This question is about a probable confusion of definitions that I may have somewhere. Also my math knowledge is not too big so I'll try not to get too abstract.
Let's say I have a vector space made of objects called Kets (${|\psi\rangle}$), the ones used in Quantum Mechanics. They have an inner product ${\langle\phi|\psi\rangle}$, and they have continuous (uncountable) dimension. Take an Orthonormal Basis of the space, for example, the eigen-kets of the position operator, ${|x_j\rangle}$, where ${x_j}$ sweeps all the real numbers (as they are all the possible positions).
–Orthonormal means (I think) that the kets satisfy:
${\langle x_a|x_b\rangle=0}$ if a ${\neq}$ b and ${\langle x_a|x_b\rangle=1}$ if a = b.
–Basis means that every ket ${|\psi\rangle}$ can be written as a linear combination of the basis kets. I'm going to take a shot and say that this can be expressed as
$${|\psi\rangle=\int_{-\infty}^\infty dx_i.C(x_i)|x_i\rangle}$$
However, if I turn that into the following expression
$${\langle x_a|\psi\rangle=\int_{-\infty}^\infty dx_i.C(x_i) \langle x_a|x_i\rangle}$$
and then use the Orthonormality condition, I am left with something akin to
$${\langle x_a|\psi\rangle=\int_{x_a}^{x_a} dx_i.C(x_i).1 }$$
which I am pretty sure yields ${\langle x_a|\psi\rangle\neq C(x_a)}$ -in fact it probably yields ${\langle x_a|\psi\rangle=0}$ since the area under a single point of finite height is zero-. This seems totally wrong for the orthonormal basis of a vector space, which makes me think that I there is something wrong somwhere.
-A possible solution to this would be to redefine the concept of orthonormality for such an uncountable vector basis, so that Orthonormal means:
${\langle x_a|x_b\rangle=0}$ if a ${\neq}$ b and ${\langle x_a|x_b\rangle=\infty}$ if a = b, where said ${\infty}$ makes sense inside an integral and yields 1 when integrated. This would indeed then return ${\langle x_a|\psi\rangle= C(x_a)}$.
However, redifining orthonormality also seems kind of iffy. Plus it would imply that ${\langle x_a|x_a\rangle=\infty}$ which is not very encouraging.
So could you tell me how do I make these statements work?
==============
Bonus: A third option could be to say that the expression is not an integral, but a summation of the form
$${|\psi\rangle=\sum_{x_j=-\infty}^\infty C(x_j)|x_j\rangle}$$
where ${x_j}$ sweeps all the real numbers.
Yet I'm sure that such an expression is not correct either.
Best Answer
As you correctly concluded, there is something wrong with your original orthonormality condiction and indeed it has to do with the fact that you are using a discrete basis to span a continuous space.
So if you have a continuous infinite dimensional Hilbert space, then your basis would also need to be continuous and infinite. (Actually, one can sometimes use a discrete infinite basis, but let's ignore this caveat for the moment.) In that case the orthonormality condition would be given by $$ \langle x_a| x_b \rangle = \delta(x_a-x_b) , $$ where $\delta(x_a-x_b)$ is a Dirac delta function.
When you now want to expand an arbitrary ket, $$ |\psi\rangle = \int |x\rangle C(x)\ dx , $$ you can apply the inner product and get $$ \langle x_a|\psi\rangle = \int \langle x_a|x\rangle C(x)\ dx = \int \delta(x_a-x) C(x)\ dx = C(x_a) , $$ which follows form the properties of the Dirac delta function.
In the case where you have a discrete basis you can replace the Dirac delta with a Kronecker delta function, which is basically what you had originally. Then the expansion would be a summation as you have shown at the end.