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Can we find uncountably many disjoint dense measurable uncountable subsets of $[0,1]$? Obviously we may as well assume all the subsets have measure $0$. If I didn't specify the subsets were uncountable, we could easily find uncountably many dense disjoint measurable subsets $X_\alpha$ (all countable) by taking a Hamel basis $H$ of $\mathbb{R}$ as a vector space over $\mathbb{Q}$, where $H$ contains a non-zero rational number, and then set $X_\alpha = (\mathbb{Q} + \alpha) \cap [0,1]$ for every $\alpha \in H$.

But what about if we want uncountable measurable dense subsets? I think it should still be that there are uncountably many that can be found that are disjoint, according to my intuition, but I don't know an "easy" way to show it like I can in the case of countable dense subsets.

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  • $\begingroup$ Have you though about similar constructions using the Cantor set? (I haven't, that's why I am just commenting) $\endgroup$ – Silvia Ghinassi Aug 28 '15 at 15:58
  • $\begingroup$ I don't have an example yet, but the density assumption can easily be dropped. Simply unite every set in your uncountable family of disjoint uncountable sets with one of your $X_\alpha$. $\endgroup$ – Dominik Aug 28 '15 at 16:02
  • $\begingroup$ @Dominik Good point. $\endgroup$ – user2566092 Aug 28 '15 at 16:11
  • $\begingroup$ @SilviaGhinassi I've tried but every time I try to think about using something like the Cantor set, I run into the problem of whether it can be assumed to be a field over which we can define ${\mathbb R}$ as a vector space and furthermore get an uncountable Hamel basis. $\endgroup$ – user2566092 Aug 28 '15 at 16:20
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    $\begingroup$ An interesting related construction: if $\gamma : [0,1] \to [0,1]^2$ is a space filling curve, the sets $A_t = \gamma^{-1}(\{t\} \times [0,1])$ partition $[0,1]$ into uncountably many disjoint uncountable closed sets. $\endgroup$ – Nate Eldredge Aug 28 '15 at 18:57
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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.

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  • $\begingroup$ I like this even more because the function is explicit here (and also intuitive, although showing that $f^{-1}(1/2)$ has measure $1$ and all other preimages have measure $0$ is a bit non-trivial. But I didn't even ask for all sets to be measure $0$ =)). So +1, and like I said before, if no one can come up with a Borel example then I'll accept this answer. $\endgroup$ – user2566092 Aug 28 '15 at 17:25
  • $\begingroup$ I also like this because it doesn't seem to use the axiom of choice, and the disjoint sets are automatically all dense without further modification. $\endgroup$ – user2566092 Aug 28 '15 at 17:38
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    $\begingroup$ I think you are correct since $f$ is the liminf of Borel measurable functions. And also an upvote for this splendid example! $\endgroup$ – Sangchul Lee Aug 28 '15 at 17:45
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    $\begingroup$ You might want to explain how we know that each $A_t$ is uncountable. $\endgroup$ – Nate Eldredge Aug 28 '15 at 19:07
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    $\begingroup$ I agree with @NateEldredge that explaining uncountability in the answer would be best, although it seems pretty intuitively clear to me that each set is uncountable. An augmented diagonalization argument should apply because changing the $n^2$th bit place in a number for each $n$ doesn't change the long-run ratio of positive bits. $\endgroup$ – user2566092 Aug 28 '15 at 19:37
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The following, which is related to the answer user24142 gave, is taken from this 21 January 2003 sci.math post:

[. . .] here's a collection of $c$-many pairwise disjoint sets with cardinality $c$ in every open interval. In fact, each of these sets will have a positive Hausdorff dimension intersection with every open interval and they are not very complicated from a descriptive set theoretic point of view (they're all $F_{\sigma \delta}$ sets, I believe), and thus are very low in the Borel hierarchy of sets.

For each real number $r$ between $0$ and $1$ (inclusive), let $P_r$ be the set of irrational numbers whose decimal expansions have $r$ for their limiting proportion of $5$'s. That is, if $N(x,5,n)$ is the number of appearances of the digit $5$ to the right of the decimal point in the first $n$ digits of the decimal expansion of the real number $x,$ then $P_r$ is the set of real numbers $x$ such that $\lim_\limits{n \to \infty}{N(x,5,n)}$ exists and is equal to $r.$

Then it is known that each $P_r$ is a Borel set (in fact, $F_{\sigma \delta}$ I think) with positive Hausdorff dimension in every open interval (see these sci.math posts: 16 June 2001 and 19 February 2003), which is quite a bit more than simply saying that each open interval has $c$-many numbers belonging to each of the $P_r$ sets.

On the other hand, these sets are small in a couple of other ways. All of them except for $P_{\frac{1}{10}}$ have measure zero, and all of them, including $P_{\frac{1}{10}},$ are first (Baire) category sets. Incidentally, the union of all the $P_r$ sets doesn't include all the real numbers. If you let $Q$ be the remaining real numbers, then $Q$ is a measure zero set with a first category complement. Hence, $Q$ is co-meager in every open interval, and so it too will have $c$-many elements in every open interval.

