1
$\begingroup$

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 $M$.

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ Commented Oct 6, 2021 at 15:46
  • $\begingroup$ Yes, that does require the assumption of AC, in general, although the statement is weaker than AC (significantly!). $\endgroup$
    – Asaf Karagila
    Commented Oct 6, 2021 at 15:52
  • $\begingroup$ I see, thanks for the answer! $\endgroup$ Commented Oct 6, 2021 at 15:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .