $k^{\oplus I}$, so let
$W=k^{\oplus I}$.
Then essentially for $l_x$ to be a basis for $V^*$, we want to know if the natural map $W\to V^*=W^{**}$ is an isomorphism. The map $W\to W^{**}$ is always injective, and if $X$ is finite, then it's easy to see that $W\to W^{**}$ is surjective as well, since $\dim W=\dim W^*=\dim W^{**}=|X|$, and injective map between vector spaces of the same finite dimension is surjective. However if $X$ is infinite, then we have a linearly independent collection of functions $\delta_x :x\in X$, the functions that take value 1 at x and 0 everywhere else and $1$, the function that is constantly 1. Then we can extend this to a basis for $k^X$, and define an element of the dual space, $\alpha$ by $\alpha(1)=1$ and $\alpha$ is zero for all other elements of the basis. Then $\alpha$ is not in the span of the $l_x$, since $\alpha(\delta_x)=0$ for all $x$, so if $\alpha = \sum a_i l_{x_i}$, we have $\alpha(\delta_{x_i}) = a_i = 0$, but $\alpha\ne 0$. Hence the map $W\to W^{**}$ is not surjective. I.e. $l_x$ do not span $V^*$ if $X$ is infinite.
A: For simplicity, let $k$ be the two-element field. Then $|k^X|=2^{|X|}$ and so the dimension of $V$ is $2^{|X|}$ by direct computation.
An element of the dual space $V^*$ is basically an indicator function on a basis of $V$, so the cardinality (and dimension) of $V^*$ is $2^{2^{|X|}}>|X|$. Hence $\{l_x:x\in X\}$ cannot be a basis.
If your claim could be proved by only using vector space axioms, it would also be provable when $k$ is the two-element field.
The claim is true when $X$ is finite. Indeed, in this case $\dim V=|X|$ and so also $\dim V^*=|X|$. It's sufficient to prove that $\{l_x:x\in X\}$ is linearly independent. Suppose
$$
\sum_{x\in X}\alpha_xl_x=0
$$
Define, for $y\in X$,
$$
f_y(x)=\begin{cases} 1 & x=y\\ 0 & x\ne y\end{cases}
$$
Then, for every $y$,
$$
0=\sum_{x\in X}\alpha_xl_x(f_y)=\alpha_y
$$
