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We have two theorems about Cartesian product:

  1. A finite Cartesian product of countable sets is countable;
  2. A countable infinite Cartesian product of countable sets is not necessarily countable.

To prove the first theorem, we can show

  • Firstly, we can prove $A\times B$ is countable,
  • Secondly, suppose it is true for $n-1$, i.e. $X=A_1\times A_2\times \ldots \times A_{n-1}$ is countable, we prove the case for $n$ by the fact $A_1\times A_2\times \ldots \times A_{n-1}\times A_{n}=X\times A_{n}$ is in fact the product of two countable sets, in above we know it is still countable.

The above method to prove the first theorem is exactly the mathematical induction, which should show the theorem is true for all $n\in\mathbb{Z}_{+}$, so it should works for countable infinite unions.

However, we may use Cantor diagonal method to prove at least $\{0,1\}^w$ is not countable, thus the second theorem is also right. Does this means that the mathematical induction is wrong?

EDIT: After the following discussions, I have the answer:

The key point is that any element $n\in\mathbb{Z}_+$ is always a finite number, thus induction principle only guarantee the theorem is true for any $n\in\mathbb{Z}_+$, i.e. for any finite number.

When discuss a countable infinite union, it is infinite number of unions, this infinite number no more belong to $\mathbb{Z}_+$, thus can not be guaranteed by induction principle.

X Leo
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    Of course this can't possibly mean that induction is wrong, because induction is right. Induction shows that the product of $n$ countable sets is countable - it simply doesn't show anything about the product of infinitely many. "For all $n\in\Bbb N$" only says something about finite values of $n$. – David C. Ullrich Aug 27 '18 at 17:31
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    No. Your understanding of induction is flawed. You prove something holds for all natural numbers $n$ by induction. It does not follow that the same thing holds for $\infty$. Or maybe you think (falsely) that ${0,1}^\omega$ is the union of the sets ${0,1}^n$. – GEdgar Aug 27 '18 at 17:32
  • @DavidC.Ullrich That make sense. Thank you. – X Leo Aug 27 '18 at 17:37
  • It is not the same question as the ones I marked as duplicate, that's true. But it's the same answer: induction proves "for all finite", but does not prove "there is an infinite object whose restriction works for all finite". – Asaf Karagila Aug 27 '18 at 17:55

2 Answers2

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The theorem that "for all finite $n$, $\mathbb{N}^n$ is countable" is true. The "theorem" that "$\mathbb{N}^{\mathbb{N}}$ is countable" is not true. Induction on $\mathbb{N}$ shows that something is true for arbitrary finite naturals, but not for $\mathbb{N}$ itself.

I understand induction as justifying that "if I can build the truth of a statement out of its truth for smaller statements, then I can climb the ladder all the way to any desired endpoint": if I can show $P(n+1)$ from $P(n)$, then I can show $P(1000)$ by showing first $P(1)$, then $P(2)$, and so on. But I can't hope to show $P(\infty)$ this way, because my building procedure will never get there. To justify full transfinite induction, you need an additional induction step, one that applies at this sort of limit.

Patrick Stevens
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    Can I understand in this way? the theorem is true for arbitrary $n\in\mathbb{N}$ just means it is true for a finite $n$ as any element of $\mathbb{N}$ is finite. While when we discuss a countable infinite union, it is infinite number of unions, this infinite number no more belong to $\mathbb{N}$, thus can not be guaranteed by induction principle? – X Leo Aug 27 '18 at 17:41
  • @XLiu Yep, that looks about right. – Patrick Stevens Aug 27 '18 at 17:54
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Your first proof shows that for any $n\in\mathbb{N}$, the porduct $A_1\times\cdots\times A_n$ is finite but, this not say nothing about $\mathbb{N}$ itself. In the last case you need a transfinite induction argument and as you see it is false for the first limit ordinal.

YCB
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