0

If we know that $\mathbb{N} \times \mathbb{N}$ is equivalent to $\mathbb{N}$ (same cardinal), can we conclude that the cartesian product $A \times B$ of two countable sets $A$ and $B$ is countable? In particular I already proved that a bijection $\phi: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ exists such that $\phi(x, y) = x +\frac{(x + y -1)(x + y - 2)}{2}$. So, presuming that I am writing a formal proof, how would I show that $A \times B$ is countable? Any assistance is much appreciated.

SupremePickle
  • 1,291
  • You’d need a lemma that if $X$ is the same cardinality of $A$ and $Y$ the same as $B$ then $X\times Y$ is the same cardinality as $A\times B.$ Then $A,B$ countable means $A\times B$ has the same cardinality as $\mathbb N\times N$ which you have shown is the same cardinality as $\mathbb N.$ This lemma is not hard to prove. – Thomas Andrews Feb 25 '21 at 19:45
  • @amwhy I disagree that the marked duplicate is a duplicate. Kraftsman specifically has trouble getting from $A$ to $\mathbb{N}$, not showing that $\mathbb{N} \times \mathbb{N}$ is countable. – Patrick Stevens Feb 25 '21 at 19:46
  • I think in my search so far https://math.stackexchange.com/a/2914553/259262 has come closest to answering the question. – Patrick Stevens Feb 25 '21 at 19:47
  • Bafflingly enough, I still haven't found a suitable answer other than mine, i.e. one that notes the reason it suffices to consider $\mathbb{N} \times \mathbb{N}$ is because we can biject $A \times B$ with it. – Patrick Stevens Feb 25 '21 at 19:49
  • I was just going to check in, but your answer does not verify that A bijects with N, it merely claims it does, so I cannot yet find your answer suitable. It's a one word sentence, followed by a question. Hardly an answer. If you want to sincerely answer, I'll reopen and upvote your answer. But as of know, your answer adds nothing. – amWhy Feb 25 '21 at 19:55
  • @amwhy I answered as I did because I believe the question is one which requires a hint, not a solution. I have provided the domain and codomain of two bijections, thereby rephrasing the question to highlight precisely the important parts of it. I justify this based on https://math.meta.stackexchange.com/questions/28969/is-it-acceptable-to-leave-hints-as-answers, and in particular on my belief that a solution will be less helpful to the asker than a hint. – Patrick Stevens Feb 25 '21 at 20:06
  • It's not helpful. You are selective in your links; only good hints can help a user move along. Yours is not a good hint.. But I've seen you have a pattern, at least today, of answering questions that need to be closed. I don't know anything to say anything more, but it does raise my concerns. – amWhy Feb 25 '21 at 20:38
  • Please post comments about your answer below your answer. If you address me again, below the question, I will repeat your address, and respond, below your answer. – amWhy Feb 25 '21 at 20:42

1 Answers1

0

Certainly. A[n infinite] countable set is one which bijects with $\mathbb{N}$, so can you build a bijection $A \times B \to \mathbb{N} \times \mathbb{N}$ and then to $\mathbb{N}$?

Patrick Stevens
  • 36,855