The following statement is false. Could someone give a counterexample?

If V is a real vector space and W is a (non-empty) subset of V such that for all vectors

w1,w2 Є W it holds that also w1 + w2 Є W, then W is a subspace of V .

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- Dec 1st 2011, 12:25 PMLotte1990Subspace of a vector space
The following statement is false. Could someone give a counterexample?

If V is a real vector space and W is a (non-empty) subset of V such that for all vectors

w1,w2 Є W it holds that also w1 + w2 Є W, then W is a subspace of V . - Dec 1st 2011, 12:26 PMDrexel28Re: Subspace of a vector space
This is not true. Consider $\displaystyle \mathbb{Z}\subseteq\mathbb{R}$. Said differently, (if you had changed $\displaystyle w_1+w_2$ to $\displaystyle w_1-w_2$) being a subgroup of a vector space's underlying abelian group is not equivalent to being a subspace of the vector space.

- Dec 1st 2011, 12:28 PMLotte1990Re: Subspace of a vector space
- Dec 1st 2011, 12:43 PMDrexel28Re: Subspace of a vector space
Well, I'll give you a hint. As I said, the problem is that $\displaystyle W$ being a subgroup of $\displaystyle V$ does not imply that $\displaystyle W$ is a subspace. Said more concretely, we have that $\displaystyle w_1,w_2\in W$ implies $\displaystyle w_1+w_2\in W$ DOES NOT imply that $\displaystyle \alpha w_1\in W$ for all $\displaystyle \alpha\in\mathbb{R}$. To see this, look at my example. For all $\displaystyle x,y\in\mathbb{Z}$ one has that $\displaystyle x+y\in\mathbb{Z}$ so that $\displaystyle \mathbb{Z}$ is closed under addition. But, look at where the problem comes in, is $\displaystyle \mathbb{Z}$ closed under arbitrary real number multiplication?

- Dec 1st 2011, 01:16 PMHartlwRe: Subspace of a vector space