# Finite dimensional subspaces of a linear space

Suppose $V$ is an infinite dimensional vector space. I do not want to assume the axiom of choice, so I will define a vector space $V$ to be infinite dimensional if there is a proper subspace $W\subseteq V$ such that $V$ and $W$ are isomorphic.

My question: Given $n \in \mathbb{N}$, is there a subspace $U\subseteq V$ of dimension $n$?

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## 3 Answers

Even without the axiom of choice there is a sense to say that a vector space is finitely generated, i.e. has a finite dimension. It is just to say that there the vector space is a finite direct sum of copies of the field.

To say that a vector space has an infinite dimension, relies on the sense of dimension which indeed breaks down when the axiom of choice does not hold; however one can say that a vector space is not of finite dimension. Much like an infinite set is just a set which is not finite. This is a broader definition than the one you give, it is possible to have a vector space without a basis (thus not of finite dimension) that every proper subspace has a finite dimension. The following arguments works regardless:

Given a vector space which is not of finite dimension, we can choose $n$ which are linearly independent their span is $U$ that you seek. We prove that by induction:

For $n=1$ it is obvious. Take any nonzero vector in $V$.

Suppose that you chose $n$ vectors, call them $v_1,\ldots,v_n$. Since $V$ is not spanned by finitely many vectors the span of $\{v_i\}_{i=1}^n$ is not the entire space $V$, therefore we can take some $v\in V$ which is not in this span. Now we have $v_1,\ldots,v_n,v_{n+1}=v$. As wanted.

Note that the induction only holds for finitely many vectors and we cannot deduce that there are countably many linearly independent vectors. It would require some choice, namely The Principle of Dependent Choice. From this principle follows that if we define something inductively then we can find an infinite sequence with the property we want.

Further reading material:

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Yes.

Induction on $n$:

For $n=0$, take $U=\{\mathbf{0}\}$.

Assume the result is true for $k$. Let $U$ be a subspace of $V$ of dimension $k$. Then $U\neq V$, because $U$ is not isomorphic to any proper subspace of $U$, but $V$ does. Therefore, there exists $v\in V-U$. Let $\beta$ be a basis for $U$ (which exists since we are assuming $U$ is of dimension $k$). Then $\beta\cup\{u\}$ is linearly independent, and spans $\mathrm{span}(U,v)$. So $\mathrm{span}(U,v)$ is of dimension $k+1$, proving that $V$ has a subspace of dimension $k+1$.

The choice of the single $v\in V-U$ does not require the Axiom of Choice, since $V-U$ is nonempty and this is a single choice.

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Yes. You can sequentially find $n$ linearly independent vectors in $V$ by the process of always picking something outside the span of the preceding ones. No finite set will generate all of $V$, because then the theory of f.d. vector spaces (where dimension can be defined without AC) prohibits the existence of $W$. Thus there will always be room for one more linearly independent vector.

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