I recently had problem set which asked to prove that if for any , we have
then is an injection. (EDITED**)
My proof started with the intention to show that if (pointwise) . And I did that, but when I got it returned, apparently I didn't use the definition of an injection since injectivity implies that for all implies .
However, I feel that my proof was valid since A and B can have very well been singleton sets with x1 and x2, respectively. Is my approach incorrect?
I had to prove the following statement:
Let be a function with domain and codomain
Prove that if for any we have then is an injection.
Sorry, I did not clarify the injection clause before.
I agree that if
f(A) = f(B) implies A = B for all A, B (*),
then
f(x1) = f(x2) implies x1 = x2 for all x1, x2 (**).
As you said, (*) implies (**) by taking A = {x1} and B = {x2}. Therefore, if you proved (*), you basically solved the problem. However, while the step from (*) to (**) can be considered negligible and omitted when talking about some complicated theorem whose proof takes many pages, it is not negligible in the problem designed to test your understanding of injective functions. Since the problem is relatively simple, the proof also has to be spelled out in great detail. So I can understand the issue your instructor had with your proof, which does not even use the definition of injection.
As Plato says, it is easier to prove (**) by considering singleton A and B from the start.