Does a vector space and its dual contain the same number of elements? I'm pretty sure if we're talking the vector space $\Bbb{R}^3$ the number of vectors in this space is infinite, but does the dual space of $\Bbb{R}^3$ have a one-to-one correspondence with its vector space?
Have mercy on my mathematical abilities in any answers please; I'm not a mathematician.
 A: If we're talking about finite dimensional vector spaces, then the answer is yes. The correspondence is actually a really common object.
In particular, given two vectors, $\langle x_1,y_1,z_1\rangle$ and $\langle x_2,y_2,z_2\rangle$, their inner product (or dot product) is
$$\langle x_1,y_1,z_1\rangle \cdot \langle x_2,y_2,z_2\rangle = x_1x_2+y_1y_2+z_1z_2.$$
This is handy in all sorts of geometrical contexts, but, in relation to your question, this map has a handy property: it is bilinear, meaning that, if we fixed either vector, the function is linear in the other. For instance
$$\langle 1,2,4\rangle \cdot\langle x_2,y_2,z_2\rangle = x_2+2y_2+4z_2$$
is clearly a linear function of $\langle x_2,y_2,z_2\rangle$ and hence defines an element of the dual space of $\mathbb R^3$. Thus, we can map vectors $\langle x_1,y_1,z_1\rangle$ to the element of the dual space that takes $\langle x_2,y_2,z_2\rangle$ to $x_1x_2+y_1y_2+z_1z_2$. One can check that this is a one-to-one correspondence; the inverse map takes a map $f:\mathbb R^3\rightarrow \mathbb R$ in the dual space of $\mathbb R^3$ to the vector $$\langle f(\langle 1,0,0\rangle),f(\langle 0,1,0\rangle),f(\langle 0,0,1\rangle)\rangle$$
which basically just evaluates $f$ on the basis elements, then records those values which can, by linearity, reconstruct what $f$ must be on every element. Otherwise said, every linear function is given as the inner product between some fixed vector and the argument.
In very broad generality, this is exactly finite case of the Riesz representation theorem (for Hilbert spaces). A Hilbert space is just a space that acts like $\mathbb R^n$ in that it has a sensible inner product and in which convergence works is a sensible way - and the argument given above extends easily to the continuous dual of such a space.
In broader generalizations, like just using vector spaces with no additional structure, the answer tends to be no. For instance, the dual of an infinite dimensional vector space has higher dimension than the original space - which is sort of surprising - and this implies that, as long as the dimension of the first space was infinite enough, the second space will not only have higher dimension (meaning no linear one-to-one correspondence exists), but will also have more elements (in the sense of cardinality - meaning absolutely no one-to-one correspondence exists).
A: Yes, and it is very general. For each basis $\{e_1,\dots, e_n\}$ in a $K$-vector space of dimension $n$ $V$, there exists a dual basis $\{\varepsilon _1,\dots,\varepsilon_n\}$ in $V^*$ defined by the conditions
\begin{cases}
\varepsilon _i(e_j)=0&\text{ if }i\ne j,\\
\varepsilon _i(e_i)=1&\text{ for all  }\,i.
\end{cases}
Mapping each $e_i$ onto $\varepsilon_i$ ($i=1,\dots, n$) defines an isomorphism from $V$ onto its dual.
A: Yes. Indeed, if $V$ is finite dimensional, then there is an isomorphism between $V$ and $V^*$ given by taking any basis $B = \{e_1,\ldots,e_n\}$ of $V$, and defining a basis $B^* = \{f_1,\ldots,f_n\}$ of $V^*$ given by $f_i(\sum a_j e_j) = a_i$, then having our isomorphism send each $e_i$ to $f_i$, and extend linearly from there. 
In the infinite-dimensional case, the dual is not isomorphic: indeed, the dimension of $V^*$ is always of higher cardinality than that of $V$. Thus, we can have $|V| = |V^*|$ only in the cases where the cardinality of the dimension of $V$ is smaller than the cardinality of the base field (though it is not guaranteed there). 
