A subset of a finite set is finite

Question

I saw this problem in a book and the solution given and thought of a simpler solution. This makes me suspicious of its validity. If someone could point out why my solution is incorrect I would appreciate it.

First, suppose that $|S|=1$ and $T\subset S$. Then $T=\emptyset$ or $S$. Hence it is finite. Assume that if $|S|=n$ and if $T\subset S$ then $T$ is finite. Now let $|S|=n+1$ and $T\subset S$. If $T=S$ then indeed it is finite. If not, $\exists a\in S$ such that $a\notin T$. Then $T\subset S\setminus${$a$}. $|S\setminus${$a$}$|=n$ (which I have proved in a previous example). Then by the induction hypothesis $T$ is finite.

Answer 1

I would guess that your book's proof is more complicated for one or more of the following reasons:

1. Cardinality has not been defined yet
2. Your book is using a definition of finiteness that requires more work than this
3. The authors intend to later address constructive set theory in which the result can fail for the usual definitions of finiteness (!), and the way in which the result is written will be used to highlight this
4. The authors wrote a worse proof

If you already know all the relevant facts about cardinality and have a definition of finiteness phrased in terms of cardinality, then you're good to go.