Consider the set:

$$I_n = \{p\in \mathbb{N}; 1<p\le n\}$$

My book says that a set is finite when it's not empty or when there exists, for some $n\in \mathbb{N}$, a bijection: $$\phi: I_n\to X$$

Then, it defines an infinite set as a set that is not finite. This can be understood as:

a set $X$ is infinite when it's not empty and for all $n\in \mathbb{N}$, there isn't any bijection $\phi: I_n \to X$

Later, my teacher gave a list of exercises, in which one of them is:

Prove that a set $X$ is infinite iff it's no empty neither has a bijection $f: I_n \to X$ no matter which $n\in \mathbb{N}$

Well, I know that:

$\to$ if there isn't a surjective function $f: I_n \to X$, then there isn't also a bijective function $f: I_n\to X$, because bijectivity is surjectivity with injectivity, and by the definition, this set is not finite, that is, infinite.

$\leftarrow$ well, since $X$ is infinite, then it's not finite, which means that there isn't any bijective function from $I_n$ to $X$, so I know that this function must be either only injective, only surjective, or none of them. I must prove now that surjectivity must always fail, but I can't see anymore assumptions to use here. Intuitively I know that there can't possibly exist a surjective function from $I_n$ to an infinite set because all members of $I_n$ must be mapped to a unique element in $X$, but there are infinite elements in $X$. Could someone help me?


Using the definition of your book, proving your teacher exercice is a simple tautology.

$\leftarrow$ Since there isn't a bijection $f: I_n \rightarrow X$ then $X$ is not finite, i.e. $X$ is infinite.

$\rightarrow$ Since $X$ is infinite, then it's not finite, which means that there isn't any bijective function from $I_n$ to $X$.

Your proof would end here.

But what you DO want to prove is that there is no surjective $f: I_n \rightarrow X$.

First things first: We can define a injective $f': I_n \rightarrow X$ for every $n \in \mathbb{N}$. Using induction:

  • For $n=2$ the set $I_n$ has one single element, so $f'$ must be injective (in order to be a function). Since $X$ is non-empty, the function $f'$ exists (and isn't surjective);
  • Now assume your function $f'$ is injective for some $n$. Since $X$ is infinite, $f'$ isn't surjective, so there is a $x \in X$ such that doesn't exists $m \in I_n$ such that $f'(m)=x$. Define $f'': I_{n+1} \rightarrow X$ as $f''(i)=f'(i)$ for all $i \in I_n$ and $f''(n+1) = x$.

So now supose you have a surjective $f$ for some $n$. Then we can define a injective function $g: X \rightarrow I_n$ associating for every $x \in X$ one of the elements of the pre-image $f^{-1}(x)$. By the Schröder–Bernstein theorem this would mean there is a bijection $h: I_n \rightarrow X$ (since we have proven that exists a injective $f': I_n \rightarrow X$ for every $n \in \mathbb{N}$). And so we conclude such a surjective function $f$ cannot exist.

PS: I'm asking myself why the book did not include the number $1$ on the definition of $I_n$.


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