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?

• @ Rafael , Typo fixed . Thank you . – Suman Kundu Apr 4 '17 at 4:49
• A quick way to do this would be Zorn's Lemma to construct a maximal subspace which doesn't contain a given non-zero vector, and then adjoin it. Seems a bit overkill though. By contradiction, you could use Dependent Choice to construct a countable linearly independent set which couldn't exist by assumption. Without any form of choice... that's the question. – user123641 Apr 4 '17 at 4:50
• @Bryan: Not without choice. – Asaf Karagila Apr 4 '17 at 5:13

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.

• In my current phase of life, I have found that the best choice function has Bordeaux on the label. – copper.hat Apr 4 '17 at 5:32
• I have found that the best choice usually has "Speyside Single Malt Whisky" on the label. – Asaf Karagila Apr 4 '17 at 5:33
• If you are willing to add an 'e' a Teeling is not bad. Prefer the grape to the grain myself... – copper.hat Apr 4 '17 at 5:42
• To me this seems to be one of the less weird results of no AC. You get this kind of behavior in groups quite easily. – DRF Apr 4 '17 at 9:49
• @DRF: You cannot possibly compare (non-abelian) groups to vector spaces. You cannot even compare modules---even free modules---to vector spaces in terms of well-behaved properties. – Asaf Karagila Apr 4 '17 at 10:26

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.

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.

Take a subspace $U$ of codimension $1$ (why exits it?), i.e. $\dim V/U = 1$. Applying the First Isomorphism Theorem to the the projection $$\pi:V\longrightarrow V/U,$$ we have $$\dim V = \dim(\ker\pi) + \dim (V/U).$$ But by hypothesis $\dim(\ker\pi)<\infty$.