Is My Proof on Finite Sets and One-to-One Correspondence Correct?

[AdrianZ]

Suppose that f is a one-to-one correspondence between two sets X and Y. Prove that if X is finite, then Y is finite too.

my proof: I've already proved that if X is infinite, then Y is infinite too. since f is a one-to-one correspondence, f^-1: Y->X exists and by applying the same theorem it can be shown that if f:X->Y and Y is infinite, then X is infinite as well.so, I can claim that if f is a one-to-one correspondence, then X is infinite if and only if Y is infinite. hence, It's possible to say that if f is a one-to-one correspondence between the two sets X and Y, then X is finite if and only if Y is finite.
Is my proof correct?

 

[micromass]

Yes, your proof is correct (once you know that infinite = not-finite). Though, it might not be the shortest proof available. That is, that you proved it by making use of infinite sets is suprising.

 

[AdrianZ]

Yes, your proof is correct (once you know that infinite = not-finite). Though, it might not be the shortest proof available. That is, that you proved it by making use of infinite sets is suprising.

Actually It's because the book I'm studying from uses Dedekind's definition of an infinite set that a set is infinite iff there is a one-to-one correspondence between the set and a subset of the set. and then it defines a finite set as a set that is not infinite.so, I tried to stay faithful to the definitions that my book suggests. surely there is a shorter way of proving this using the other definition that says a set A is finite iff it's in one-to-one correspondence with N_k. then I can say X~N_k and X~Y, hence Y~N_k. I guess you meant I could use the second approach and It would be shorter. Is that you what you mean?

 

[micromass]

Actually It's because the book I'm studying from uses Dedekind's definition of an infinite set that a set is infinite iff there is a one-to-one correspondence between the set and a subset of the set. and then it defines a finite set as a set that is not infinite.so, I tried to stay faithful to the definitions that my book suggests. surely there is a shorter way of proving this using the other definition that says a set A is finite iff it's in one-to-one correspondence with N_k. then I can say X~N_k and X~Y, hence Y~N_k. I guess you meant I could use the second approach and It would be shorter. Is that you what you mean?

Aah, that explains things! Yes, when working with Dedekind-finite things then you need to do it the way you do it.
Also remark that Dedekind-finite is not the same as finite in the other definition. You need the axiom of choice for that. So it's best to stay close to the definitions!

 

[blue_raver22]

Yes, your proof is correct. You have used the definition of a one-to-one correspondence and the theorem about finite sets being equivalent to each other under a one-to-one correspondence. Your explanation is clear and concise, and your conclusion follows logically from your proof. Good job!