Perfect set without rationals [closed]

Give an example of a perfect set in $\mathbb R^n$ that does not contain any of the rationals.

(Or prove that it does not exist).

Solution 1:

An easy example comes from the fact that a number with an infinite continued fraction expansion is irrational (and conversely). The set of all irrationals with continued fractions consisting only of 1's and 2's in any arrangement is a perfect set of irrational numbers.

Solution 2:

Consider the set of reals x whose binary expansion, if you look only at the even digit places, is some fixed non-eventually-repeating pattern z. This is perfect, since we have branching at the odd digits, but they are all irrational, since z is not eventually repeating.

You can draw a picture of this set, and it looks something like the Cantor middle third set, except that you divide into four pieces, and take either first+third or second+fourth, depending on the digits of z.

Another solution: Begin with an interval having irrational endpoints, and perform the usual Cantor middle-third construction, except that at stage n, be sure to exclude the n-th rational number (with respect to some fixed enumeration), using a subinterval having irrational endpoints. By systematically excluding all rational numbers, you have the desired perfect set of irrationals.

(Hi François!)