Dual Vector Space Proof I am trying to prove the following:
Let $\phi: V \rightarrow (V^*)^*$ be defined by $v \to (f \to f(v))$, where $v \in V$ and $f \in V^*$. Show that $\phi$ is an injective map which is an isomorphism if and only if $V$ is of finite dimension. Note that $V$ is a vector space. 
I am not too sure where to start on this at all, any help or hints would be appreciated. 
 A: Suppose for $v, v' \in V$ we have $\phi(v)=\phi(v')$, which means that these are equal as elements of $(V^\ast)^\ast$. So $$\forall f \in V^\ast: \phi(v)(f)= \phi(v')(f).$$
By definition of $\phi$, we know
$$\forall v \in V: \forall f \in V^\ast: \phi(v)(f)=f(v).$$
So $$\forall f \in V^\ast: f(v)= f(v').$$
If $v \neq v'$, find $f \in V^\ast$ such that $f(v) \neq f(v')$. 
It then follows that $v=v'$ and $\phi$ is injective.
If $\dim(V)=n$, then $V \simeq \mathbb{K}^n$ then $\dim(V)=\dim(V^\ast)$ so $V^\ast \simeq \mathbb{K}^n$. Repeat for $V^\ast$. This shows one direction. 
A: Here is a general way to think about $V$ and its dual $V^{\ast}$.  To make life easy, let's assume that the dimension of $V$ is countably infinite.  You can extrapolate what is going on here to the case where $V$ is finite dimensional or when $V$ has uncountable dimension, and use the general principle indicated here to solve your problem.
To say that $V$ has countably infinite dimension means that there exists a countable basis $e_1, e_2, ...$ of $V$.  This means that every $v \in V$ is equal to unique sum of the form
$$v = \sum\limits_{n=1}^{\infty}c_ne_n \tag{1}$$
where all but finitely many scalars $c_n$ are zero.  In other words, the sum (1) is finite.
For each basis element $e_i$, define an element $e_i^{\ast}$ of $V^{\ast}$ as follows: $e_i^{\ast}$ is the linear map which sends an element $\sum\limits_{n=1}^{\infty}c_ne_n$ to the scalar $c_i$.
It is not too hard to see that the elements $e_1^{\ast}, e_2^{\ast}, e_3^{\ast}, ...$
of $V^{\ast}$ are linearly independent in $V^{\ast}$.  However, they will not be a basis of $V^{\ast}$.  If they were a basis of $V^{\ast}$, then every $f \in V^{\ast}$ could be written as 
$$\sum\limits_{n=1}^{\infty} d_ie_i^{\ast} \tag{2} $$
where all but finitely many scalars $d_i$ are zero.  But you can construct an element of $V^{\ast}$ which cannot possibly be of the form (2).  Indeed, let $f \in V^{\ast}$ be the linear functional defined by
$$f(\sum\limits_{n=1}^{\infty}c_ne_n) = \sum\limits_{n=1}^{\infty} c_n$$
This is well defined (the sums are finite), and $f(e_j) = 1$ for every $j$.  However, elements of the form (2) do not satisfy this last property.
A more abstract, but maybe less insightful way to formulate all this: By definition, $V^{\ast}$ is the vector space $\operatorname{Hom}_F(V,F)$ of linear transformations from $V$ to $F$, where $F$ is your field of scalars.  Since $V$ has countably infinite dimension, we have 
$$V \cong \bigoplus\limits_{n=1}^{\infty} F$$
as vector spaces, and therefore
$$\operatorname{Hom}_F(V,F) \cong \operatorname{Hom}_F(\bigoplus\limits_{n=1}^{\infty}F,F) \cong \prod\limits_{n=1}^{\infty} \operatorname{Hom}_F(F,F) \cong \prod\limits_{n=1}^{\infty} F$$
and this contains the direct sum $\bigoplus\limits_{n=1}^{\infty} F$ as a proper subspace.
A: Clarifications: 


*

*I will replace your vector $v$ with $x$.

*To the function $f\mapsto f(x)$ I will denote it by $\operatorname{ev}_x(f)=f(x)$.

*Also, the dual space of $\textsf{V}^*$ I will denote it by $\textsf{V}^{**}$.


I hope everything that follows is helpful.
Lemma : Let $\textsf{V}$ a finite-dimensional vector space and let $x\in \textsf{V}$. If $\operatorname{ev}_x(f)=0$ for all $f\in \textsf{V}^*$, then $x=0$.
Or equivalently, if $x\neq 0$ then $\operatorname{ev}_x(f)\neq 0$ for some $f\in \textsf{V}^*$.
Proof : Let $x\neq 0$. Let $$\beta =\{v_1,v_2,\dots,v_n\}$$ be a basis for $\textsf V$ such that $v_1=x$. Now, let $$\beta^* =\{f_1,f_2,\dots,f_n\}$$ the dual basis of $\beta$. Then $f_1(v_1)=1\neq 0$. That is $$\operatorname{ev}_x(f_1)\neq 0$$ 
Finally, pick $f:=f_1$. $\blacksquare$
Now, with this lemma, suppose that $\textsf V$ is finite-dimensional and we will prove that $\operatorname{ev}_x$ is an injective map from $\textsf V$ to $\textsf{V}^{**}$.
Theorem : Let $\textsf V$ a finite-dimensional vector space and define the function $\phi : \textsf{V} \to \textsf{V}^{**}$ by $$\phi (x)=\operatorname{ev}_x$$
Then, $\phi$ is an isomorphism.
Proof : First, we see that $\phi$ is linear : 
Let $x,y\in \textsf V$ and $a$ an arbitrary scalar. For any $f\in \textsf{V}^*$ we have
$$\phi(x+y)(f)=f(x+y)=f(x)+f(y)=\phi(x)(f)+\phi(y)(f)=\left( \phi(x)+\phi(y)\right)(f)$$
and $$\phi(ax)(f)=f(ax)=af(x)=\left(a\phi(x)\right) (f)$$
So, $\phi$ is linear. Next, $\phi$ is one-to-one : 
Suppose $\phi(x)$ is the zero functional on $\textsf{V}^*$ for some $x\in \textsf V$ (it is the same as saying that $\phi(x)$ is the zero vector on $\textsf{V}^{**}$). Then $\phi(x)(f)=\operatorname{ev}_x(f)=0$ for all $f\in \textsf{V}^*$. By the previous lemma we conclude that $x=0$. 
That is $\phi$ is injective.
And since $\textsf V$ and $\textsf{V}^{**}$ has the same dimension, $\phi$ is also surjective. $\blacksquare$
