A set of vectors in $R^2$ for which $x + y$ stays in the set but $\frac12x$ may be outside. These questions refer to subspace.  
One requirement can be met while the other fails. Show this by finding
(a) A set of vectors in $R^2$ for which $x + y$ stays in the set but $\frac12x$ may be outside.
(b) A set of vectors in $R^2$ (other than two quarter-planes) for which every $cx$ stays in the set but $x + y$ may be outside.
these are the answers my textbook gives,
(a) The vectors with integer components allow addition, but not multiplication by $\frac12$
This is saying that you can't multiply a vector by a non-integer? why not?  
(b) Remove the x axis from the xy plane (but leave the origin). Multiplication by any c is allowed but not all vector additions.
If you remove the x-axis then you are only left with vectors on the y-axis, so shouldn't it still be closed under addition?
 A: (a) If you multiply a nonzero vector with integer components by a noninteger, the result is not in "the set of vectors with integer components" anymore.
(b) if you remove the $x$-axis, you still have points like $(1,1)$ and $(1,-1)$ even though you don't have their sum.
A: *

*$\frac12(1, 1)$ is not an integer vector.

*The plane with the $x$-axis removed contains all points with the first coordinates non-zero.

A: They are asking for sets that satisfy some property of vector spaces, but that are not vector spaces.
(a) the only operation they are allowing is addition, thus clearly vectors with entire componends only add to vectors with entire components, it's not a vector space since $\mathbb{Z}$ is not a field (you don't have $n^{-1}$.
(b) You are assuming it's a vector space, it is not, thats why the set doesn't collapses to only the y-axis.
A: b)
imagine a line that is exactly on the y-axis.
you have a vector v = (0,1) , so cv always on the y-axis but for example 
(0,1) + (1,1) = (1,2) wich is not on the line (y-axix) 
