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If $A$ is a set of odd numbers and $B$ is the set of even number, then is their product = set of natural numbers ?? In other word, how can we define a map $f: A\times B \rightarrow\mathbf{N}$ and prove the bijection of that function?

learner
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Danish Butt
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  • The answer to the first question is no. This renders the second question useless. – Git Gud Nov 16 '13 at 11:07
  • I think Danish must be talking about cardinality, because the question makes no sense otherwise. In which case the answer to the first question would be yes, and my only comment is that it's easier to find two injections that it is a single bijection. –  Nov 16 '13 at 11:10
  • Even even if he means cardinaly instead of equality, the answer to the first question is still no. Consider $A={1}$ and $B={2}$. – Git Gud Nov 16 '13 at 11:12
  • Again, I think that's just down to Danish writing the question out incorrectly. –  Nov 16 '13 at 11:14
  • Simply i need to prove : AxB ~ N is bijective. – Danish Butt Nov 16 '13 at 11:20
  • @DanishButt Read my comment above. – Git Gud Nov 16 '13 at 11:22
  • @GitGud No there exist the bijectivity because it has used in order to solve the problem of cardinal numbers. – Danish Butt Nov 16 '13 at 11:28
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    If we take the question literally, then $A$ is a subset of the odd numbers, whereas $B$ are all of the even numbers. Hence, the answer is "yes" iff $A \ne \emptyset$. – gerw Nov 16 '13 at 11:48

2 Answers2

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Hint: There is standard theorem witch says that for infinite set $X$, $|X\times X|=|X|$. Try to find bijections $B\rightarrow N$, $A\rightarrow N$. Use theorem (or proof of this theorem if you want bijection not only equality of cardinalities).

user52045
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  • Doesn't that rely on the axiom of choice in general? –  Nov 16 '13 at 11:54
  • I guess it does. At least i would prove this well ordering X. – user52045 Nov 16 '13 at 11:58
  • If i remember correctly equivalence ($|X|\le|Y|$ and $|X|\ge|Y|$) iff $|X|=|Y|$ require AC too. – user52045 Nov 16 '13 at 12:02
  • I'm not sure if that's right or not, but I did find this question: http://math.stackexchange.com/questions/69774/is-the-class-of-cardinals-totally-ordered

    Nevermind, it's not true. http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem

     –  Nov 16 '13 at 12:12
  • Zermelo's theorem says that fact that every set can be well ordered is equivalent to AC. – user52045 Nov 16 '13 at 12:17
  • It is interesting tho how strong theory ZF + Cantor theorem is. – user52045 Nov 16 '13 at 12:20
  • @user52045: Cantor's theorem is provable in ZF; for infinite well-ordered sets $A$ it is true in ZF that $|A|=|A\times A|$. Countable sets are well-ordered by definition. – Asaf Karagila Nov 16 '13 at 12:34
  • Ofc but can you tell something about ZF + Cantor or is this for some reason not interesting. – user52045 Nov 16 '13 at 12:36
  • Your question doesn't make any sense. Adding to ZF a theorem of ZF doesn't add any new consequences. Cantor's theorem is a theorem of ZF, so adding it as an explicit axiom doesn't add any new consequences to ZF. – Asaf Karagila Nov 16 '13 at 12:39
  • I know that cantor works in ZF if you apply it only to well ordered sets. But i dont think it does if we cant well order set $X$. My question is if its stronger if i have Cantor for all sets without having them well ordered. I dont think cardinals are well ordered without AC. – user52045 Nov 16 '13 at 12:44
  • No, Cantor's theorem about the size of power sets is true for every set in ZF. – Asaf Karagila Nov 16 '13 at 12:45
  • Sorry i meant Cantor–Bernstein–Schroeder theorem i thought its obvious form the previous comments. – user52045 Nov 16 '13 at 12:46
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    That theorem is also provable for every two sets in ZF. What is not true anymore is that we can use surjections instead if injections. – Asaf Karagila Nov 16 '13 at 12:48
  • Thanks for the clarification. I guess i have to look at this one more time. It was a long time ago. – user52045 Nov 16 '13 at 12:51
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We know that N~E where N is set of natural no and E is set of even no.

Its mean that there exist a map such as 

     f:N--->E   (which is bijective)

similarly N~O where N is set of natural no and O is set of Odd no. Such as g:N--->O (which is bijective) So there exist a map such as:

     h:NxN ---> ExO
     h(m,n)=(f(m),g(n)) for all m,n belongs to N.

This is bijective. So NxN ~ ExO And N ~ NxN which implies that N ~ ExO

Danish Butt
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