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I'm having an extremely hard time figuring out how to prove this, would you have to start from one side of the proof and move to the other or prove with induction?

For any $n\in\mathbb{N}$, prove that the cartesian product $A^n = A\times,\dotsc,\times A$ ($n$ times) is countably infinite if $A$ is countably infinite.

Thanks so much for the help!

MickG
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Sydney Welch
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  • Do you know of a theorem that Cartesian product of two countable sets is countable? – Wojowu Jan 18 '16 at 21:16
  • Induction will work, but the case $n=2$ requires non-trivial work, unless you know already that the cartesian product of two countably infinite sets is countably infinite. – André Nicolas Jan 18 '16 at 21:17
  • I haven't done much of anything to try to prove it. I have a few notes about cardinality from class, but never on more than two sets in a cartesian product – Sydney Welch Jan 18 '16 at 21:21
  • Can you use the fact that $\mathbb{N}$ is of the same cardinality as $\mathbb{N} \times \mathbb{N}$? – AnotherPerson Jan 18 '16 at 21:27
  • If your notes on two sets in a cartesian product include $\Bbb N \times \Bbb N \simeq \Bbb N$, that's all you need. – BrianO Jan 18 '16 at 22:58
  • I suppose I could, but I'm still not sure where to go from there – Sydney Welch Jan 19 '16 at 01:55

2 Answers2

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You want to show that $A$ countable $\implies A^n$ countable. If $A$ is countable there is $f : A \to \mathbb N$ injective. We can create a new injection $g : A^n \to \mathbb N^n$ for any $n$, defined $$(a_1, a_2, \dots, a_n) \mapsto (f(a_1), f(a_2), \dots, f(a_n)).$$ To show $A^n$ countable it will suffice to find an injection $h : \mathbb N^n \to \mathbb N$; then the composition $h \circ g$ will give the desired map $A^n \to \mathbb N$.

The famous injective example $\mathbb N \times \mathbb N \to \mathbb N$ is sometimes called the Cantor pairing function. Rather than duplicating the explanation here I'd refer you to this Math Stack Exchange thread and this Wikipedia article. The idea is to represent $\mathbb N \times \mathbb N$ as a grid, and count the elements, starting at $(0,0)$ and weaving along. (This is the basis for showing $\mathbb Q$ and $\mathbb N$ are of the same cardinality.)

Now, if $k : \mathbb N\times \mathbb N \to \mathbb N$ is our injection, consider $\hat k : (\mathbb N \times \mathbb N) \times \mathbb N \to \mathbb N$ defined $((n,m),l) \mapsto k(k(n,m),l)$. The argument extends inductively.

Community
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amcerbu
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Your question is basically asking if the equality $|\Bbb N|^n=|\Bbb N|$ holds.

This is true, using the lemma below we have that

$$|\Bbb N^{n+1}|=|\Bbb N^{n}\times\Bbb N|=|\Bbb N^n|\cdot |\Bbb N|=|\Bbb N|\cdot|\Bbb N|=|\Bbb N|_\square$$

Lemma. $|\Bbb N\times \Bbb N|=|\Bbb N|$

Let $f:\Bbb N \to \Bbb N^2, x\mapsto (x,0)$. This is injective, thus $|\Bbb N|\le |\Bbb N^2|$.

Let $g:\Bbb N^2 \to \Bbb N, (x,y)\mapsto 2^x3^y$, if this is injective, we're done, is it?

YoTengoUnLCD
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