# Well-definedness of vector space dimension

In Kreyzig's Functional Analysis with Applications, the dimension of a finite-dimensional vector space is defined to be $$\dim(X) = n$$ if there exists a linearly independent set of $$n$$ vectors in $$X$$ and no set of $$n+1$$ vectors is L.I.

I'm trying to prove that for all $$n$$ the following are equivalent, so that dimension becomes a well-defined value:

1. $$\dim(X) = n$$.
2. $$X$$ has a Hamel basis (an L.I. spanning set), and every L.I. set of $$n$$ vectors in $$X$$ is a basis.
3. There exists a spanning set of $$n$$ vectors, and no set of $$n-1$$ vectors spans $$X$$.

So far I've got (1) and (2) are equivalent because I've shown every maximal L.I. set is a basis, and I've also got that (3) implies a basis exists (because minimal spanning sets are such), but I'm having trouble showing (2) implies (3) and finishing proving (3) implies (2).

Separately I also have that the representation of a vector $$x$$ as a linear combination of the basis vectors is unique (but I'm not sure if that's useful here).

How should I proceed from here?

For 3 implies 2, consider a l.i. list of $$n$$ vectors $$u_1,...,u_n$$ and a Hamel basis (the one you showed exists) $$v_1,...,v_n$$. You can make a list of $$n+1$$ vectors $$u_1,v_1,...,v_n$$ which will necessarily be l.d.; you can throw away one of the $$v_j$$ and still have a spanning set. Iterate to show that the $$u_j$$ are also a spanning set.
For 2 implies 3, you already have a spanning set of $$n$$ vectors, so you only have to show no set of $$n-1$$ vectors spans the space. I have a vague idea that you can force a contradiction by representing the hamel basis of $$n$$ vectors in terms of the $$m < n$$ l.i. subset of the spanning set of $$n-1$$ vectors. I mean, you'd have
$$a_1u_1 + ... + a_nu_n$$
You'd have some degrees of freedom in the choice of the $$a_i$$ so as to make the original hamel basis l.d.