If $A$ is finite and $B\subset A$, then $B$ is finite

Question

I am having trouble completing the proof of this statement.

If $A$ is finite then every $B \subset A$ is finite, where "finite" means "empty or in bijection with $\{1,...,t\}$ for some $t \in N$"

Suppose that $A\neq \emptyset$ and $A$ is not a singleton. It suffices to prove that if $B \subset \{1,...,m\}$ for some $m \ge 2$, then $B$ is in bijection with some $\{1,...,t\}$ for some $t \ge 1$. Let $a=\min B$.

For $x\in B$, consider:

$$S(x) = \begin{cases} \min\{t\in B: t>x\}, \text{ if $x < \max B$} \\ \max B, \text{ otherwise} \end{cases}$$

It's clear that there exists $r\in N$, such that $S^n(a) =S^r(a)$ for all $n\ge r$. Where $S^n(a)$ is $S(S(...S(a)..))$.

Let $t = \min\{r\in N: S^n(a) = S^r(a), \forall n\ge r\}$. We claim that: $B=\{a,...,S^t(a)\}$.

Proof: (here is where I am having trouble)

It is clear that $\{a,...,S^t(a)\}\subset B$. Now if $x\in B$, then if $x=a$ we are done. Suppose that $x > a$. If $L_x:= \{k\in B: a<k<x\}=\emptyset$, then clearly $x = S(a)$.

If $L_x \neq \emptyset$, how to prove that there must be some $r\in \{1,...,t\}$ such that $x = S^r(a)$?

Thanks

Have you looked at http://math.stackexchange.com/questions/239566/subset-of-a-finite-set-is-finite? — Asaf Karagila, Sep 21 '16 at 04:37
@AsafKaragila yes, thanks. I am interested in completing my approach. — Banach Manifold, Sep 21 '16 at 04:38
@BanachManifold It seems a bit fishy to me to use the order properties of $\mathbb{N}$ and "concrete" notions like $\max$ and $\min$ to prove a general set theory problem. Note that the set ${1, 2, 3...n}$ does not have any order, or any structure at all. It's isomorphic to any set containing $n$ objects, like for example ${1,2}$ is isomorphic to ${ \emptyset, { \emptyset } }$ — MathematicsStudent1122, Sep 21 '16 at 04:58
@BanachManifold In any case, I'm probably wrong since Asaf has uses those notions in his answer too :P But yes you are right that every bounded nonempty subset of $\mathbb{N}$ has a max and a min — MathematicsStudent1122, Sep 21 '16 at 05:00
@MathematicsStudent1122 any subset of $N$ has the order induced by the usual order on $N$. — Banach Manifold, Sep 21 '16 at 05:00

score 0 · Answer 1 · answered Sep 21 '16 at 04:40

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Prove the following by induction on $n$: If $A$ is in bijection with $\{1,...,n\}$ then $B$ is finite.

Step Case: Suppose $A$ is in bijection with $\{1,...,n+1\}$. Let $\phi$ be this bijection. Then $A - \{\phi(n+1)\}$ is in bijection with $\{1,...,n\}$. Apply the IH to conclude that $B - \{\phi(n+1)\}$ is in bijection with $\{1,...,t\}$ for some $t$. Then it is easy to show that $B$ must be finite.

answered Sep 21 '16 at 04:40

James

989

Yes, I know this approach. I would like to see how my proof can be completed. – Banach Manifold Sep 21 '16 at 04:46

If $A$ is finite and $B\subset A$, then $B$ is finite

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