Infinite dimensional vector space, and infinite dimensional subspaces. I'm having trouble with this question:
Let  $V$ be an infinite dimensional vector space.
Prove that there exist subspaces $U_1,U_2,\dots$ of $V$ with $U_{n+1}\neq U_n$ for all $n\in\mathbb{N}$, such that $U_1\supset U_2\supset\cdots$ and $U_n$ is infinite dimensional for all $n\in\mathbb{N}$.
Thanks for your help.
 A: Let $v_1,v_2,\dots$ be infinitely many linearly independent vectors in $V$. 
Define
$U_n:={\rm span}(v_n,v_{n+1},v_{n+2},\dots)$.
A: We assume the axiom of choice, of course.
Since $V$ is infinitely dimensional, it has an infinite basis, and therefore there is a countably infinite set of linearly independent vectors, write then as $\{v_n\mid n\in\Bbb N\}$.
Now you can easily find a strictly decreasing sequence of infinite sets, $A_n\subseteq\Bbb N$. Let $U_n=\operatorname{span}(\{v_k\mid k\in A_n\})$. Show that from the reverse inclusion of the $A_n$'s, the sequence of $U_n$'s is decreasing, and it is strictly decreasing because we took the $v_n$'s to be independent.
Note that by choosing your $A_n$'s carefully you can control whether or not $\bigcap U_n$ is the trivial subspace, finite dimensional subspace, or an infinite dimensional subspace of $V$.
Without the axiom of choice it is consistent that there is a vector space which is not spanned by a finite set, but every proper subset is of finite dimension, so you really have to use the axiom of choice for this.
