# Homework Help: Union of Proper Subspaces Problem

1. Jul 20, 2008

### e(ho0n3

The problem statement, all variables and given/known data
Let V be a vector space over an infinite field. Prove that V is not the union of finitely many proper subspaces of V.

The attempt at a solution
Suppose V is the union of the proper subspaces U1, ..., Un. Let ui be a vector not in Ui. If u1 + ... + un is in the union, then there must be some subspace that contains it. But then that subspace contains a sum where one of the terms doesn't belong to it. I'm hoping this isn't possible but I can't think of anything contradictory.

In any case, I think my approach is wrong because I haven't really used the fact that V is defined over an infinite field and that the union is a finite one.

2. Jul 20, 2008

### n_bourbaki

Of course that isn't contradictory: 0 is in any subspace and 0=u-u for any u irrespective of what u is.

Just consider two subspaces, X and Y. When is XuY even a vector space?

3. Jul 20, 2008

### e(ho0n3

Let x belong to X and y belong to Y. If the union of X and Y is a vector space, then x + y must be in the union and so x + y must be in either X or Y. If y were solely in Y, then there would be a problem: If x + y is in X, then x + y - x = y is in X which is impossible.

So for the union of X and Y to be a vector space, one has to be a subset of the other?

4. Jul 21, 2008

### n_bourbaki

Correct.

5. Jul 21, 2008

### HallsofIvy

You can prove a general statement (the union of two subspaces is a subspace) is NOT true by giving a single counterexample.

Consider R2 where U= {(x,y)| y= x} and V= {(x,y)|y= -x}. It is easy to show that U and V are subspaces of R2 but that their union is not.

6. Jul 21, 2008

### e(ho0n3

Say V is a two-dimensional vector space defined over Z2. V has four elements and the span of each of those elements is a proper subspace of V. The union of the spans is V.

To be more precise (in post #3), if X and Y are vector spaces defined over an infinite field, then their union is a subspace if one is a subset of the other.

7. Jul 21, 2008

### Dick

You have a really good example there. As people have pointed out before, the union of two subspaces is only a subspace if there is a containment relation between them. This is true even for finite fields. The union of any two nontrivial subspaces of your space over Z2 is not a subspace. The union of all of them is. I actually don't know how to solve this problem yet. I'm just thinking out loud. You have to figure out how what happens over Z2 doesn't happen over an infinite field.

Last edited: Jul 21, 2008
8. Jul 22, 2008

### n_bourbaki

Here's the 'explanation' for you, Dick.

Take 1 subspace, there are infinitely many elements in the complement. Picking another subspace cannot hit all of the things in the complement, and there will still be infinitely many things left over. Picking finitely many subspaces can never get to hit all of these infinitely many things that are left over.

The measure theoretic statement of that fact (i.e. one for R^n, C^n etc and one the OP shouldn't use, so it's ok to give), is that a proper subspace of R^n has measure zero. So the union of finitely many of them has measure zero (in fact the union of countably many of them). So no finite, or even countable, union of proper subspaces of R^n can equal R^n.

9. Jul 22, 2008

### Dick

Sure. The measure theory picture is what I have in my head. I'm just having trouble trying to think of a way to write the 'infinitely many things' argument.