Proof that a subset of a finite set is finite Can someone please tell me if this proof is rigorous? I have a hunch it's circular, and I am suspicious at how short it is. However, I'm also having trouble convincing myself that it's incorrect.
Suppose $B$ is a finite set and let $A \subseteq B$.
There exists $k \in \mathbb{N}$ such that $\{1, 2, \dotsc, k\} \sim B$.
Set $k' = |B \setminus A|.$
Then $A$ is isomorphic to $\{1, 2, \dotsc, k - k'\}$, so it is finite.
Finite sets are countable, so $A$ is countable.
Thank you!
 A: It's not exactly circular, but when you use a symbol like $|B\setminus A|$, you are implicitly assuming that $B\setminus A$ is finite, which is premature.  Also, you can't just jump to saying that $A$ is isomorphic to a finite set -- you have to actually establish a bijection for that to happen.
The challenge of cleaning up this proof depends on how many tools you have at this point and how many tools you're being asked to build as homework.  There is clearly a bijection from $B$ to $1..n$ which you are allowed to choose.  You can manipulate that mapping to demonstrate an injection from $A$ to $1..n$.  Depending on your definitions, that may be enough to directly state that $A$ is finite (if, for instance, your definition is cardinality is the minimum $k$ such that there is an injection from $A$ to $1..k$).  
On the other hand, if you need to find a bijection to prove finiteness, then you get to look forward to writing a lemma (hint: by induction on the maximum member of the set) that there exists a surjection from every finite subset of $\mathbb N$ to some set of the form $1..n$.
