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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.

Starship
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    What about the union of the Cantor set and the set of rationals in $[0,1]?$ – mfl Jan 13 '16 at 00:31
  • That's so simple! Thank you! –  Jan 13 '16 at 00:34
  • 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?

     – Vinícius Ferraz Jun 22 '16 at 13:19

5 Answers5

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

Leandro
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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.

md2perpe
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    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. – md2perpe Jun 15 '23 at 05:03
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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)

charlestoncrabb
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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 $aX+b\subseteq I.$

bof
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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.

Francis Adams
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