Prove that V is infinite dimensional if and only if there is a sequence v_1, v_2,... of vectors in V such that (v_1,...,v_n) is linearly independent for every positive integer n.
A vector space is finite dimensional if some list of vectors in it spans the space.
The Attempt at a Solution
We need to prove two directions. For the forward direction, we assume V is infinitely dimensional (and therefore is not finite dimensional). It's really giving me a headache. All of the theorems in my book involve finite-dimensional vector spaces, and none of the proofs seem to give me any information pertaining to the forward direction of this problem. Do I want to use induction somehow? For example, the infinite dimensional vector space F^(infinity) has elements that can be written in the form a_1e_1+a_2e_2+...+a_ne_n+... with the a_i's scalars in the field and the e_i's the vector with all zeroes in every spot except for a 1 in the ith place. That is a linearly independent list, that works for all n, but i don't know how to explain it for every infinite dimensional vector space.
For the backward direction, first I wanted to assume V is finite dimensional and use contradiction. Then V has a basis that is linearly independent spans V, and has a dimension, lets say m=n-1. Then consider the linearly independent list (v_1,...,v_n). It has dimension n, but every linearly independent list of vectors in a finite-dimensional vector space can be extended to a basis of the vector space, so since dim(V) < dim(v_1,...,v_n) we have a contradiction and V must be infinite dimensional.
I think the backward direction is almost complete, but i'm having trouble with the forward. Any hints, please?