Show that, there are uncountably many disjoint dense subsets in $\Bbb{R}$.
I know $\Bbb{Q}$ is a dense subset in $\Bbb{R}$. But other than this and disjoint to $\Bbb{Q},$ I have no idea.
Please help me. Thank you.
Uncountably many disjoint dense subsets in $\Bbb{R}$
real numbersreal-analysis
Related Solutions
No.
Since $\mathbb{Q}$ is dense in $\mathbb{R}$, within each interval $[a_i, b_i]$, we can select a single $q_i \in \mathbb{Q}$ with which to "label" the interval. The set of $q_i$'s is a subset of $\mathbb{Q}$, which is countable.
Sure.
Let $f:[0, 1] \to [0, 1]$ be such that $$f(0. b_1 b_2 b_3\ldots) = \liminf_{N\to \infty} \frac{1}{N}\sum_{n=1}^N b_n$$ where $\{b_n\}$ is the usual binary representation of a number in that interval. We have that $f$ is well defined, measurable and integrable and the Birkhoff ergodic theorem implies that $\lambda(f^{-1}(1/2)) = 1$. For all other values $t\in [0, 1]$, $A_t = f^{-1}(t)$ has measure $0$.
Furthermore $A_s\cap A_t =\emptyset$ for $s\neq t$, and for any $N$ and $0\leq n < 2^N$, $A_t \cap [\frac{n}{2^N}, \frac{n+1}{2^N}] \neq \emptyset$ (the initial part of the expansion doesn't really matter, so $A_t$ has elements in every interval), and so is dense in $[0, 1]$ for all $t$.
The collection $\{A_t\}_{t\in[0, 1] \setminus 1/2}$ is what you were looking for.
I just read your comment to the other answer. I'm not sure that the sets I have here are Borel. Let me think about that for a bit.
As mentioned in the comments, we have the $\liminf$ of Borel functions so we have a Borel function and it's preimages of singletons are therefore also Borel. I'm satisfied that this is all correct, but I don't have a specific reference, sorry.
Regarding uncountability, there are at least two arguments that come to mind. The first, which I had in mind as I wrote, relates to a class of measures on $\prod_{n\in \mathbb{N}} \{0, 1\}$, namely those measures that correspond to an infinite product of the measure where $\mu(\{0\}) = 1-t$ and $\mu(\{1\}) = t$. This is a non-atomic (for $t\neq 0\text{ or }1$) standard probability space, and so a set with measure $1$, which $A_t$ corresponds after a natural map, must be uncountable. (I really like this argument because it highlights that having measure $1$ is more about conforming to the expectations of a measure than being large).
More directly, assume that we have $A_t = \{a_1, a_2, a_3, \dots\}.$ For some $t$, and that $a_i = 0. b_{i, 1} b_{i, 2} b_{i, 3}\ldots$ where this is the usual binary representation of $a_i$. Using a standard diagonalisation argument would be problematic as we would lose control of the assymptotic density of $1$s in its representation. However, if we form the number $a$ which is such that $a= 0.b_1 b_2 b_3\ldots$ where $$b_i = \begin{cases} 1-b_{n,n^2} &\text{if } i = n^2 \text{ for some } n\in\mathbb{N} \\ b_{1, i}& \text{otherwise.} \end{cases}$$ then I claim that $f(a) = f(a_1)$ as the assymptotic density of the squares is $0$, so $a\in A_t$ but $a\neq a_n$ for all $n$ as their binary representations differ at the $n^2$ place. So $A_t$'s elements cannot be listed and $A_t$ is uncountable.
Best Answer
Consider an equivalence relation $\sim$ on $\Bbb R$ given by $x\sim y$ iff $x-y\in \Bbb Q$. Using axiom of choice, from each equivalence class take a representative and consider the corresponding translate of $\Bbb Q$. Note that any translation of $\Bbb Q$ is dense in $\Bbb R$.
Actually, any equivalence class can be written as $i+\Bbb Q$ for some $i\in \Bbb R\backslash \Bbb Q$, in particular any equivalence class is countable. So, there are uncountably many equivalence classes, as countable union of countably many sets is countable, but $\Bbb R$ is uncountable.
Note another fact, $i+\Bbb Q=(i+r)+\Bbb Q$ for any $r\in \Bbb Q$ and any $i\in \Bbb R\backslash \Bbb Q$. In other words, two distinct irrational translates of $\Bbb Q$ may give same equivalence class. That's why we need to consider the equivalence relation $\sim$ to get uncountably many disjoint dense subsets of $\Bbb R$.