Completeness relation of coherent spaces In a set of notes it is stated that: Given coherent states of a harmonic oscillator $$| \alpha \rangle = \pi^{-\frac{1}{2}} \text{exp}(-\frac{1}{2}|\alpha|^2)\sum_{n = 0}^{\infty} \frac{\alpha^n}{(n!)^{\frac{1}{2}}}|n \rangle~~~~~~~~~~(1)$$where $|n \rangle$ is the nth energy eigenstate of the oscillator. We can define the completeness relation as the integral $$\int d^2 \alpha | \alpha \rangle \langle \alpha | = \sum_{n=0}^{\infty}| n \rangle \langle n| = 1~~~~~~~~~~~~~~~~(2)$$ But these states are not orthonormal. They do not span the Hilbert space in the two-dimensional $\alpha$ space.
Question: How is the "two dimensional $\alpha$ plane" defined? If $\{ | \alpha \rangle \}$ satisfies completeness relation (2) then what space does it span?  
 A: So let's start a step back because your coherent states are not normalized as I would normalize them.
Coherent states
The coherent states come from their response to the bosonic annihilator, $$\hat b |x , y\rangle = (x + i y) |x,y\rangle.$$From this one can derive that any particular one's representation among the number states must satisfy, $$\hat b~\sum_n c_n |n\rangle = \sum_n c_n \sqrt{n} |n-1\rangle=(x + i y) \sum_n c_n |n\rangle,$$giving the recursive relation that $c_n = \frac{x+iy}{\sqrt n}~c_{n-1}.$ Starting from $c_0$ we then find indeed the relation that $$|x,y\rangle = c_0~\sum_n \frac{(x + i y)^n}{\sqrt{n!}} |n\rangle.$$The remaining $c_0$ with the proper normalization gives $$\langle x,y|x,y\rangle = 1 = |c_0|^2 \sum_n \frac{(x-iy)^n(x+iy)^n}{n!} = |c_0|^2 \exp\big(x^2 + y^2\big).$$Choosing these to all have the same complex phase for their vaccum component finally yields,$$|x, y\rangle = \exp\left(-\frac12(x^2 + y^2)\right)\sum_n \frac{(x + i y)^n}{\sqrt{n!}}~|n\rangle.$$
So the question is, why does your expression have a leading $\pi^{-1/2}$ in it? That's because they resolve the identity in a somewhat weird way. What does that mean?
Resolving the identity
Suppose you have an expression for some average $\langle A \rangle.$ QM is very clear that this expression may be written based on its quantum state $|\psi\rangle$ as $\langle \psi|\hat A|\psi\rangle.$
But using the fact that $1 = \sum_n |n\rangle\langle n|,$ for example, we can insert these sums ad-hoc into that expression to find that in fact this expectation value also reads, $$\langle A \rangle = \sum_{mn} \langle\psi|m\rangle\langle m|\hat A|n\rangle\langle n|\psi\rangle = \sum_{mn} \psi^*_m~A_{mn}~\psi_n.$$ So that is the value of resolving the identity; it means that you can define this matrix $A_{mn}$ which fully specifies the action of $\hat A$ on the Hilbert space, recovering every single expectation value from the matrix.
Well we see something very similar when we look at the operator, $$\hat Q = \int_{-\infty}^\infty dx~\int_{-\infty}^\infty dy~|x,y\rangle\langle x, y| = \sum_{mn} \iint dx~dy~e^{-x^2-y^2}\frac{(x-iy)^m(x+iy)^n}{\sqrt{m!n!}} |m\rangle\langle n|.$$
At this point it is useful to shift to polar coordinates where $x + i y = r e^{i\theta},$ yielding $$\hat Q = \sum_{mn}\int_{0}^\infty dr~\int_0^{2\pi} r~d\theta~e^{-r^2}~\frac{r^{m+n} e^{i(n-m)\theta}}{\sqrt{m!n!}} |m\rangle\langle n|.$$ Note that the angle over $\theta$ integrates a sinusoid over one or more full periods and therefore vanishes if $m\ne n$; it is $2\pi$ if $m = n$, so we
must get:$$\hat Q = \pi\sum_{n}\int_{0}^\infty dr~2r~e^{-r^2}~\frac{r^{2n} }{n!} |n\rangle\langle n|.$$Substituting $u=r^2, du=2r~dr$ we find that this is:$$\hat Q = \pi\sum_{n}\frac1{n!}~|n\rangle\langle n|~\int_{0}^\infty du~e^{-u}~u^n.$$If you've never seen the gamma function before, the integral on the right hand side is $n!$ and in fact it is the canonical way to extend the factorial function to non-integers to find e.g. that $(-1/2)! = \sqrt{\pi},$ though of course we only need the integers here. After cancelling that through we find out that in fact, $$\hat Q = \pi,$$ or in other words we recover this property of resolving the identity even though not all of these functions are orthogonal, because the way that they're non-orthogonal just comes down to a constant multiplicative factor. We can therefore state unequivocally, $$1 = \iint dx~dy~\frac1\pi~|x,y\rangle\langle x,y|.$$ Your expression absorbs a $1/\sqrt{\pi}$ term into each of these kets, and writes $\pi^{-1/2} |x, y\rangle = |\alpha\rangle$ (where $\alpha = x + i y$) for short, both of which help in writing these expansions. One then finds similarly to the above expression with $A_{mn}$, that $$\langle A \rangle = \iint d^2\alpha~d^2\beta~\psi^*(\alpha)~A(\alpha,\beta)~\psi(\beta).$$The only cost to this notation is that we then have to express the above integrals with the more clumsy $\int d^2\alpha$ which is short for something like $d\alpha_x~d\alpha_y$ where $\alpha = \alpha_x + i \alpha_y.$
A: They do not span a 2-dimensional space; you can see this because $\vert\alpha\rangle$ is a linear combination of harmonic oscillator states.  Thus state $\vert\alpha\rangle$ live in the $\infty$-dimensional space.  They span that space because any state can be written as a linear combination of the $\vert \alpha\rangle$'s and the expansion coefficient is unique and completely fixed by $\vert \alpha\rangle$, just like you would expand over the set $\{\vert n\rangle,n=0,1,\ldots\}$.  

Edit: It seems that part of the difficulty is connected to the observation that the eigenvalues of $\hat b$ are continuous and complex rather than real and discrete.  Thus a sum of the type
$$
\sum_n \vert n\rangle \langle n\vert = \mathbb{I}
$$
that would hold for discrete eigenstates must be replaced by a continuous sum (i.e. an integral) in the continuous case.  Since the eigenstates are labelled by complex numbers $x+iy$, and since the closure involves a continuous sum over all eigenstates, 
$$
\sum_n \vert n\rangle \langle n\vert = \mathbb{I}\propto
\int dxdy \vert x,y\rangle\langle x,y\vert = \int d^2\alpha 
\vert\alpha\rangle\langle\alpha\vert
$$
with $d^2\alpha=dxdy$ if $\alpha=x+iy$. 
