1. Nov 8, 2007

### mrroboto

I don't understand this, can someone help?:

What is an example of a subset of R^2 which is closed under vector addition and taking additive inverses which is not a subspace of R^2?

R, in this question, is the real numbers.

Thanks!

2. Nov 8, 2007

### JasonRox

Can you think of a subset of R^2 such that it is a subspace?

3. Nov 8, 2007

### ZioX

Well you can add things but you can't subtract things. This should be a big enough hint.

4. Nov 8, 2007

### slider142

He mentioned additive inverses and closure under addition (implying 0 being an element), so he has no problem subtracting things. The key lies in the one part of the definition of a vector space that is left out.

5. Nov 8, 2007

### ZioX

I misread it. I thought the question asked to find something that fails to be a subspace because it's not closed under additive inverses.

6. Nov 8, 2007

### mrroboto

To JasonRox: Yes, I can think of a subset, V, of R^2 that is a subspace. V= {(0,0)}. But I still don't understand how to approach this problem.

7. Nov 9, 2007

### dalle

first try to solve a simpler problem.
are there nontrivial subgroups of the group $$(\mathbb{R},+,-,0)$$ ? that means is there a set $$A$$ with $$\{0\} \subset A \subset \mathbb{R}$$ such that $$A$$ is closed under addition and substraction?
once you have found such a subgroup $$A$$, is $$A^2$$ a set with the desired property?

8. Nov 9, 2007

### HallsofIvy

In order that a subset be a subspace, it must be closed under addition, additive inverses, and scalar multiplication. Since your subset is required to be closed under addition and additive inverse, there's only one place left to look!