Today after lunch I was hungry for math problems so I started begging for some at the department and finally someone threw me this: Can $\mathbb{R}$ be partitioned into two non-countable dense subsets? It was a good starter, after a few minutes I got: Take the irrationals less than $0$ and the rationals greater than $0$, this is one subset, the complement of course works. Then the following question came into my mind: Can $\mathbb{R}$ be partitioned into two locally-non-countable dense subsets?

I'm still hungry


The answer is yes. I think that one way to do it is to iteratively construct Cantor sets in each interval $[n,n+1]$, $n\in\mathbb{Z}$, i.e., for a given interval, construct a Cantor set, then construct a Cantor set in each of the (countably many) disjoint segments whose union is the complement of that Cantor set intersected with the original interval, and repeat this process. In the end you would consider the union of those Cantor sets, and the complement of that union.

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    $\begingroup$ An easier way to achieve the same is to enumerate the rationals and take a scaled copy of the Cantor set in each $[q_i, q_j]$ such that $q_i \lt q_j$ and take the union $U$. This will result in a dense set $U$ of measure zero whose intersection with every open set is uncountable. The intersection of the complement $U^c$ with every interval has full measure, hence will be uncountble. $\endgroup$ – t.b. Apr 29 '11 at 4:06
  • $\begingroup$ It's funny that you should mention that; there is a problem in Rudin's Real & Complex about constructing a Borel set $E\subset\mathbb{R}$ such that $m(E)<m(E\cap I)<m(I)$ for every interval $I$. The method that you outlined is similar to the solution I came up with (using "fat Cantor sets" though), and the answer I gave above is similar to a solution everyone I spoke with thought was simpler! $\endgroup$ – Nick Strehlke Apr 29 '11 at 4:11
  • $\begingroup$ I guess simplicity is in the eye of the beholder. Probably my main point is that the same construction works for any uncountable dense zero set of $[0,1]$ regardless of its fine structure. $\endgroup$ – t.b. Apr 29 '11 at 4:19
  • $\begingroup$ Oops, I meant $0<m(E\cap I)<m(I)$. $\endgroup$ – Nick Strehlke Apr 29 '11 at 5:51
  • $\begingroup$ Ah, now I know why I was having this uncanny déjà-vu feeling when reading your most recent answer... $\endgroup$ – t.b. Aug 16 '11 at 14:07

$\mathbb{Q}$ has $2^{\aleph_0}$ many cosets in the additive group $\mathbb{R}$. Therefore $\mathbb{R}$ can be partitioned into 2 sets each of which is a union of $2^{\aleph_0}$ many cosets of $\mathbb{Q}$. Each coset is dense, so each of these sets is locally uncountable.


Updated in light of the comment: How about we express all numbers in base 3. For set $A$ take all the numbers that eventually have $1$'s in all the even places behind the base 3 point-things like $0.x1x1x1x1x1\ldots$ (where the $x$'s can be any digit and need not be the same), $0.xxx1x1x1x1x1x1\ldots$, $0.xxxxx1x1x1x1x1\ldots$ and for $B$ take everything else.

  • $\begingroup$ How can a sequence from $A$ converge to $1$? $\endgroup$ – Mr. Hobeiche Apr 29 '11 at 4:01
  • $\begingroup$ @Mr. Hobeiche: good point. I have updated it. $\endgroup$ – Ross Millikan Apr 29 '11 at 4:11

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