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As this question clearly shows that the countable many product of countable sets is uncountable. However, I do not understand why the below proof is wrong:

(False) Proof:

Let $A$ be a countable set. We use induction to show that the countably many product of $A$ with itself is countable. We use induction. When $n=1$, the theorem is true by our hypothesis.Let assume it is true for $A^n = A\times ... A : n$ times.

Since $A^{n+1} = A^n \times A$, which is the finite product of countable sets, it is also countable by this question. Hence, by induction $A^n$ is countable for all $n \in \mathbb{N}$. QED

Question:

Why is the above proof wrong ? Where is the flaw ?

Edit:

I'm trying to show that $A^{|\mathbb{N}|}$ is countable.

Andrés E. Caicedo
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Our
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    To assert that every finite product of countable sets is countable implies that an infinite product of countable sets is countable is a non sequitur. – Angina Seng Oct 10 '18 at 06:20
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    To put it another way, there is an important distinction to be drawn between arbitrarily large (but finite) $n$ and infinitely large $n$. – Brian Tung Oct 10 '18 at 06:23
  • @LordSharktheUnknown I used induction. Isn't that a valid move ? – Our Oct 10 '18 at 06:23
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    @onurcanbektas You successfully proved that $A^n$ is countable for any $n\in\Bbb N$. Well done! But your "proof" does not even consider infinite products. – Angina Seng Oct 10 '18 at 06:25

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The proof you give proves that $A^n$ is countable for all $n \in \mathbb{N}$ however what you wish to prove is $A^\omega$ is countable. That is the propitiation is true for the infinite case. However $\omega$ isn't in $\mathbb{N}$ so the induction you use doesn't work.

Q the Platypus
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