Can the set of natural numbers have a injection with a finite set? [closed]

My intuition says that is not possible, but i am not achieving a great formal proof

• You can't have a formal proof without proper definitions. For instance, what does "finite" actually mean to you? – Arthur May 27 at 14:32
• finite set is a set with a bijection between itself and a set I={1,2,3...,n} for some n. This is the definition i have here – Jordan May 27 at 14:36
• An injection is a bijection with the image. A set that injects onto a finite set bijects with the image, which is also finite. Hence, the set is finite. – Don Thousand May 27 at 14:38
• @DonThousand That assumes you know that a subset of a finite set must be finite. – fleablood May 27 at 15:27
• I'd like to see this question reopened, but I can't vote for it in its present form. I suggest (1) incorporating the question already in the title into the body of the ... er ... question (this is always required in Maths.SE - the body of the question is expected to be self-contained, because the title is only a summary), and (2) incorporating the definition of finiteness given in a comment into the body of the question. I don't think anything more is needed, although I'm open to corerection ... er, correction. – Calum Gilhooley May 27 at 15:48

A function $$f: \mathbb{N} \to A$$ is injective if $$n_1 \ne n_2 \Rightarrow f(n_1) \ne f(n_2)$$. If the images of different natural numbers must be different, $$A$$ cannot be finite.
You could apply the pigeonhole principle, which is that if you have $$n$$ holes and $$n + 1$$ items you have more than $$1$$ item in at least one hole.