Probability of a rational nearest neighbor of an irrational Consider the following distribution over $(X, Y) \in [0,1] \times \{0, 1\}$:
$\mathbb{P}(Y = 0) = \mathbb{P}(Y = 1) = 1/2$
If $Y = 0$ then $X$ is uniformly distributed over the interval $[0, 1]$
If $Y = 1$ then $X$ is distributed over the rationals in $[0, 1]$ such that each rational has positive probability (for example, consider $Z = \min(A, B)/\max(A,B)$ for $A,B$ with geometric distribution)
Show that for any irrational $x \in [0, 1]$, the nearest neighbor (call it $NN_n(x)$) to $x$ in a set of $n$ draws of $X_i$ from the distribution above is such that as $n \rightarrow \infty$:
$$\mathbb{P}(NN_n(x) \, \text{is rational}) \rightarrow 0$$
What I've Tried
Listing the probabilities for increasing values of $n$:
For $n = 1$ it is obvious:
$$\mathbb{P}(NN_1(x) \, \text{is rational}) = \mathbb{P}(X_i \, \text{is rational}) = 1/2$$
For $n = 2$ it is more difficult. My initial thought is to consider all orderings of the $n$ samples. The orderings can't be equally likely since then we have a probability of $1/2$ for all $n$. I think there is some measure theoretic concept I am missing.
It's also clear to me that when $Y=0$, $X$ is irrational wp 1, since the set of rationals have measure $0$.
EDIT: The following has some issues (see the discussion in the comments)
Another thought I had is that
$$\lim_{n \rightarrow \infty}\mathbb{P}(NN_n(x) \, \text{is rational}) \leq \mathbb{P}(NN_n(x) \, \text{is rational infinitely often})$$
We can show the probability on the RHS is $0$ by showing that $\|NN_n(x) - x\| \rightarrow 0$ almost surely:
First note that the event $\{\|NN_n(x) - x\| > \varepsilon\}$ is equivalent to the event $\{1/n \sum_i^n\mathbb{I}(\|X_i - x\| \leq \varepsilon) = 0\}$. By the strong law of large numbers,  wp $1$:
$$ \frac 1 n \sum_i^n\mathbb{I}(\|X_i - x\| \leq \varepsilon) \rightarrow \mathbb{P}(\|X_i - x\| \leq \varepsilon)$$
Since the ball around $x$ has positive support, the event $\{1/n \sum_i^n\mathbb{I}(\|X_i - x\| \leq \varepsilon) = 0\}$ has probability $0$ as $n\rightarrow \infty$. Given that this event is equivalent to $\{\|NN_n(x) - x\| > \varepsilon\}$, then $\mathbb{P}(\lim_{n \rightarrow \infty} \|NN_n(x) - x\| > \varepsilon) = 0$ as was desired.
 A: This is not true.  For the case of $Y = 1$ I'll define the law of $X$ as follows: with probability $1/2$ it is equal to some distribution taking each rational with positive probability and with $1/2$ it is equal to $Z$ where $$P(Z = 1/n^n) = 2^{-n}$$ for each $n \geq 2$.  I claim that for $x = 0$ and $n$ large enough, the nearest neighbor will be a rational number with high probability.
I'll just show it for, e.g. $n = 3^m$ for $m$ large.  There will with high probability at least, say, $3^{m-1}$ samples with $Y = 1$ and $3^{m-2}$ with $X$ sampled from $Z$; let $N$ be this number.  The number of samples of $Z$ equal to $1/m^m$ is a binomial variable with success parameter $2^{-m}$ and $N \geq 3^{m-2}$ samples and so with high probability there is some $Z = 1/m^m$.
However, even if you sample all $3^m$ uniformly from $[0,1]$ the probability that some sample lies in $[0,1/m^m]$ is bounded above by $3^m/m^m$ which tends to zero.  Thus, the point closest to $0$ must be some sample from the case of $Y = 1$, i.e. a rational.

EDIT: How to generalize this to arbitrary $x \in [0,1]$.  For each $n \geq 2$ find rational number $x_n$ so that $|x - x_n| \leq 1/n^n$.  Then, define the law $Z$ to be $$P(Z = x_n) = 2^{-n}$$ for all $n \geq 2$.  The argument then goes through exactly the same way: for $n = 3^m$ with high probability there is some sample of $x_m$ and no samples from the uniform distribution in $[x - m^{-m}, x+ m^{-m}]$.
