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 .
Well, I'll give you a hint. As I said, the problem is that being a subgroup of does not imply that is a subspace. Said more concretely, we have that implies DOES NOT imply that for all . To see this, look at my example. For all one has that so that is closed under addition. But, look at where the problem comes in, is closed under arbitrary real number multiplication?