Is it Possible to Conclude that g(T) = T in this Proof?

[SithsNGiggles]

This is the conclusion of a proof (any subset of a finite set is finite, which was worked on in this thread: https://www.physicsforums.com/showthread.php?t=592636) I recently presented, but I was told that I couldn't immediately make it:

$S = T \cup (S \setminus T)$ (where T is a subset of S)
$f(S) = g(T) \cup h(S \setminus T)$

I showed earlier in the proof that $f(S) = S$ and $h(S \setminus T) = S \setminus T$.
I say that it immediately follows that $g(T) = T$ because

$f(x) = \left\{ \begin{array}{l} g(x) & : x \in T\\ h(x) & : x \in S \setminus T \end{array} \right.$

which shows that $g(T) \cap h(S \setminus T) = \emptyset$.

What could I be missing? Thanks for any help.

 

[lippyka]

In order to conclude that g(T) = T, you need to show that g(T) is a subset of T. In other words, you need to show that for all x in T, g(x) is in T as well. You have not yet done that, so you cannot make the conclusion that g(T) = T.