Suppose every proper subspace $(\neq V)$ of a vector space $V$ is finite dimensional. Prove that $V$ is finite dimensional.
This question just popped into my head when I was reading about inner product space, so I can't guarantee how legitimate the question is.
My try:
Assume V is an inner product space. Take any proper subspace $U$ of $V$. Then $V=U\oplus U^{\bot}$. Both $U$ and $U^{\bot}$ are finite dimensional. So they both have a basis of finite dimension and we will be done.
So for the inner product space $V$ the statement holds true.
What about other cases? Does the statement hold true for every vector space? Does there exist a characteristic to identify vector spaces with this property?
Best Answer
You need to use the fact that every linearly independent set can be extended to a basis.
Then you can take any nonzero vector $v$, extend $\{v\}$ to a basis $B$, and consider the span of $B\setminus\{v\}$. Being finitely dimensional, it means that $B\setminus\{v\}$ is finite, so $B$ is finite, and so $V$ has a finite dimension.
It might be relevant to point out that the fact "every linearly independent set can be extended to a basis" is equivalent to the axiom of choice. Of course we need only a small fraction of choice for this specific proof (although that would alter the formulation a bit).
Nevertheless, it is consistent without the axiom of choice that there is a vector space which is not finitely dimensional, but every proper subspace does in fact have a finite dimension. Weird.