30 July 2019

Let’s define two real numbers to be equivalent (denoted by $\sim$) if their difference is a rational number:

Now consider the resulting equivalence class, which, for any real number $x$, is the set of real numbers that are equivalent to $x$ under $\sim$:

For example, all the rational numbers trivially fall into $[0]$.

How many such equivalence classes are there in $\mathbb{R}$? The answer to this question is a key (if small) step along the way to a proof about the nonexistence of a universal measure (i.e., a measure defined on all subsets) on the real numbers, which I came across recently while watching this intro video about Measure Theory. The answer, which may seem obvious to some, took me a while to figure out, so I figured I’d share my thought process:

First, note that each equivalence class is countable. In fact, each class is exactly as large as the set of rational numbers, which is of course countable. This becomes obvious if we rewrite the class like so:

Next, notice that the union of all these equivalence classes needs to cover the real numbers. After all, every real number belongs to (exactly) one of these classes.

It follows that there must be an uncountable number of these classes, since otherwise, we would have the union of a countable number of countable sets, which cannot possibly cover $\mathbb{R}$ (which is uncountable). Neat!