For a homework assignment, I am asked to show that two sets A and B are equinumerous. I am wondering, if showing that the cardinality of A is equal to the cardinality of B, is enough to say a bijection exists which implies they are equinumerous? It is important to note that in the problems, the sets A and B are finite and countable. One of the problems is shown to give you a feel:

Prove that the following couples of sets are equinumerous by finding a one-to-one relation between them.

A is the fingers of one hand, B is the set of vowels

P.S. I don't want a solution, this is homework, I just want to know if I am on the right track on tackling the problem, thank you.

  • $\begingroup$ You mean you only finitely-many fingers? Weird :). $\endgroup$ – Passing By Dec 7 '14 at 1:44

Two sets have the same cardinality if and only if there is a bijection between them.

  • $\begingroup$ So am I correct in my thinking? This is a history of math class, I have taken discrete a long time ago. $\endgroup$ – Daniel Lopez Dec 7 '14 at 1:26
  • $\begingroup$ @DanielLopez Your thinking is correct. $\endgroup$ – Matt Samuel Dec 7 '14 at 1:27
  • $\begingroup$ Thanks, I have to wait 10 minutes to mark your answer. $\endgroup$ – Daniel Lopez Dec 7 '14 at 1:28
  • $\begingroup$ @DanielLopez You're welcome. $\endgroup$ – Matt Samuel Dec 7 '14 at 1:28

It is true that two sets have the same cardinality if there is a bijection between them. While in the finite case it can be straightforward to construct the actual bijections, this is not always easy to do when the sets are infinite. In this case there are other tools you may use, like Cantor-Schroeder -Bernstein's theorem http://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem

which says that if, given two sets $A,B$ , if you can find an injection $f: A \rightarrow B$ and an injection into $g: B \rightarrow A$, then you can conclude the two sets have the same cardinality. Of course, if $f,g$ as in the theorem exists, then you can construct an actual bijection, but this is not necessary if you can come up with $f,g$ as in the theorem.


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