Is there an intuitive way of justifying why the square of an infinite cardinal is itself?

Question

By no means I am an expert in this subject, but I do have some knowledge of ZFC. While there are many proofs which are difficult to recollect, I feel like I have enough knowledge that if I am given a statement about ordinals or cardinals, then I can prove it myself. However, there is one particular thing which didn't seem intuitive to me at all and whose proof is also difficult to remember or recreate, which is:

If $\kappa$ is any infinite cardinal, then $\kappa\cdot\kappa = \kappa$.

Does anyone know of an intuitive proof of this? The fact that $\aleph_0\cdot\aleph_0 = \aleph_0$ is very intuitive, because this is just saying that cartesian product of two countable sets is countable. If $c = |\mathbb{R}|$, then the fact that $c\cdot c = c$ is also more or less intuitive: that $|(0 , 1)| = c$, we can use decimal expansions of numbers to justify this.

Can we visualise this fact somehow, or if not do you know of seeing this more easily?

Answer 1

You're wrong about $\aleph_0$. The fact that the product of two countable sets is countable is not "the intuitive reason" for $\aleph_0\cdot\aleph_0=\aleph_0$. It is the very definition of the equality. You've merely recast the symbols into words. It only becomes intuitive once you've seen the proof a few times.

The general fact that $\kappa\cdot\kappa=\kappa$ for any infinite cardinal is equivalent to the axiom of choice, so we may as well assume it. In this case, the intuition is sort of the same from the case of $\aleph_0$: we can order the set $\kappa\times\kappa$ in a well-order satisfying the property that every proper initial segment has size $<\kappa$. Much like how we order $\Bbb{N\times N}$ in a way that every proper initial segment is finite.

If you want to talk about sets like $\Bbb R$, then you can actually talk about this without the axiom of choice: if $A$ is an infinite set and $|A\times\{0,1\}|=|A|$, then we can use a bijection witnessing that to prove that $|\mathcal P(A)\times\mathcal P(A)|=|\mathcal P(A)|$. Since the real numbers satisfy $|\Bbb R|=|\mathcal P(\Bbb N)|$, the result follows. Alternatively, you can verify that for any set $X$, setting $A=X^\Bbb N$, we have $|A\times A|=|A|$, and again apply this to $X=\Bbb N$ and $\Bbb{|R|=|N^N|}$.

Or, more generally, if $|A\times\{0,1\}|=|A|$, then $|X^A\times X^A|=|X^A|$ for any set $X$.

But now we see why "intuitive" is the wrong term here. This is not true for every infinite set without assuming the axiom of choice: if AC fails, not every infinite set can be well-ordered, and certainly not every infinite set is a power set, or equipotent to a set of the form $X^A$ with the needed properties.