# A subspace of a finite-dimensional vector space $V$ is finite-dimensional (Alternative proof)

I have looked at several proofs of the following theorem:

Let $$W$$ be a subspace of a finite-dimensional vector space $$V$$, then $$W$$ is finite-dimensional.

But all of them use a similar argument to prove it:

Proof. Let $$dim(V)=n$$. If $$W =\{0\}$$, then $$W$$ is finite-dimensional. Otherwhise, W contains a nonzero element $$x_1$$; so $$\{x_1\}$$ is a linearly independent set. Continuing in this way, choose $$x_1,...,x_k$$ in $$W$$ such that $$\{x_1,...,x_k\}$$ is linearly independet. Since no linearly independent subset of $$V$$ can contain more than $$n$$ elements, this process must stop at a stage where $$k\leq n$$ and $$\{x_1,...,x_k\}$$ is linearly independet but adjoining any other element of $$W$$ produces a linearly dependent set, this implies that $$\{x_1,...,x_k\}$$ generates $$W$$, and hence it is a basis for $$W$$.

I'm fine with this proof, but also I'm looking for a proof that doesn't use the argument of "continuing in this way a finite number of times" or "perform this algorithm", my first attempt was to prove the following inductively:

P(n) : If $$W$$ is a subspace of a n-dimensional vector space $$V$$, then $$W$$ is finite-dimensional.

but I'm not sure how to prove that $$P(n)\rightarrow P(n+1)$$.

EDIT: At this point, I can't use the fact that every vector space has a basis.

• It would make much more sense if you'd state what we can use at this point. For example do you know at this stage that every vector space admits a basis? Sep 17, 2022 at 5:08
• @SeverinSchraven Thanks, I forgot to mention it Sep 17, 2022 at 5:45

First, the base case $$n = 0$$ is very easy (and this is a usable base case, we don't even need to do $$n = 1$$). Next, as you say, if $$W = 0$$ then the statement is clear. Otherwise $$W$$ contains some nonzero vector $$w$$. Then $$W/w$$ is a subspace of the $$(n-1)$$-dimensional quotient $$V/w$$, so by the inductive hypothesis $$W/w$$ is finite-dimensional, hence has some basis. Let $$w_1, \dots w_k \in W$$ be any lift of this basis to $$W$$; then $$\{ w, w_1, \dots w_k \}$$ is a basis of $$W$$, so $$W$$ is also finite-dimensional.
(If the statement that $$V/w$$ is $$(n-1)$$-dimensional also needs proof, the easiest argument is to extend $$\{ w \}$$ to a basis of $$V$$. If you want to avoid that argument then I guess we can talk about that too.)