# The union of finite subspaces

• Sep 10th 2011, 09:02 AM
dwsmith
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?
• Sep 10th 2011, 03:28 PM
Drexel28
Re: The union of finite subspaces
Quote:

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 $\mathbb{F}_q$ which it isn't ( $\mathbb{F}_2^2=\{(0,0),(0,1)\}\cup\{(0,0),(1,0)\}\ cup\{(0,0),(1,1)\}$.

Suppose that $V$ is a $k$-space for some infinite field $k$ and $V=U_1\cup\cdots\cup U_n$ $n\geqslant 2$. Now, we may assume WLOG that $U_j\not\subseteq U_k$ for any $k\ne j$. It follows that there exists $x\in U_1$ with $x\notin U_2,\cdots,U_n$ and that there is some $v'\notin U_1$. So, consider the set $S=v'+kv$ (vectors of the form $v'+\alpha v$ for $\alpha\in k$). By assumption we have that $S$ is infinite. That said, note that $S\cap U_1=\varnothing$ (since otherwise $v'\in U_1$) and $\text{card}(S\cap U_j)\leqslant 1$ for $j\geqslant 2$, for if $v+\alpha v'$ and $v+\beta v'$ were both in $U_j$ then their difference $(\alpha-beta)v\in U_j$ and so $v\in U_j$. It follows then that $U_1\cup\cdots\cup U_n$ can contain at most $n$ elements of $S$ which is ridiculous since $S$ is infinite and $S\subseteq V=U_1\cup\cdots\cup U_n$.