Example of an uncountable dense set with measure zero

As stated in the title, I am trying to find an example of an uncountable dense subset of $[0,1]$ that has measure zero. My intuition is that such a subset cannot exist, but I do not have a proof of this.

Currently, I can construct an uncountable dense subset that has arbitrarily small measure. Also, it is easy to construct an uncountable subset that has zero measure.

Thanks in advanced!

Solution 1:

Consider the union of $\mathbb{Q}\cap[0,1]\cup K$, where $K$ is the ternary Cantor set.

Solution 2:

You can even construct a set $S \subset \mathbb{R}$ such that $S \cap U$ is uncountable for every open $U \subseteq \mathbb{R}$ and still $m(S \cap U) = 0$ where $m$ is the Lebesgue measure.

To do this, we start with the Cantor set $C \subset [0, 1]$ and create "denser" sets by gluing together scaled down copies of $C$:

$$\begin{align} S_n & := \bigcup \{ 3^{-n} (x+k) \, | \, x \in C, \, k \in \mathbb Z \} \\ S & := \bigcup_{n=0}^{\infty} S_n \\ \end{align}$$

Since $S_n$ is a countable union of nullsets (sets of measure $0$), also $S_n$ will be a nullset. In the same way $S$ will be a nullset.

The numbers in $S$ will have a ternary "decimal" expansion with only a finite number of ones.

Solution 3:

Or even (without just taking an uncountable set of measure zero and throwing in the rationals) union rational translations of the Cantor set (mod 1)