2
$\begingroup$

If $X$ is countable, then the cartesian product is countable. However, what about general cases? The googling suggests that the answer is the cardinality of $X$. But why? Could anyone please provide me with the proof?

$\endgroup$

marked as duplicate by Asaf Karagila cardinals May 2 at 15:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Have you learned about the axiom of choice and the well-ordering theorem yet? If not, the proof will be too complicated to explain. It's a lot harder that proving that the product of two countable sets is countable. $\endgroup$ – bof May 2 at 11:07
  • 4
    $\begingroup$ There is one exception on this. Also the empty set is countable and $X\times\varnothing=\varnothing$. $\endgroup$ – drhab May 2 at 11:09
  • 1
    $\begingroup$ |X| <= |X|×|Z| <= |X|×|X| = |X| $\endgroup$ – William Elliot May 2 at 12:47
  • $\begingroup$ @WilliamElliot The claim $|X|\times |X|=|X|$ is far more difficult to prove than $|X|\times |N|=|X|$ for countable $N$. $\endgroup$ – freakish May 2 at 14:43
  • $\begingroup$ Of course, this follows easily from $|X\times X|=|X|$, and that can be found in many other threads on the site. $\endgroup$ – Asaf Karagila May 2 at 15:25
1
$\begingroup$

Assume the Axiom of Choice and let $N$ be a non-empty countable set.

Lemma. $|X\times N|=|X|$.

Proof. Indeed, put a well ordering on $X$. I will call an element $x\in X$ initial if it has no direct predecessor. Direct successors will be denoted by $x+n$ for natural $\mathbb{N}$. Let

$$I(x)=\{y\in X\ |\ y=x+n\text{ for some }n\in\mathbb{N}\}$$ $$J=\{x\in X\ |\ x\text{ is initial}\}$$

It is easy to see that

$$X=\bigcup_{x\in J}I(x)$$ $$I(x)\cap I(y)=\emptyset\text{ for }x, y\in J, x\neq y$$

Finally each $I(x)$ is order isomorphic to either $\mathbb{N}$ or some of its subset. Note that there is at most one $x\in J$ such that $I(x)$ is finite. Denote that element by $f$. In case it doesn't exist we assume $I(f)=\emptyset$.

So now we have

$$X\times N=\bigcup_{x\in J}I(x)\times N=\big(\bigcup_{x\neq f}I(x)\times N\big)\cup \big(I(f)\times N\big)$$

Now since $I(x)$ is infinite countable for $x\neq f$ we get that $I(x)\times N\simeq I(x)$ by the standard snake proof. Thus we have

$$X\times N\simeq\bigcup_{x\neq f}I(x)\cup \big(I(f)\times N\big)$$

Note that the set $I(f)\times N$ is at most infinite countable (and so it does not have to be equinumerous with $I(f)$). But it does contain a copy of $I(f)$ inside. So we can move that copy to the left side of the union and so for $Y=(I(f)\times N)\backslash (I(f)\times\{0\})$ we get

$$X\times N\simeq X\cup Y$$

where $Y$ is at most infinite countable. This finally leads to $X\times N\simeq X$ because you can always remove a sequence from an infinite set without changing its cardinality (which again I believe requires the Axiom of Choice). $\Box$

Side note: it is also true that $|X\times X|=|X|$. But can we replace $N$ with $X$ in that proof? Sort of. We calculate

$$X\times X=\bigcup_{x\in J}I(x)\times\bigcup_{x\in J}I(x)=\bigcup_{(x,y)\in J\times J}I(x)\times I(y)$$

Now $J\times J$ is the problematic set. Note however that for infinite $X$ we have that $J$ is stricly less then $X$ in the well order ordering. Meaning there is an injective order preserving function $J\to X$ but not vice versa. So by reducing the problem from $X$ to $J$ we can prove the result. And this is done by the transfinite induction. Although I haven't worked out the details yet.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.