Partition of set of size more than $2^{\aleph _0}$

Question

1. Can every set of size more than $2^{\aleph _0}$ be partitioned into subsets, such that each is non-singleton and each has size at-most $2^{\aleph_0}$?

2. Can every set of size more than $2^{\aleph _0}$ be partitioned into infinite subsets , each of size at-most $2^{\aleph_0}$?

3. Can every set of size more than $2^{\aleph _0}$ be partitioned into subsets, each of size exactly $2^{\aleph_0}$?

(without axiom of choice ) ?

Answer 1

You may want to rephrase the first question. Given any set of more than three elements, pick two $\{x,y\}$ and consider the partition $\{\{x,y\}\}\cup\{\{z\}\mid z\in X\setminus\{x,y\}\}$. This is a non-singletons partition, all the elements are finite. What you want to ask is that there are no singletons in the partition. And if you make this change then the answer is negative in the absence of the axiom of choice, to all three questions.

Suppose that there exists an amorphous set, namely an infinite set $A$ such that whenever $B\subseteq A$ either $B$ is finite or $A\setminus B$ is finite. You can show that if $A$ is amorphous, any function between $A$ and $\Bbb R$ (in either direction) has a finite range. We say that $A$ is a strongly amorphous set if every partition of $A$ has finitely many non-singleton parts. The existence of such set is consistent with the failure of the axiom of choice, relative to the rest of the axioms of $\sf ZF$.

Consider now $X=\Bbb R\cup A$, where $A$ is a strongly amorphous set. Then certainly $|X|>2^{\aleph_0}$, since $\Bbb R$ is a subset of $X$. But if we partition $X$ into non-singletons, either one of the part contains infinitely many points from $A$, in which case one of the part has a cardinality which is not "at most $2^{\aleph_0}$"; or there is a partition of $\Bbb R$ which selects a single point from $A$ for infinitely many points from $A$, and this defines a function from $\Bbb R$ into $A$ that has an infinite range, which is impossible.

So now you have a set that any partition of it either contains singletons, or a set whose cardinality is not less or equal to $2^{\aleph_0}$.

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Using the axiom of choice this is an easy consequence of the fact that $|X\times\Bbb R|=|X|$ for any $X$ whose cardinality is at least $2^{\aleph_0}$.

Answer 2

Well, every $ A = \bigcup\limits_{a \in A} \{a\}$