# Is "fattening" countable sets a good way to compare their relative density within an uncountable set?

In a recent conversation, someone posed the following argument:

The cardinality of the natural numbers and the rational numbers is the same, since they're both countable. So the probability of randomly selecting a natural number from the rational numbers is the same as selecting a non-natural number.

This is obviously false for any finite interval (there are a finite number of natural numbers but an infinite number of rational numbers in that interval, after all), but it wasn't immediately clear to me how to properly extend to the entire real line.

Cardinality is, of course, not a good way to think about probability on infinite sets, and ordinarily I would go to measure theory as the way to compare the sets' relative sizes (where, for the improper analogue of a uniform distribution, relative size is proportional to relative probability of selection). But the problem is that both sets are countable, and therefore both have measure zero using the usual measure on the real line. So I tried constructing something like a measure on the rationals, but a quick search of the literature suggested that such a thing wasn't possible, or at least wasn't likely to be at all well-behaved if it was.

So, the solution I came upon was the following procedure: for any countable set $$X$$, construct the "fat" version of $$X$$, which we will call $$X_d$$, by replacing each element of $$x$$ with an open* ball of some (small) radius $$d$$, centered at $$x$$; then, $$X_d$$ is the union of all such open balls. In other words, we define:

$$X_d=\bigcup_{x\in X}B(x,d)$$

where $$B(x,d)$$ is the open ball centered at $$x$$ of radius $$d$$.

This allows us to measure-theoretically compare the densities of countable subsets of an uncountable set. The set $$\mathbb{Z}_d$$, or indeed the set $$\mathbb{N}_d$$, has gaps for all $$d<1$$. On the other hand, since the rationals are dense in the reals, we have that $$\mathbb{Q}_d=\mathbb{R}$$ for any $$d$$. So the ratio of the sizes of $$\mathbb{Z}_d$$ and $$\mathbb{Q}_d$$ is roughly $$d$$ ("roughly" because you could choose endpoints within a distance $$d$$ of an integer), regardless of the size of the interval. Sending $$d$$ to zero essentially recovers the original sets, along with the result that you will randomly select an integer from among the rationals with probability zero (thus, the same result holds for the natural numbers).

Is this a reasonable way to get around the problems inherent in working with probability on the rationals? More generally, is it a reasonable way to compare the relative densities of countable subsets of uncountable sets?

*As far as I can tell, the choice of open or closed ball doesn't matter for this procedure; I could be wrong on this, though.

• There is no uniform distribution on the rational or natural numbers, though there are discrete probability distributions which are not uniform. A cardinality or density argument then fails to answer the question. It is possible to construct a discrete distribution on the rationals which gives the integers overall a probability of $\frac12$ of being chosen, though many other distributions are possible too. Oct 5, 2021 at 19:44
• As you can see with dense subsets of $\mathbb{Q}$, this is not a good way. There is a surrogate "uniform distribution" on the natural numbers called natural density, it is the limit of uniform distributions on finite segments, and a finitely additive probability measure. You can transfer it to other countable sets, like integers or rationals, but the transfer depends on how they are corresponded to $\mathbb{N}$. "Natural" enumerations of rationals (going by increasing numerators and denominators in lowest terms incrementally) make integers rare. Oct 5, 2021 at 20:10
• What you are suggesting is similar to conditioning on zero probability events, but it does not work so well for dense discrete subsets like $\mathbb{Q}$. Oct 5, 2021 at 20:23

I think your demonstration that the fat rationals satisfy $$\mathbb{Q}_d = \mathbb{R}$$, for all $$0, is just another way of saying that $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$.
Also this analysis won't always work. For example, consider the set $$Q =\mathbb{Q}+\sqrt{2}$$. Then $$Q$$ is countable, and in some sense "more dense than $$\mathbb{Q}$$ in $$\mathbb{R}$$", since there are twice as many elements. But $$Q_d = \mathbb{Q}_d$$, so these countable sets cannot be compared by "which one is more dense"