# 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.

• What about the union of the Cantor set and the set of rationals in $[0,1]?$
– mfl
Commented Jan 13, 2016 at 0:31
• That's so simple! Thank you!
– user304718
Commented Jan 13, 2016 at 0:34
• There is a book "my numbers, my friends" which define the Liouville numbers L and state they are U-numbers: uncountable, dense, measure zero. The Cantor set C is uncountable, not dense, measure zero. I'm searching for an isomorphism between L and C. Do you know? Commented Jun 22, 2016 at 13:19

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

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.

• Another way is to just take $C + \mathbb{Q} := \{ c + q \mid c \in C, q \in \mathbb{Q} \}.$ It's a countable union of uncountable null sets and thus an uncountable null set. And it's uncountable on every interval. Commented Jun 15, 2023 at 5:03

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)

Let $X$ be the uncountable measure zero subset of $[0,1]$ which you constructed. Let $Y$ be the union of all sets of the form $aX+b$ where $a,b$ are rational numbers, $a\gt0.$ Then $Y$ is a set of measure zero (countable union of measure zero sets) and is "uncountably dense" in the sense that every interval $I$ of the real line has uncountable intersection with $Y,$ because there are rational numbers $a,b$ with $a\gt0$ such that $aX+b\subseteq I.$

There can even be an example of a set with measure zero which is a dense $$G_\delta$$ in $$[0,1]$$. Sets containing a dense $$G_\delta$$ are called comeager, and in particular are dense and uncountable. If a set's complement is comeager it is said to be meager. This is another notion of smallness that is different than having measure zero.

The example: let $$q_1,q_2,\ldots$$ be the rationals in $$[0,1]$$. Let $$O_{n,m}$$ be an interval around $$q_n$$ with diameter $$2^{-(m+n)}$$. Then define $$\displaystyle{B_m=\bigcup_{n\geq1}O_{n,m}}$$ which is open dense. Then also define $$\displaystyle{C=\bigcap_{m\geq1}B_m}$$. This $$C$$ is comeager, but because the measures of the $$B_m$$ are decreasing to zero, $$C$$ has measure zero.

So this set has measure zero, and is dense and uncountable in a very strong way.