I have two questions. First of all, I have defined the set $\mathbb{N}$ using the axioms of ZFC, and I have also defined arithmetic on it. I have also proven the induction principle and I am free to use it.
I am now trying to prove the following: a set is finite if and only if it is in bijection with a natural number. I have shown by induction that every natural number is finite, so only one direction remains. For your information, in my framework a set is defined to be infinite if it is in bijection with a proper subset of it.
I have also proven the following fact: if a set contains an infinite subset, then it is infinite. So here is my proof: (the set is denoted by $S$)
If $S$ is empty then there is nothing to prove. So, let $S$ be a non-empty finite set. For the sake of contradiction, suppose $S$ is not in bijection with a natural number. We will show that there is an injective map $f:\mathbb{N}\to S$, and hence $S$ will contain an infinite subset, which will imply that $S$ is itself infinite, and hence a contradiction will be obtained.
First, we show that for every non-zero $n\in\mathbb{N}$, there is a subset of $S$, which we denote by $S_n$, such that $n$ is in bijection with $S_n$. Moreover, we will show that $S_{n}\subset S_{n + 1}$ for every $n\in\mathbb{N}$. This we show by induction.
For the base case, consider $1 = \{0\}$. Choose any $s_0\in S$, and consider the map $0\mapsto s_0$. So, $1$ is in bijection with the set $S_1 = \{s_0\}$. Now, suppose $n\in\mathbb{N}$ is in bijection with some subset $S_n\subset S$, and let this bijection be $\theta$. Clearly, $S - S_n$ is non-empty, otherwise it would contradict our assumption that $S$ is not in bijection with a natural number. So, there is some $s_{n}\in S - S_n$. Consider the map that sends any $m < n$ to $\theta(m)$, and maps $n$ to the element $s_n$. Let $S_{n + 1} = S_n\cup\{s_n\}$. So this map is a bijection from $n + 1$ onto $S_{n + 1}$, and also $S_n\subset S_{n + 1}$ (where the inclusion is proper). So by induction, this statement is true for all $n\in\mathbb{N}$.
Now, we construct an injection from $\mathbb{N}$ into $S$ as follows. By our hypothesis, every set $S - S_n$ is non-empty. Let $f:\mathscr{P}(S)\to S$ be a choice function as guaranteed by the axiom of choice. Then consider the following map: $0\mapsto s_0$ and $n\mapsto f(S_{n + 1} - S_{n})$ for every $n\ne 0$. This is easily seen to be a one-one map from $\mathbb{N}$ into $S$. This completes the proof.
Now I have two questions: I wanted to confirm whether the induction step is valid or not. And second and more importantly, can this proof be written without invoking the Axiom of Choice?
I have been thinking hard about it, but still no progress.
