I'm trying to show that the map $\Phi :V\to V^{**}$ given by $\Phi (v)(f):=f(v)$ is well-defined. To show that I need to prove that for all $v\in V$ there's $F\in V^{**} $ such that $F(f)=f(v)$ for all $f\in V^*$. However I don't know how to prove that in the infinite dimensional case!
My question: Suppose that $V$ is an infinite dimensional vector space. How can I show that for all $v\in V$ there's $F\in V^{**} $ such that $F(f)=f(v)$ for all $f\in V^*$?
Below there's a lemma that I'm trying to use to prove the question above.
Lemma: Let $V$ be a vector space. Suppose that $\{v_i\}_{i\in I}$ is a base of $V$. Then there's a unique linearly independent subset $\{v^i\}_{i\in I}$ of $V^*$ such that $v^i(v_j)=\delta _{ij}$ for all $i,j\in I$ in which $\delta $ is the Kronecker delta.
EDIT:
I'll clarify my question:
Given two sets $A$ and $B$, we say that $f\subseteq A\times B$ is a map from $A$ to $B$ (which we denote by $f:A\to B$) if $(\forall x\in A)(\exists !y\in B)\left((x,y)\in f\right)$ is a true proposition.
So, my question is to prove that $\Phi :=\big\{(v,F)\in V\times V^{**}:(\forall f\in V^*)\left(F(f)=f(v)\right)\big\}$ is a map from $V$ to $V^{**}$. That is, I want to prove that $(\forall v\in V)(\exists! F\in V^{**})\left((v,F)\in \Phi \right)$ is a true proposition.
Suppose that $V$ is a vector space over a field $k$. Take $v \in V$. Then, if you define a map $\Phi(v) : V^* \to k$ by $$(\forall f \in V^*) \quad \Phi(v)(f) = f(v)$$ note that $\Phi(v)$ is well-defined ($f \in V^* \implies \Phi(v)(f) \in k$) and is linear since if $f_1,f_2 \in V^*$ and $\lambda \in k$, $$\Phi(v)(\lambda f_1+f_2) = (\lambda f_1+f_2)(v) = \lambda f_1(v)+f_2(v) = \lambda \Phi(v)(f_1)+\Phi(v)(f_2).$$ Thus, $\Phi(v) \in (V^*)^* =: V^{**}$, and then $\Phi : V \to V^{**}$ is well-defined!