This is a bit of a silly question, but I can't seem to find the answer anywhere.
I feel like $X\sim \mathcal{N}(\mu,\sigma^{2})$ means that $\sim$ is a relation, but if it is a relation, what precisely is this relation?
If this is a relation, could I instead write $X\in \mathcal{N}(\mu,\sigma^{2})/ \sim$?
One definition I think would be reasonable is to say that $X \sim Y$ if $X$ and $Y$ have the same characteristic or moment generating function, but it seems a little heavy handed to be the "right" definition.
Another definition I would guess is to put a topology on something which characterizes the random variables and say $X\sim Y$ if they satisfy a homeomorphism?
From the homeomorphism viewpoint, I would want to conclude that if $X$ is a random variable whose image is always non-negative, then $X^{2}$ could be normally distributed since you can take the positive square root and exhibit a bijection between $X$ and $X^{2}$ without going into $\mathbb{C}$, but I know this to not be true since it is Chi-Squared, so this can't be a reasonable definition.
I agree with Emily that most people just understand this notation to be shorthand for the corresponding sentence in words.
But if you want to think of it as having formal meaning, I would say that $X \sim \nu$ means:
$X$ is a random variable on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$, taking its values in some measurable space $(E, \mathcal{B})$;
$\nu$ is a probability measure on $(E, \mathcal{B})$;
For every $A \in \mathcal{B}$, we have $\mathbb{P}(X \in A) = \nu(A)$. That is, $\nu$ is the distribution (or law) of $X$.
So under that interpretation, you should interpret $\mathcal{N}(\mu, \sigma^2)$ as meaning "the probability measure $\nu$ on $(\mathbb{R}, \mathcal{B}_\mathbb{R})$ defined by $\nu(A) = \int_A \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-(x-\mu)^2/2\sigma^2}\,dx$", where $\mathcal{B}_\mathbb{R}$ is the Borel $\sigma$-algebra on $\mathbb{R}$.