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In Royden's book Real Analysis, page 13, he writes that "We call two sets A and b equipotent provided there is a one-to-one mapping $f$ from A onto B Equipotence defines a equivalence relation among sets, that is, it is reflexive, symmetric, and transitive." He then uses equipotence to $\mathbb{N}$ define countably infinite.

I find out another definition of countably infinite using injection which is a one-to-one and left-total relation, not necessarily onto. But if it's not onto, how can the equipotence be symmetric?

When proving that the Cartesian product $\mathbb{N}\times\mathbb{N}$ is countably infinite, he defines a mapping $g$ by $g(m,n)=(m+n)^2+n$. This mapping from $\mathbb{N}\times\mathbb{N}$ to $\mathbb{N}$ is one-to-one but not onto.

casablancahp
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    Did you prove that it is not onto? – Julian Rachman Dec 27 '14 at 07:44
  • @JulianRachman It isn't onto. There are no $a$ and $b$ such that $g(a,b)=3$. It isn't symmetric, $g(1,2)=11$ but $g(2,1)=10$. – Suzu Hirose Dec 27 '14 at 08:02
  • Leave alone the second definition for a second. My argument is that if equipotent relation is symmetric he then needs to prove not only that NxN is equipotent to N, which he did, but also that N is equipotent to NxN. Then we need a one-to-one and left-total mapping f from N to NxN. In other words, he needs a bijective mapping, not just a injective mapping. – casablancahp Dec 27 '14 at 08:15
  • @casablancahp Are you familiar with the way that it is proved that the rational numbers are countable? You can apply the same trick to show that $N\times N$ is countable. – Suzu Hirose Dec 27 '14 at 08:23
  • @SuzuHirose Here's a useful link which helps me understand the Cantor-Schroeder-Bernstein Theorem http://math.stackexchange.com/questions/54158/the-cartesian-product-mathbbn-times-mathbbn-is-countable – casablancahp Dec 27 '14 at 09:36

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The function isn't onto since there is no $a$, $b$ such that $g(a,b)=3$. You can see this by considering a few cases: $g(0,1)=2$, $g(1,0)=1$, $g(0,0)=0$, and $g(1,1)=5$.

However, if $g(a,b)$ is unique in $\Bbb N$ then it's possible to define a bijective mapping $f$ from the range of $g$ to $\Bbb N$ such that $f \circ g$ is onto. (I don't know whether $g(a,b)$ actually is unique in $\Bbb N$.)

Regardless of whether this function $g$ "works" or not, we can make a bijective map from $\Bbb N\times\Bbb N$ to $\Bbb N$ using the same trick as the one for proving that the rational numbers are countable. The general result is that the cross-product of two countable sets is countable.

Suzu Hirose
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  • As you said, there are no such a and b that g(a,b)=3. Then there's no bijective mapping from NxN to N. The proof is fine if it's based on the second definition. But Royden's definition of countably infinite using equipotence confuses me.Especially the symmetric relation he mentioned. – casablancahp Dec 27 '14 at 08:32
  • To answer your question, g(a,b) is unique (one-to-one) in N cause m=a and n=b if g(m,n)=g(a,b). |m+n+a+b||m+n-a-b|=|b-n|. So b=n otherwise the natural number _m+n+a+b_both divides and greater than the natural number |b-n|. It's the proof in Royden's book. – casablancahp Dec 27 '14 at 08:57
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I'm not sure where Royden gives the definition of countably infinite using injection. If this is in reference to Theorem 3 on page 13, I think it only says "A subset of a countable set is countable."

Consider $A \subseteq B$, with B a countable set.

In the proof he discusses two cases, one where B is finite, and the other when B is countably infinite.

In the countably infinite case, he gives a selection process that forms a one-to-one correspondence between ℕ and A. If A is not finite, we can show that not only is A countable (finite or countably infinite), it is countably infinite.

Thus, as long as B is countably infinite, A a subset of B and A not finite, we can use the selection process he mentions to create a bijection between ℕ and A.

When proving that the Cartesian product ℕ×ℕ is countably infinite, he creates an injection from ℕ×ℕ to ℕ. Thus there is a bijection from ℕ×ℕ to a subset of ℕ. Since ℕ is countably infinite, we can use the selection process mentioned above to create a bijection from this subset of ℕ to ℕ.

Jeff N.
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