On Liouville numbers and the Continuum Hypothesis

Question

Collecting some theorems from the book Making Transcendence Transparent by its authors, there is some inconsistency, I think... :

i. $L= \sum_{n=1}^{\infty} 10^{-n!}$ is transcendental.

ii. Numbers of the form $\sum_{n=1}^{\infty} a_n 10^{-n!}$ in which $a_i \in {\{0,1}\}$ are Liouville numbers and thus transcendental. So, by Cantor diagonalization argument there are uncountably many Liouville numbers.

iii. The collection of all Liouville numbers has measure zero. The set of all sequences of zeros and ones (not all zero) are in 1-1 correspondence with $(0,2)$ and this is in 1-1 correspondence with $\mathbb{R}$.

How an uncountable subset of $\mathbb{R}$ has measure zero?

Which inconsistency do you see (" there is some inconsistency, I think"). — Dietrich Burde, Dec 17 '18 at 12:56
There is no inconsistency in an uncountable set of measure $0$. — Tobias Kildetoft, Dec 17 '18 at 13:02
@TobiasKildetoft, I had never seen an uncountable set of measure $0$ that's why I am confused. — , Dec 17 '18 at 13:03
Incidentally, I disagree with the downvote (and have upvoted) - this is a perfectly reasonable confusion to have at first, and what is MSE for if not dealing with perfectly reasonable confusions? — Noah Schweber, Dec 17 '18 at 15:24

score 5 · Accepted Answer · answered Dec 17 '18 at 13:13

5

How about the Cantor subset of $\mathbb{R}$: it is compact, uncountable, with no isolated points yet it has measure zero.

answered Dec 17 '18 at 13:13

Aphelli

34,439

You beat me to it. It is probably the easiest to understand example of an uncountable set of measure zero. – badjohn Dec 17 '18 at 13:18

On Liouville numbers and the Continuum Hypothesis

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