I was watching a video of Alain Connes in french. It's a conference aout Quantum mechanics. And at the begining of the conference he says that
"Continuous variables cannot coexist with discret variables"
He defines real variable as a function $f:X\to \Bbb{R},$ nothing about the nature of $X.$
Here is the reason given by Alain Connes in French
"Si $f$ varie continûment l'ensemble $X$ doit être non dénombrable. Tout autre variable réelle sur $X$ dont l'image est dénombrable prend alors le même valeur une infinité de fois. Ainsi dans ce formalisme les variables continues ne peuvent coexister avec les variables discrètes."
In English:
If $f$ varies continuously the set $X$ must be uncountable. Any other real variable on $X$ whose image is countable will take the same value an infinity of times. Thus in this formalism, continuous variables cannot coexist with discrete variables.
Can someone explain what it means ? I don't understand why $X$ must be uncountable. There is certainly a tacit assumption about $X$ but I don't see it (otherwise we can take $X$ with the discrete topology for example). The second sentence is as obscure as the first one!
It is possible that on a probability space $(X,\mathcal A,P)$ two random variables $C$ and $D$ are defined such that $C$ is by definition a continuous random variable and $D$ is a discrete one.
So actually I do not agree or do not correctly understand the first statement.
It is true however that the existence of a continuous random variable on $(X,\mathcal A,P)$ implies uncountability of $X$. In fact a weaker condition is already enough for that: the existence of a random variable on $(X,\mathcal A,P)$ that is not discrete (weaker than being continuous) implies that $X$ is uncountable.
This can be rephrased as: if $X$ is countable then random variables defined on $(X,\mathcal A,P)$ are discrete.
For completeness let me mention that the opposite statement is not true. It can happen that all random variables defined on $(X,\mathcal A,P)$ are discrete while $X$ is uncountable. To achieve this situation you could do it for instance with $\mathcal A=\{X,\varnothing\}$ where $X$ is uncountable.
Countability of $X$ implies for any random variable $U$ on $(X,\mathcal A,P)$ that the set: $$S:=\{u\in\mathbb R\mid\{U=u\}\neq\varnothing\}$$ is a countable set with $P(U\in S)=1$ and the existence of such a set means exactly that $U$ is a discrete random variable.