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I want to prove that infinite (yet countable) cartesian product of countable sets is countable.

Here's what I tried:

Step 1:

I proved that for 2 countable sets $ A_1,A_2 $ , the product $ A_{1}\times A_{2} $ is countable.

Step 2:

I proved by induction that for any $n\in \mathbb{N} $ if $ A_{1},...,A_{n} $ are countable sets, then

$ A_{1}\times A_{2}\times,...,\times A_{n} $ is countable.

Now, I want to show that any countable infinite cartesian product would be countable.

How do i show that $ A_{1}\times,....\times A_{\aleph_{0}} $ is countable?

Thanks in advance.

Andrés E. Caicedo
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FreeZe
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  • In general it's not countable... https://math.stackexchange.com/questions/500849/infinite-cartesian-product-of-countable-sets-is-uncountable – ECL Jun 25 '20 at 15:27
  • Isn't true. $[0,1]^{\mathbb N}$ isn't countable and if you consider a decimal expansion of a real number as an infinite-tuple of the digits $0$ to $9$ it's clear it is false. – fleablood Jun 25 '20 at 16:14

3 Answers3

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Here are some links that should answer your question:

Infinite Cartesian product of countable sets is uncountable

Countable collection of countable sets and Axiom of choice

DodoDuQuercy
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You can't prove it because it's false.

The countable product of two element sets is (essentially) the uncountable real unit interval when you think of elements of the product set as binary expansions.

Infinite Cartesian product of countable sets is uncountable

Ethan Bolker
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  • When your answer is "This is impossible, see this answer for details", maybe the right approach is to post a comment and maybe vote to close as a duplicate. – Asaf Karagila Jun 25 '20 at 15:32
  • @AsafKaragila Perhaps. The second paragraph outlines a quick counterexample, so I decided to post it. The nice accepted answer says essentially the same thing. The question could still be closed as duplicate. – Ethan Bolker Jun 25 '20 at 15:57
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I guess you know that $[0,1)$ is uncountable. If your statement was true, then $\mathbb{N}^\mathbb{N}$ would be countable. But there is a surjection of $\mathbb{N}^\mathbb{N}$ in $[0,1)$ since every $x\in[0,1)$ can be represented by a countable string of digits (writing it's decimal part in base 10 for instance). Then $[0,1)$ would be countable, which is absurd. So a countable cartesian product in general is not countable.

ECL
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