Existence of uncountable set of uncountable disjoint subsets of uncountable set

Question

"Can you to any uncountably infinite set $M$ find an uncountably infinite family $F$ consisting of pairwise disjoint uncountably infinite subsets of $M$?"

Intuitively, I feel like it should be possible for the real numbers at least: you simply split the real numbers into two intervals, and since there are uncountably many points where you can do that, there are uncountably many ways to split the reals into two disjoint subsets. But this question isn't asking specifically about the real numbers, so how can I prove this more generally?

Answer 1

Your approach does not work even for $\Bbb R$: you’ve merely shown that there are uncountably many different ways to split $\Bbb R$ into two disjoint uncountable sets. Your task is to split $M$ into uncountably many pairwise disjoint uncountable sets.

HINT: $|M\times M|=|M|$ (assuming the axiom of choice).

Answer 2

I think this should work; maybe it is what Brian Scott had in mind, this works for sets of the same cardinality as $\mathbb R$. Use the fact that $|\mathbb R \times \mathbb R|= |\mathbb R|$. Now, $\mathbb R^2$ can be split into an uncountable set of copies of $\mathbb R $ , almost by definition, since $\mathbb R^2=\mathbb R \times \mathbb R $ (setwise). Now, for any set of the same cardinality as $\mathbb R$ , use a bijection between $\mathbb R^2$ and any set $S$ with $|S|=\aleph_1$ , to do a partition for $S$.

Answer 3

It will be too much work to keep typing "uncountably infinite", so I will just type "uncountable" which means the same thing.

Plainly, if $M$ is an uncountable set, the Cartesian product $M\times M$ contains an uncountable family of pairwise disjoint uncountable sets.

The set $\omega_1$ of all countable ordinals is an uncountable set; its cardinality is called $\aleph_1$. Thus, the set $\omega_1\times\omega_1$ contains an uncountable family of pairwise disjoint uncountable sets. The cardinality of $\omega_1\times\omega_1$ is $\aleph_1\cdot\aleph_1=\aleph_1^2=\aleph_1$. (None of these equalities depends on the Axiom of Choice.) Therefore, every set of cardinality $\aleph_1$ contains an uncountable family of pairwise disjoint uncountable sets.

To conclude from this that every uncountable set contains an uncountable family of pairwise disjoint uncountable sets, we need a weak form of the Axiom of Choice, namely: "Every uncountable set has a subset of cardinality $\aleph_1$." In other words, this says that $\aleph_1$ is the smallest uncountable cardinal.

It can also be shown without the Axiom of Choice that $(2^{\aleph_0})^2=2^{\aleph_0}$, i.e., there is a bijection between $\mathbb R\times\mathbb R$ and $\mathbb R$. Hence the general result follows from the still weaker axiom: "Every uncountable set has a subset whose cardinality is either $\aleph_1$ or $2^{\aleph_0}$."