Consider the position and momentum vector sets
$$X= \{|x\rangle |\ x \in \Bbb R^3\}$$
$$P= \{|p\rangle |\ p \in \Bbb R^3\}$$
By the assumption of quantum mechanics, both $X$ and $P$ are total orthonormal sets. In addition to that, $\langle x|p\rangle = e^{ip \cdot x} \neq 0$, but this conflicts with Kreyszig's functional analysis Lemma 3.5-3 which states: a vector in an inner product space can only has countably many non-zero Fourier coefficients with respect to an orthonormal family.
How's this possible?
Is the position/momentum vector representation in quantum mechanics incorrect?
An arbitrary basis $\{v_\alpha\}_{\alpha\in A}\subset \mathscr H$ may be uncountable, but $S=\{\alpha\in A:h\in \mathscr H;\ \langle h,v_\alpha\rangle \neq 0 \}$ is countable, and in this case, $h=\sum_{\alpha\in S}\langle h,v_\alpha\rangle v_\alpha. $
I am not a physicist, but isn't this fact exactly why $p$ and $x$ are quantized?