Def: A Hamel Basis for V is a linearly independent subset that spans V.
Def: If V has a finite Hamel Basis, F, then the dimension of V is equal to the number of elements in F, if not V is infinite dimensional.
Theorem: If V has a finite Hamel Basis then every Hamel Basis has the same number of elements.
So far I have supposed V has a finite Hamel Basis F, and E is an infinite linearly independent subset of V, I want to prove this does not hold by contradiction.
I have shown that if E spans V, F is not a finite Hamel Basis.
I think that it now suffices to show when E does not span V, F is once again not a finite Hamel Basis.
However, I am unsure of how to prove this, any help or alternative proofs would be greatly appreciated.
Best Answer
If $E$ (infinite) and $F$ (finite) are both Hamel bases, then you can write the basis elements of $F$ using the basis elements of $E$. Each basis vector of $F$ by definition uses only finitely many elements of $E$, and since $F$ is finite, you can write all basis elements of $F$ using only finitely many elements of $E$. But $F$ is a basis, so any remaining basis element of $E$ that has not been used can be written in terms of elements of $F$, which can then be written in terms of finitely many elements of $E$, a contradiction, as the elements of $E$ are supposed to be linearly independent.