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

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  • $\begingroup$ You can't have a formal proof without proper definitions. For instance, what does "finite" actually mean to you? $\endgroup$ – Arthur May 27 at 14:32
  • $\begingroup$ 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 $\endgroup$ – Jordan May 27 at 14:36
  • $\begingroup$ 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. $\endgroup$ – Don Thousand May 27 at 14:38
  • $\begingroup$ @DonThousand That assumes you know that a subset of a finite set must be finite. $\endgroup$ – fleablood May 27 at 15:27
  • $\begingroup$ 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. $\endgroup$ – Calum Gilhooley May 27 at 15:48
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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.

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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.

Apply that to the concept of an injection and you're well on the way to something formal.

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