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This question already has an answer here:

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

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marked as duplicate by Andrés E. Caicedo, Noah Schweber, Aloizio Macedo, Asaf Karagila elementary-set-theory Nov 18 '15 at 20:44

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.

  • $\begingroup$ no - they are different $\endgroup$ – cameron Nov 18 '15 at 18:00
  • $\begingroup$ @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. $\endgroup$ – Noah Schweber Nov 18 '15 at 18:20
  • $\begingroup$ 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. $\endgroup$ – cameron Nov 18 '15 at 18:26
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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).

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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 $R$.

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  • $\begingroup$ 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/… $\endgroup$ – cameron Nov 18 '15 at 17:56
  • $\begingroup$ 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. $\endgroup$ – Hua Nov 18 '15 at 18:01
  • $\begingroup$ Ah yes of course - easy. Thank you $\endgroup$ – cameron Nov 18 '15 at 18:09
  • $\begingroup$ The preferred term is "countably infinite". I prefer "countable" to mean "not uncountable" so countable means "finite or countably infinite" but some people use "countable" to mean "countably infinite." A good little book I recommend is Stories About Sets by N. I'a. Vilenkin. $\endgroup$ – DanielWainfleet Nov 18 '15 at 18:23

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