Does there exist an interval on the reals, however small, in which every number is irrational?

Question

Since the irrationals are uncountably infinite, I've always imagined them "filling the space" between the rational numbers, but does that make sense? Or is it the case that any two irrational numbers has a rational number between them?

Answer 1

Geometry and cardinality aren't really related. The irrationals "fill in" the rationals, but not in the way you imagine: while between every two (distinct) rationals there is an irrational, it's also true that between any two distinct irrationals there is a rational.

• To see this, think about going "far enough" in the decimal expansions of the two irrationals until you see a difference; you can now "slot an irrational" in between them. For example, maybe my irrational numbers begin "$0.163849\color{red}{{\bf 3}}23498...$" and "$0.163849\color{red}{{\bf 8}}86047...$" - then the number $0.163849\color{red}{{\bf 5}}$ (for example) is rational and strictly in between the two irrationals. (It's a good exercise to turn this into a rigorous proof, and to use a similar argument to show that between any two rationals there's an irrational.)

Things can get even weirder: the Cantor set is uncountable, but nowhere dense: for any (nonempty) open interval $(a,b)$, there is a (nonempty) subinterval $(c,d)\subseteq(a,b)$ such that the Cantor set has no points in $(c,d)$. And as we dive further into set theory and real analysis, even stranger things can happen.

Basically, you just need to divorce in your mind geometry and set theory; while they certainly do interact - e.g. every countable set has Lebesgue measure zero - this interaction is much weaker than you might reasonably expect.