# Partitioning Infinite Set $M$ to Countable Sets Each with the Same Cardinality as $M$

I'm studying Moretti's introduction to Spectral Theory and Quantum Mechanics in which he makes the following claim: (as far as I understand, I have reworded this to avoid discussing the irrelevant parts)

Suppose $$M$$ is an infinite set. Then $$\text{card } M = \text{card } M \times \mathbb{N}$$.

Is this true? I realized that this is equivalent to stating that all infinite sets can be partitioned into a countable number of sets, whose partitions all have the same cardinality as the set itself. How can this be proven?

The claim above is easily verifiable on sets such as $$\mathbb{N}$$ (partition it into multiples of primes) or in $$\mathbb{R}$$ (partition it into reals in the interval $$(n, n+1]$$ for $$n \in \mathbb{Z}$$).

This is because then, we can easily generate a surjection from $$M$$ to $$M \times \mathbb{N}$$ by mapping the $$i$$th partition of $$M$$ (which has a cardinality of $$M$$) trivially to the elements in $$M \times \mathbb{N}$$ of the form $$(., i)$$.

A surjection from $$M \times \mathbb{N}$$ to $$M$$ is trivial. So by Schroder-Bernstein they must have the same cardinality.

Possible idea: I think for any infinite set $$M$$ we know $$\text{card } M = \text{card } M \times M$$ by the axiom of choice. source

Since $$M \times M$$ can be easily partitioned into two parts each with cardinality $$M$$, then we now $$M$$ can be partitioned into two infinite sets with the same cardinality as M$. We pick one of these two partitions and partition it again to two parts. We keep doing this and construct a countably infinite partition of sets with the cardinality of $$M$$. ## 1 Answer You've missed a point. Since $$M$$ is infinite, it contains a countably infinite subset. Moreover, $$M\times M$$ can be partitioned into $$|M|$$ parts, each with cardinality $$|M|$$, which is way more than $$2$$, and unless $$M$$ was countable, way more than $$\aleph_0$$ as well. What is true, however, is that even without the axiom of choice we have $$|M\times\Bbb N|=|M|$$ if and only if $$|M\times\{0,1\}|=|M|$$. This is accomplished by fixing a bijection between $$M$$ and $$M\times\{0,1\}$$, and then "iterating it" to define the necessary partition into $$\aleph_0$$ different parts. • I see, so what we need to prove is$|M \times \{0, 1\}| = |M|\$, I was just confusing myself. Wouldn't that require the axiom of choice though? The only way I can think of a bijection between the two sets is to use the well-ordering theorem which I think relies on the axiom of choice. Sorry if my reasoning doesn't make sense as my background in the field is very weak. Oct 6, 2021 at 15:46
• Yes, that does require the assumption of AC, in general, although the statement is weaker than AC (significantly!). Oct 6, 2021 at 15:52
• I see, thanks for the answer! Oct 6, 2021 at 15:54