Uncountable sets of transcendental numbers

Question

As a sort of follow up question to a previous question found here, besides the Liouville numbers, are there any other uncountable collections of transcendental numbers that are known? Clearly you could union the Liouville numbers with some other transcendental number and get an uncountable collection. I am seeking to find a distinct set from the Liouville numbers.

There are many examples of countable collections of transcendental numbers:

$e^a$ if $a$ is algebraic and nonzero (by the Lindemann–Weierstrass theorem).

$a^b$ where $a$ is algebraic but not 0 or 1, and $b$ is irrational algebraic (by the Gelfond–Schneider theorem)

$\sin(a), \cos(a)$ and $\tan(a),$ for any nonzero algebraic number $a$ (again by the Lindemann–Weierstrass theorem)

But, as all of these are based on the algebraic numbers, none of these collections are uncountable.

Answer 1

A simple non-Liouville example would be Fredholm numbers, i.e.,

$$\sum_{n=0}^{\infty} \beta^{2^n}$$

for $|\beta|$ in $(0, 1)$.

EDIT: This doesn't work; we only know it's transcendental when $\beta$ is also algebraic, making the family countable.

Answer 2

Real numbers with irrationality measure larger than 2 are transcedentals and uncountably many - If $a$ is algebraic irrational, then its irrationality measure is 2 - Liouville transcedentals have infinite irrationality measure.

Answer 3

It is conjectured that there are no algebraic irrational numbers in the standard ternary Cantor Set. So if you remove the rational numbers (like 1/3 and 1/4) from the Cantor Set, you have an uncountable collection of only transcendentals. This conjecture depends on the normality of algebraic irrationals, which is unknown. But I've never heard of anyone (who is knowledgeable) who believes that 1/sqrt(2) is a member of the Cantor Set.

• Barnaby Finch