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We know for $A_1, A_2, \cdots, A_n$, where each $A_i$ is countable, that $A_1 \times A_2 \times \cdots \times A_n$ is countable.

We also know that $\mathbb{N}$ is countable.

Let $N_i = \mathbb{N} \times \mathbb{N} \times \cdots \times \mathbb{N}$ ($i$ times).

I am wondering if $\bigcup_{i=1}^\infty N_i$ is countable?

This is bothering me because for any given $i$, we know that $N_i$ is countable, so we have a countable union of countable sets, therefore, $\bigcup_{i=1}^\infty N_i$ should be countable.

But then we have that as $i\to\infty$, $N_i$ is countable. This implies that an infinite cartesian product of countable sets is countable, which is not true. It is uncountable.

Where am I going wrong and which is correct?

Jac Frall
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  • "This implies that an infinite cartesian product of countable sets is countable," is not true. For any $i$, however large, $n_i$ is countable. You are saying $\lim_{n\to\infty}\aleph_0>\aleph_0$. – saulspatz Oct 30 '19 at 21:52
  • @saulspatz right, that is what I thought was wrong. So the limit as $n\to\infty$ is not uncountable? – Jac Frall Oct 30 '19 at 21:54
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    No, it's countable. A countable union of countable sets is countable. The proof requires the axiom of choice (or perhaps some weaker variant, I'm not an expert.) – saulspatz Oct 30 '19 at 21:55
  • @saulspatz Thank you. I have seen the proof and it makes sense, I just was not sure about this since is seemed like an infinite cartesian product which we know is uncountable. – Jac Frall Oct 30 '19 at 21:58
  • You're more than welcome. – saulspatz Oct 30 '19 at 22:00
  • Taking the limit as n approaches infinity requires a topology for a collection of sets and that topology is missing. – William Elliot Oct 30 '19 at 22:49
  • @WilliamElliot I don't have a great understanding of topology. I guess I just fundamentally misunderstand this but it seems to me that $N_\infty$ is the same thing as an infinite cross product which we know is is uncountable. Isn't $N_\infty \subset \bigcup_{i=1}^\infty N_i$. But we know that $N_\infty$ is uncountable while $\bigcup_{i=1}^\infty N_i$ is countable – Jac Frall Oct 31 '19 at 03:34
  • @JacFrall The last subset relation you wrote doesn't hold. Left side consists of infinite sequences, right side consists of finite sequences (of arbitrary length, but still finite). – Ned Oct 31 '19 at 09:18
  • @Ned that makes sense. Thank you, I thought I found some sort of paradox for a bit – Jac Frall Oct 31 '19 at 13:33

1 Answers1

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Yes, $\bigcup_{i=1}^\infty N_i$ is countable.
It is the collection of all finite sequences of integers,
a union of countablely many countable sets.
It is disjoint from N$^{\aleph_0}$,
the set of all infinite sequences of integers.

William Elliot
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