Example 1.5 gives a subset of R2 that is not a vector space, under the obvious operations,

Question:

Example 1.5 gives a subset of R2 that is not a vector space, under the obvious operations, because while it is closed under addition, it is not closed under scalar multiplication. Consider the set of vectors in the plane whose components have the same sign or are 0. Show that this set is closed under scalar multiplication but not addition.