# Is the Cartesian Product of a finite number of infinitely countable sets countable? [duplicate]

Let $S=A_1 \times A_2 \times ...\times A_n$

Is S countable? And how do I prove it? I think the answer is yes because $A_1 \times A_2$ creates an infinite table so for n sets we would have an infinite table - just much bigger?

This is not a duplicate because no other question is for infinitely countable sets only

## marked as duplicate by Andrés E. Caicedo, Noah Schweber, Aloizio Macedo♦, Asaf Karagila♦ elementary-set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 18 '15 at 20:44

• no - they are different – cameron Nov 18 '15 at 18:00
• @cameron No, this is a duplicate - the fact that you ask only about infinite countable sets just means this question is a special case of the other, more general question. – Noah Schweber Nov 18 '15 at 18:20
• But the answer that cam from that page is not what I will use to answer my question so it did not answer my question. I wanted to prove it with an infinite table. – cameron Nov 18 '15 at 18:26

Yes, basically this comes from the fact that $$\mathbb{N} \times \mathbb{N} \simeq \mathbb{N}$$ which is pretty easy to prove.
Things change if you consider an infinite product of countable sets (you get $\mathbb{R}$'s cardinality) This can easily be seen considering the decimal representation of $\mathbb{R}$. You can easily look at any real number as an element of $\displaystyle \prod^\infty C$ if $C=\{0,\dots,9\}$ (excluding infinite $9$s, but that doesn't change the cardinality).
Yes, S is countable. You know $A_{1}\times A_{2}$ is countable, so you just need repeat the procedure $n-2$ times more. But you need note that if n goes to infinity, then S is uncountable. In this case, it is of the same cardinality as $R$.
• How can I show this in the form of a proof? I wanted to use the infinite table. For example here we can see it for $A_1 \times A_2$: proofwiki.org/wiki/… – cameron Nov 18 '15 at 17:56
• You can prove $A_{1} \times A_{2}$ is countable, right? Next you treat $A_{1} \times A_{2}$ as a new countable set denoted by $B$, and then apply the same proof on $B \times A_{3}$ and get $A_{1} \times A_{2} \times A_{3}$ is countable. You just repeat the procedure until $A_{n}$ is considered. – Hua Nov 18 '15 at 18:01