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My work so far:

I think I need to prove that N x N is either countable or countably infinite. My first step:

A set A is countably infinite if there is a bijection (1-1 and onto) function f : N → A. If a function is bijective, it is also surjective.

Let A represent N x N. The set N × N = {(n, m) : n, m ∈ N} is countable.

PROOF of the above statement

Is this the correct way to prove the original question?

rs479173
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1 Answers1

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There are many ways to construct functions in order to show that $\Bbb N$ and $\Bbb N\times\Bbb N$ have the same cardinality. A bijection can be constructed, but not by using the correspondence $(n,m)\leftrightarrow2^n3^m$. That correspondence doesn't give us a bijection, because not all integers are represented on the right side, indeed, anything with a prime factor greater than $3$ does not occur here.

One can make a bijection according to the strategy outlined at this MSE question: How does one get the formula for this bijection from $\mathbb{N}\times\mathbb{N}$ onto $\mathbb{N}$?, but that's not really necessary. Another way to show that two sets have equal cardinality is to construct injections in both directions.

An injection from $\Bbb N$ to $\Bbb N\times\Bbb N$ is trivial: $n\mapsto (n,0)$ works, for example. In the other direction, sending $(n,m)\mapsto p^mq^n$ for any primes $p$ and $q$ is great.

G Tony Jacobs
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  • I need to construct a surjection. How would I go about doing that? – rs479173 Dec 04 '17 at 00:13
  • @rs479173: the opposite directions give a surjection. $(n,0)\to n$ is a surjection, now just send all the other pairs somewhere. – Ross Millikan Dec 04 '17 at 00:30
  • A surjection from $\Bbb{N}$ to $\Bbb N\times\Bbb N$ could be constructed in the same spirit. Just send any integer of the form $2^n3^m$ to the pair $(n,m)$, and send every other integer to $(0,0)$ or something. – G Tony Jacobs Dec 04 '17 at 00:33