## Puzzle corner: Solution to Sylvy’s puzzle #4

In this post I want to discuss the solution to Puzzle #4. Let’s recall the puzzle: it was to find an uncountable chain in .

#### Solution

First we note that and have the same cardinality, i.e. . There are several well-known bijections that are often used to show this, but as I was writing this post I was reminded that none of the bijections that I’m familiar with are *explicit* in the sense of being given by some sort of `formula’. Here is a stackexchange discussion of this point (in particular, see the answer by Thomas Andrews).

Anyway, for the sake of this puzzle, we just need some bijection .

Now, for each , let be the preimage (under ) of the set of rational numbers in the interval .

**Proposition ** is an uncountable chain

This doesn’t need too much proof. Let be real numbers. If is such that then also . Thus .

Furthermore, there exists a rational number such that [why?]. Therefore , and so . This shows that , as required.

## Reply