# Why is the proof of “countable many product of countable sets is countable” wrong? [duplicate]

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.

## marked as duplicate by Asaf Karagila♦ 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(); } ); }); }); Oct 10 '18 at 7:53

• 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. – Lord Shark the Unknown Oct 10 '18 at 6:20
• 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 6:23
• @LordSharktheUnknown I used induction. Isn't that a valid move ? – onurcanbektas Oct 10 '18 at 6:23
• @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. – Lord Shark the Unknown Oct 10 '18 at 6:25

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.
• i.e $A^{|\mathbb{N}|}$ is countable. – onurcanbektas Oct 10 '18 at 6:30
• @onurcanbektas $|\Bbb N|\notin\Bbb N$. – Lord Shark the Unknown Oct 10 '18 at 6:32
• $|\mathbb{N}| = \aleph_0 = \omega$ – Q the Platypus Oct 10 '18 at 6:34