Thread: The union of finite subspaces

Let F be a an infinite field. Prove that a vector space, V, over F cannot be a finite union of proper subspaces.

I have already shown that the union of two proper subspaces is a subspace iff. one is contained in the other. Let's call this lemma 1.

Can I use lemma 1 and prove this by induction?

P(2): proves the union may not even be a subspace.

Assume P(k) is true for a fixed by arbitrary integer k that is greater than or equal to n, i.e. P(k) isn't a subspace.

Then, by lemma 1 and the inductive hypothesis, P(k+1) is just the union of P(k) and the k+1 subspace which may not even be a subspace. Therefore, since the union may not be a subspace, the finite union of proper subspaces cannot be a vector space over F.

Does this work?

Originally Posted by dwsmith

Let F be a an infinite field. Prove that a vector space, V, over F cannot be a finite union of proper subspaces.

I have already shown that the union of two proper subspaces is a subspace iff. one is contained in the other. Let's call this lemma 1.

Can I use lemma 1 and prove this by induction?

P(2): proves the union may not even be a subspace.

Assume P(k) is true for a fixed by arbitrary integer k that is greater than or equal to n, i.e. P(k) isn't a subspace.

Then, by lemma 1 and the inductive hypothesis, P(k+1) is just the union of P(k) and the k+1 subspace which may not even be a subspace. Therefore, since the union may not be a subspace, the finite union of proper subspaces cannot be a vector space over F.

Does this work?

But, you don't know that the union is the union of two subspaces. Also, that proof is true for finite fields, so it can't apply here otherwise the theorem would be true of vector spaces over which it isn't ( .

Suppose that is a -space for some infinite field and . Now, we may assume WLOG that for any . It follows that there exists with and that there is some . So, consider the set (vectors of the form for ). By assumption we have that is infinite. That said, note that (since otherwise ) and for , for if and were both in then their difference and so . It follows then that can contain at most elements of which is ridiculous since is infinite and .