# If a subspace of a finite-dimensional vector space. Then the subspace is finite dimensional?

I have difficulty in understanding the proof of this statement: Let W be a subspace of a finite-dimensional vector space V. Then W is finite dimensional.

The proof goes like this. (Linear algebra, Friedberg, theorem 1.11)

Let dim(V) = n. W contains a nonzero vector v1, and {v1} is a linearly independent subset of W. We continue choosing vectors v2,v3,...,vk, if possible, so that {v1,v2,...,vk} is linearly independent, and adjoining another vector from V results in a linearly dependent set. Since no linearly independent subset of V has more than n vectors, this process stops at k ≤ n. Then {v1,v2,...,vk} is a basis for W, so dim(W) ≤ dim(V ).

The part I cant understand is in the bold text. That part seems seems really vague to me. W is likely to be an infinite set and we may need infinite steps to choose such an independent set! Then it may take forever to choose such set.

I think the bold part needs more explanation. How do you think?

The apparently infinite process must stop with $k$ at most $n$ because by the assumption that $\dim{V} = n$, no subset of $V$ containing more than $n$ vectors is linearly independent.
• More formally, you can **specify ** the sequence $v_1, v_2 \ldots, v_i$ of elements of $W$ with the required property and prove that such a sequene exists for $i \le n$. The word "choose" is being used to give the proof a more computational flavour (if you were presented with $W$ and $V$ represented as matrixes of basis vectors say and if you were working over a computable field, then the computations could be carried out effectively). – Rob Arthan Feb 10 '16 at 2:34
• But all we need to be able to do is to prove that there exists a $v_2$ that is not in the span of $v_1$. That's what the proof means when it says "choose", (In fact, finite-dimensional linear algebra is quite computational and if you were given all the data effectively you really could carry out the computations, but that's not important here.) – Rob Arthan Feb 10 '16 at 2:57