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  • $\begingroup$ I think that these sets are indeed Borel, and $F_\sigma$ or $G_\delta$ or what have you. The Hausdorff dimension fact is quite interesting! $\endgroup$ – user24142 Aug 28 '15 at 18:50
  • $\begingroup$ Thank you for the enlightening answer including Hausdorff dimension. +1 from me. Higher technical details aside, this seems the same as the other answer that used binary expansion, so I accepted that answer, especially because it came first. Thank you again though for the additional enlightening details. =) $\endgroup$ – user2566092 Aug 28 '15 at 19:46
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It's enough to do this for $\mathbb R,$ since $(0,1)$ and $\mathbb R$ are homeomorphic. Let $c$ denote the cardinality of $\mathbb R.$

Claim: There are $c$ many disjoint countable dense subsets of $\mathbb R.$

Proof: Let $K$ be the classical Cantor set. By "a Cantor set" I'll mean any set of the form $a+bK,$ where $a\in \mathbb R$ and $b>0.$

Let $I_1,I_2,\dots$ be the open intervals with rational end points. Then we can inductively choose disjoint Cantor sets $K_n\subset I_n$ for $n=1,2,\dots.$ (This is pretty straightforward. I'll leave it for now; ask questions if you like.)

For each $n,$ write $K_n = a_n + b_nK,$ and let $f_n: K \to K_n$ be the map $t\to a_n +b_nt.$ For $t\in K,$ define the sets $D_t = \{f_n(t): n=1,2,\dots \}.$ Then the sets $D_t$ are disjoint. Each $D_t$ is dense in $\mathbb R,$ simply because $D_t$ contains a point in each of the intervals $I_n.$ Because $K$ has cardinality $c,$ the claim follows.

To prove the full result, observe that there is a continuous surjective map $g:K\to [0,1]\times [0,1].$ (For example $g =P\circ C,$ where $P$ is the Peano space filling curve, and $C$ is the Cantor function.) For $t\in K,$ let $K_t = \{t\} \times [0,1].$ Then $\{ g^{-1}(K_t): t \in K\}$ is a disjoint collection of $c$ many sets, each of which is compact and of cardinality $c.$

Finally, consider the sets $D_t\cap (\mathbb R\setminus K) \cup g^{-1}(K_t)$ for $t\in K.$ These sets are disjoint, and each such set is dense in $\mathbb R$ and has cardinality $c.$ Since there are $c$ many such sets, we're done.

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I found a construction based on this idea.

We need the following lemma:
If $M$ is an uncountable set, then there is an uncountable family of pairwise disjoint uncountable subsets of $M$.

To prove this assertion, observe that $|M \times M| = |M|$ and let $f$ be a bijection between the two sets. The family $\{f(\{x\} \times M) \;|\; x \in M\}$ satisfies the conditions of our lemma.

Now we only apply this lemma to the Cantor set to get an uncountable family of disjoint uncountable subsets. Since the Cantor set is a nullset, every subset is a nullset and therefore Lebesgue-measurable. Now we only need to join every set with one of the sets $X_\alpha$ from the starting post to get an uncountable family of uncountable disjoint dense subsets of $[0, 1]$.

Edit: I think the sets can be chosen Borel with the help of Zorn's Lemma, but I can't quite finish the proof. Let $B$ be the set of all uncountable Borel-nullsets and define $$S = \{M \in \mathcal{P}(B) \;|\; \text{the sets in $M$ are pairwise disjoint}\}.$$

$S$ is obviously nonempty and we can define a partial order on $S$ via inclusion. Now every chain in S has an upper bound, namely the union of all elements in the chain. Therefore by Zorn's lemma there is a maximal element $T \in S$. We are done if we can prove that $T$ is uncountable.

Let's assume that $T$ is countable and that $N$ is the union of all sets in $T$. Then $N$ is Borel-measurable and a nullset. We now only need to prove that there is an uncountable Borel-nullset in $[0, 1] \setminus N$, but I can't find an argument to prove this assertion.

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  • $\begingroup$ Are the subsets obtained from the lemma necessarily measurable just because the original set is measurable? I actually don't know the answer, which is why I'm asking. I guess the question is, if we have a bijection between two measurable sets, can we insist that the IMAGE of every measurable set is measurable? At least, perhaps using some additional axioms besides axiom of choice? $\endgroup$ – user2566092 Aug 28 '15 at 17:10
  • $\begingroup$ In general, no. But all subsets of nullsets are (Lebesgue)-measurable and our sets are subsets of the Cantor set. If you actually wanted Borel-measurability, this construction won't work. $\endgroup$ – Dominik Aug 28 '15 at 17:12
  • $\begingroup$ Ok I guess this issue of BOREL measurability basically addresses my concern. I should have specified that I wanted Borel measurability if that's what I wanted, but honestly I didn't even think about it. I'll give this +1 and if no one can come up with Borel-measurable disjoint uncountable subsets in the near future then I'll accept. Thanks! $\endgroup$ – user2566092 Aug 28 '15 at 17:15
  • $\begingroup$ I've tried to prove the assertion for Borelsets in an elementary way, but I couldn't quite finish the proof. I'm pretty certain that the last remaining assertion is correct, I just can't find a proof for it. $\endgroup$ – Dominik Aug 28 '15 at 18:39

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