The question from a Stanford assignment requests:
Example of a nonempty subset U of $R^2$ such that U is closed under addition and under
taking additive inverses but U is not a subspace of $R^2$.
The answer says:
Proof. Consider the subset $Z^2$. It is closed under addition; however, it is not closed under scalar multiplication. For example $\sqrt2$ (1, 1) = ($\sqrt2$, $\sqrt2$) ∈/ $Z^2$.
However, doesn't the scalar multiple need to be a member of the subset?
Best Answer
No! The scalar multiple is from the field over which the vector space is defined. In this case you have a real vector space so the scalar can be any real number.