If $\phi(v)=0, \, \forall \phi \in V'$ then $v=0$ Suppose that $V$ is a vector space over the field $\Bbb{F}$. $V'=\mathcal{L}(V,\Bbb{F})$ is  the vector space of all linear maps from $V$ into $\Bbb{F}$. Then I want to prove that

If $\phi(v)=0, \, \forall \phi \in V'$ then $v=0$.

My main concern is to prove it for finite dimensional vector spaces $V$, however, it will be useful to see how it works in infinite dimensions too.
This problem happened to me while trying to  solve the part $(c)$ of the exercise $3.F-34$ of the well-known book Linear Algebra Done Right. In fact, I wanted to prove that $\Lambda$ is injective and then I encountered this problem. 
Any hint or clue is appreciated.

 A: To prove the statement, we prove its contrapositive.  In particular, we want to show that for any $v \neq 0$, there exists a $\phi \in V'$ such that $\phi(v) \neq 0$.
By the axiom of choice, we may select a basis $\{v_\alpha\}_{\alpha \in I}$ of $V$ (where $I$ is some indexing set; in the finite dimensional case, we can take $I = \{1,2,\dots,n\}$) where one of the basis elements is $v$.
We can define a transformation that acts as follows on a basis: we take
$$
T(v_\alpha) = \begin{cases}
1 & v_{\alpha} = v\\
0 & \text{otherwise}
\end{cases}
$$
From there, it suffices to note that any function on a basis of $V$ extends to a linear transformation on the entirety of $V$.  In particular, we have
$$
T\left[\sum_{i=1}^n c_{\alpha_i}v_{\alpha_i}\right] = \sum_{i=1}^n c_{\alpha_i}T(v_{\alpha_i})
$$
where we note that any element of $V$ can be expressed (uniquely) as a sum $\sum_{i=1}^n c_{\alpha_i}v_{\alpha_i}$ for some choices of (finitely many) $\alpha_i$ and constants $c_{\alpha_i}$.
Thus, by providing an example of a suitable transformation, we prove the contrapositive.
A: Assuming that $V$ is finite dimensional, then this is a way of showing this:
Let $B = (b_1, ..., b_n)$ be a base of $V$, and let $B^* = (b_1^*, ..., b_n^*)$ be the according dual base of $V^*$. Then you have $b_i^*(v) = 0$ for all $i=1,...,n$ and this gives you by definition of the dual base 
$$\begin{align}
b_i^*(v) &= b_i^*(\sum_{j=1}^{n}c_jb_j) =\sum_{j=1}^{n}c_jb_i^*(b_j)= \sum_{j=1}^{n}c_j \delta_{ij} = c_i=0, & i=1,...,n
\end{align}$$
and hence
$$v = 0\cdot b_1 + ... + 0\cdot b_n = 0 \in V$$
