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The Cartesian product of a family $(A_i\mid i\in I)$ is defined as $$\prod\limits_{i\in I}A_i=\{f:I\to\bigcup A_i\mid f(i)\in A_i\}$$

Let $I_n=\{i \in \mathbb N \mid i \le n\}$ for $n\in \Bbb N$ and $A_i$ be countable for all $i\in I_n$. Then $\prod\limits_{i\in I_n}A_i$ is countable.

My attempt:

Lemma 1: $(\prod\limits_{i\in I_n}A_i) \times A_{n+1}$ and $\prod\limits_{i\in I_{n+1}}A_i$ are equinumerous for all $n \in \mathbb N$. (I presented a proof here)

Lemma 2: If $A$ and $B$ are countable, then $A \times B$ is countable. (I presented a proof here)

I will prove the theorem by induction on $n$. The theorem is trivially true for $n=0$. Assume that the theorem is true for $n=k$, then $\prod\limits_{i\in I_k}A_i$ is countable. Since $\prod\limits_{i\in I_k}A_i$ is countable (by inductive hypothesis) and $A_{k+1}$ is countable, then by Lemma 2 $(\prod\limits_{i\in I_n}A_i) \times A_{n+1}$ is countable. Furthermore, by Lemma 1 $(\prod\limits_{i\in I_k}A_i) \times A_{k+1}$ and $\prod\limits_{i\in I_{k+1}}A_i$ are equinumerous, then $\prod\limits_{i\in I_{k+1}}A_i$ is countable too. Thus the theorem is true for $n=k+1$. This completes the proof.

Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!

Community
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Akira
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  • I don't understand why you need Lemma 1 at all. You could finish with "then by Lemma 2 (...) is countable." I think you need to prove lemma 2 (this is basically the heart of the original question) – Yanko Sep 12 '18 at 16:01
  • @Yanko While it's very intuitive, I found quite embarrassed If I can not give a proof for it ^^ – Akira Sep 12 '18 at 16:03
  • This is correct assuming you when you say $I_n={i \in \mathbb N \mid i \le n}$ you mean $n\color{red}{\in\Bbb N}\mbox{ s.t. }I_n={i \in \mathbb N \mid i \le n}$ – ℋolo Sep 12 '18 at 16:03
  • @LeAnhDung This is not so easy at all. Look here https://proofwiki.org/wiki/Cartesian_Product_of_Countable_Sets_is_Countable – Yanko Sep 12 '18 at 16:05
  • @Yanko how from $(\prod\limits_{i\in I_n}A_i) \times A_{n+1}$ is countable you get $\prod\limits_{i\in I_{n+1}}A_i$ is countable without Lemma 1? – ℋolo Sep 12 '18 at 16:06
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    @Holo: that's not right! By your definition, every $I_n$ is equal to $\mathbb N$. – TonyK Sep 12 '18 at 16:06
  • @TonyK Yes, thanks for notifying me, I should have write this before the definition of the set(I edit it) – ℋolo Sep 12 '18 at 16:08
  • @Yanko Did you mean my proof is not correct? :) – Akira Sep 12 '18 at 16:11
  • Could some of you please have a check on my attempt? – Akira Sep 12 '18 at 16:14
  • @LeAnhDung Like I said, it is correct assuming the $n$ you are talking about are natural – ℋolo Sep 12 '18 at 16:15
  • @Holo, of course $n \in \Bbb N$ :) – Akira Sep 12 '18 at 16:16
  • @Holo I have edited my post to reflect that. – Akira Sep 12 '18 at 16:18
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    @LeAnhDung you would be surprised, "of course" is a dangerous statement :), but with the edit it is correct – ℋolo Sep 12 '18 at 16:19

3 Answers3

4

Let $A$ and $B$ be countable, i.e. we have the enumerations

$$a_1,a_2,a_3,\cdots$$ and $$b_1,b_2,b_3,\cdots$$

Then we can enumerate $A\times B$ as

$$(a_1,b_1),(a_2,b_1),(a_1,b_2),(a_3,b_1),(a_2,b_2),(a_1,b_3),\cdots$$

(notice the sums of the indexes, $2,3,3,4,4,4,\cdots$). You can easily check that all pairs $(a_n,b_m)$ are cited.

Then by induction,

$$\prod_{i=1}^n A_i=\left(\prod_{i=1}^{n-1} A_i\right)\times A_n.$$ is countable.

  • Hi @Yves, I have been thinking about your approach. Recently I have formalize it into a proof at https://math.stackexchange.com/questions/2918643/bbb-n-times-bbb-n-is-countably-infinite. Could you please have a look at it? Thank you so much :) – Akira Sep 16 '18 at 05:06
  • @LeAnhDung: sorry, I haven't had the courage to scrutinize it. The first part of my answer shows an easy bijection. Is there a need for more sophistication ? –  Sep 16 '18 at 09:33
  • It's ok :) In my proof, I use a mapping $f$ such that $f(i,j)=i+\sum\limits_{k=0}^{i+j}k$. I'm curious about your mapping. Please show it! – Akira Sep 16 '18 at 09:42
  • @LeAnhDung: on the sixth line. –  Sep 16 '18 at 09:54
  • It seems that your mapping is $f(i,j)=i+\sum\limits_{k=0}^{i+j}k$. – Akira Sep 16 '18 at 09:58
  • @LeAnhDung: possibly, but I don't see a need for a formula. It is obvious that all sums of indexes are enumerated, and for a given sum, all possible pairs of indexes. –  Sep 16 '18 at 10:21
  • Did you mean: All sums of indexes are countable. And all pairs of indexes for a given sum are also countable. – Akira Sep 16 '18 at 10:26
  • @LeAnhDung: no, I said enumerated. This is a constructive answer. –  Sep 16 '18 at 10:33
  • Honestly, I think that if I wrote Then we can enumerate $A\times B$ as $(a_1,b_1),(a_2,b_1),(a_1,b_2),(a_3,b_1),(a_2,b_2),(a_1,b_3),\cdots$. I may receive such criticism from other users as You are sloppy at that point or The general idea is good. You may want to give further justification for some statements. How do you know that there exists such a bijection :) – Akira Sep 16 '18 at 10:39
  • @LeAnhDung: a matter of point of view. The case is simple enough that any reader can generalize and extend the sequence at will (in fact the notation defines an algorithm). You might consider that a plain summation is better, but it's just the same rigorous information presented differently. –  Sep 16 '18 at 10:46
  • I got it. Thank you so much! – Akira Sep 16 '18 at 10:49
  • @LeAnhDung: but you are right, some people won't be open to this point of view. –  Sep 16 '18 at 10:51
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    I did meet some people of that kind in MSE :) – Akira Sep 16 '18 at 11:01
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It's easier using prime factorization.

Map the j-th element of $A_i$ to $p_i^j$, where $p_i$ is the i-th prime And multiply these representations to get the unique integer corresponding to any particular subset.

marty cohen
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  • I got your solution @Marty. My main concern is the correctness of the above proof. Could you please verify it? – Akira Sep 12 '18 at 16:10
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With this inductive method, the heart of the matter boils down to proving $A\times B$ is countable -- your Lemma 2. While Lemma 1 and the actual induction are necessary (and it was smart to notice that Lemma 1 needs proving), a proof of Lemma 2 is still essential.

Standard arguments for Lemma 2 include primes like @Marty used or enumerating from a corner when considering the cartesian product as a table of pairs.

Hans Musgrave
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