Splitting the Real Line By definition a $\mathfrak c$-dense subset of $\mathbb{R}$ has $\mathfrak c$-sized intersection with every non-empty open set. Using transfinite recursion it is quite easy to prove that every $\mathfrak c$-dense subset of $\mathbb{R}$ can be partitioned into $\mathfrak c$-many $\mathfrak c$-dense subsets.
I am wondering if there are other (more elementary) proofs of this, perhaps one that doesn't use transfinite recursion, but some raw form of AC?
 A: You can prove this using Zorn's lemma.
Fix a $\frak c$-dense set $A$. Now consider the following partial order $\Bbb P$ whose elements are $S$ such that:


*

*$S$ is a partition of a subset of $A$.
and for every $X\in S$, we have:

*$|X|=|S|$.

*$X$ is $|X|$-dense.

*$A\setminus X$ is $\frak c$-dense.


We say that $S_1\leq S_2$ if and only if the following is true:


*

*For every $X\in S_1$ there is a unique $Y\in S_2$ such that $X\subseteq Y$, and

*For every $Y\in S_2$ there is at most one $X\in S_1$ such that $X\subseteq Y$.


Now suppose that $\{S_i\mid i\in I\}$ is a chain. Then there is an obvious upper bound, since each $X_i\in S_i$ has a unique extension in every other $S_j$ for $i<j$, we simply take the union over these unique extensions. Let $S$ be the obtained partition.
It is easy to see that if indeed $S\in\Bbb P$, then $S_i\leq S$ for every $i\in I$. So it suffices to show that $S\in\Bbb P$.


*

*$S$ is a partition of $\bigcup S\subseteq A$ is trivial.

*The cardinality of $S$ is $\sup|S_i|\cdot |I|$, and each $X\in S$ is the $|I|$-union of $X_i$'s whose cardinality is $|S_i|$ as well. So we get that $X$ has the wanted cardinality as well.

*If $X\in S$ and $(a,b)$ is an open set, then $X\cap(a,b)=\bigcup X_i\cap(a,b)$, and since each $X_i$ is $|X_i|$-dense, this union gives us again that $X$ is $|X|$-dense.

*Finally the last part is also easy, if $|X|<\frak c$, then of course that $A\setminus X$ is $\frak c$-dense by obvious cardinal arithmetic. If $|X|=\frak c$, then $|S|=\frak c$, and every $Y\in S$ is already $\frak c$-dense, so $\bigcup S\setminus\{X\}$ is a subset of $A\setminus X$ witnessing that it is $\frak c$-dense.


Now comes the easiest part. Using Zorn's lemma we obtain a maximal element $S$. If $|S|=\frak c$, then we are done, since this gives us the wanted partition. Otherwise $S$ cannot be maximal, since $A\setminus\bigcup S$ is $\frak c$-dense, pick from every interval with rational endpoints a subset of $A\setminus\bigcup S$ of size $|S|$, and this allows extending $S$ by another part so it is not maximal.
A: Use your favorite method to pick (say) an almost disjoint family $A$ of infinite sets of integers. For each $x \in A$, let $W_x$ be the set of all reals whose binary expansion eventually agrees with $x$ on every other coordinate.
-hq
p.s. Forgot to say, $A$ has size continuum.
