A vector space is finite dimensional if all of its proper subspaces are finite dimensional

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?

Solution 1:

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.

Solution 2:

Let $V$ be an infinite dimensional vector space, let $v$ be a vector of $V$. Then in particular there exists a countable subset $\{v_1,\cdots, v_n,\cdots\}$ of $V$ such that $\{v_1,\cdots, v_n,\cdots\}$ spans a subspace of $V$. We may assume $\{v_1,\cdots, v_n,\cdots\}$ is a basis, so in particular $\{v_2,\cdots, v_n,\cdots\}$ spanns a proper subspace $U$ of $V$ which is of countable dimensionality. The finite dimensional case is automatic.

Solution 3:

Your proof for inner product spaces can be generalized (in some sense) in the following manner:

Let $U$ be a (nonzero) subspace of $V$, and let $\ \mathcal U=\{u_1,\ldots, u_n\}$ be a basis for $U$. Using Zorn's lemma we can extend $\ \mathcal U$ to a basis $\mathcal V$ of $V$. Let $U'$ be the subspace spanned by $\mathcal V \setminus \mathcal U$. Then $V=U\oplus U'$, and since both $U$ and $U'$ are finite-dimensional it follows that $V$ is finite-dimensional.