Could someone please walk me through line by line what is going on in this proof.

The proof for a finite set having a finite subset has been shown to me as follows

Proposition: If $S$ is a nonempty finite set and $A \subseteq S$ then $A$ is also finite.

Proof: If $A = \emptyset$ there is nothing to prove (since the $A$ would be finite).

Since $S$ is finite there exists a bijection $f: S \to [n] $ for some $n \in \mathbb{N}$ consider the restriction of $f$ to $A$, $f|_A : A \to [n] $. This is injective since $f$ is injective. Let $S =\{k\in[n] : f(a) = k \text{ for some }a \in A\}$. Write $S$ in increasing order $s_1 , s_2 , s_3 , s_4 , \dots , s_m $. Let $a_1 ,\dots,a_m $ be elements of $A$ where $f(a_1) = s_1$ , $ f(a_2) = s_2$, $\dots$ , $f(a_m) = s_m$. Since $f$ is 1-1 these elements are unique and since $f$ is surjective, $A = \{ a_1 ,a_2 ,\dots,a_m\}$. Define $g:A\to [m]$ by $g(a_k) = k$ from above we see that $g$ is a bijection, so $A$ is finite.

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    $\begingroup$ What exactly don't you understand? What's the first bit that trips you up? $\endgroup$ – Noah Schweber Mar 18 '17 at 21:43
  • $\begingroup$ @NoahSchweber I understand that the bijection exists from $S \to \mathbb{N}$ I think Im getting tripped up when S is redefined and it appears to me that we are using the same mapping from $S \to \mathbb{N}$ as from $A \to \mathbb{N}$. I guess Im lost from how we show that $A$ is finite by using the same subscripts that we had in the S sequence in the A sequence. Thank you for your help. $\endgroup$ – user372382 Mar 18 '17 at 21:52
  • $\begingroup$ Also this: math.stackexchange.com/questions/239566/… $\endgroup$ – Asaf Karagila Mar 19 '17 at 4:13

The proof is indeed confusing.

I assume your definition of finite set is “having a bijection with $[n]$, for some natural number $n$”, where $[n]=\{1,2,\dots,n\}$ (and $[0]=\emptyset$).

We'll do induction over $n$. The case $n=1$ is obvious, as $S=\{f^{-1}(1)\}$ and its subsets are $\emptyset$ and $S$.

Assume $n>1$ and that the statement is true for sets in bijection with $n-1$.

Suppose $f\colon S\to[n]$ is a bijection and that $A\subseteq S$. Set $s_n=f^{-1}(n)$ and $S'=S\setminus\{s_n\}$.

If $A=\emptyset$, there is nothing to prove. If $s_n\notin A$, then $A$ is a subset of $S'$ and the map $f$ induces a bijection $f'\colon S'\to[n-1]$. By the induction hypothesis, we are done.

Suppose $s_n\in A$. Then $A'=A\setminus\{s_n\}$ is a subset of $S'$. If $A'=\emptyset$, we have $A=\{s_n\}$ and we can find a bijection $h\colon A\to[1]$. Suppose $A'\ne\emptyset$. Again, $f$ induces a bijection $f'\colon S'\to[n-1]$ so, by the induction hypothesis, there exists $m$ and a bijection $g\colon A'\to[m]$. Then, defining $h\colon A\to[m+1]$ by $$ h(a)=\begin{cases} g(a) & \text{if $a\ne s_n$}\\[4px] m+1 & \text{if $a=s_n$} \end{cases} $$ we obtain a bijection.

  • $\begingroup$ Thank you for your answer, So in essence what the proof is saying lets assume that for any $[n] \in \bbmath{N}$ there exists an element back in S which we will map to. S' is the set of us removing these elements which we map back to through the inverse one at a time from S. From there we create two scenarios where either A is the subset of S' or A' is the subset of S'. By removing those elements from S we don't want because they aren't in A. And since in these two scenarios we can make a bijection either back to the original set or [m] the set is countable. $\endgroup$ – user372382 Mar 18 '17 at 22:53
  • $\begingroup$ @user372382 Yes, the idea is to reduce to a set $S'$ having one element less than $S$, so the induction hypothesis can be applied. The proof you have goes to some length into avoiding induction (but uses it implicitly). They basically “order” $A$ using as indices afforded by $f$, but do so in a disputable way: at some point. $\endgroup$ – egreg Mar 18 '17 at 23:03
  • $\begingroup$ I think thats where I got confused originally where the indices are changing between steps but with induction it seems much simpler as the induction moves through the set in a sequential order. Its very interesting so by removing one element we arrive at two cases where either this element we choose to remove is in A or not in A but in any case they are finite. Its interesting how such a simple concept of taking finitely many things from a finite collection has such a complicated argument. Thank you for your help you have really cleared things up for me a lot. $\endgroup$ – user372382 Mar 18 '17 at 23:09

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