Subset of a vector space which isn't a subspace The content of my exercise says: Show an example of subset U of vector space $\Bbb R^{2} = \Bbb R^{2 \times 1} $ which isn't a subspace of $\Bbb R^{2}$, under conditions:
a) $\forall \space u,v \in U: \space u+v \in U $ ,
$\space \space \space \space \forall \space u \in U: -u \in U $
b) $ \forall  \space \alpha \in \Bbb R \space \land  \forall  \space u \in \Bbb U: \space \alpha*u \in U $
Since these two assumptions together make a definition of a subspace, so in:
a) I will try to find an example which doesn't fullfill a b) condition.
Since $u \in U  \subset \Bbb R$, I can use $\alpha \in \Bbb R \land \alpha\notin U$ to show that.
Let be $U = \Bbb Z, $ For a $\forall \space u \in U \space $ and $ \alpha = \sqrt{2} \space $ Then: $ \sqrt{2}*u\notin \Bbb Z$.
Since it doesn't fullfill second condition of a subspace, $U$ isn't a subspace.
b) I will try to find an example which doesn't fullfil an a) condition.
So I have to show that $ \exists \space (u,v) \in U \subset \Bbb R: \space u+v \notin U. \space $
Then there I'm stuck. My initial thought was set $U$of negative integer numbers, but then for not every $\alpha\in\Bbb R, \space \alpha*u \in U $  - $\alpha$ can be negative too, then outcome will be positive and wouldn't contain in negative numbers set, or $\alpha$ is a root (as I did in a) ), or it's in form of $\frac{1}{x}$, then outcome will not be integer.
My another thought is that since it's $U\in \Bbb R^{2} $, I should look at $u$ as at $(a,b)$ instead but it really didn't help me. Maybe if $U=\{u,v\}$ where $u=(a,-b)$ and $v=(-a,b)$ then: $u+v=(a-a,-b+b)=0$ and $(0,0) \notin U$ but then again I've problem with an $\alpha$.
My intution is telling my to play around the $0$ because  $ \forall  \space \alpha \in \Bbb R: \alpha*0=0 \space$. But honestly I can't produce anything more than I showed there. Thanks in advance, and sorry if something isn't understandable, I'm fairly new to the english maths' nomenclature.
 A: An easy example where a) fails but b) is satisfied is the set of points in $\Bbb R^2$ where at least one coordinate is $0$. Multiplying by a constant certainly cannot make that coordinate nonzero, so it remains within this set. On the other hand $\binom03+\binom20=\binom23$ shows that addition, starting with vectors in the set, can get you out of this set. Sets closed under (only) scalar multiplication are called cones, and indeed a usual (double) cone in $\Bbb R^3$ with it apex at the origin is another example.
For the other kind of example (where a) is satisfies but b) fails) you can indeed take $\Bbb Z$ inside $\Bbb R$. Indeed you generally get something where the definition involves some kind of integrality (but not positivity, as you need to be close under taking opposites). Something fancier like $\{\, (x,y)\in\Bbb R^2\mid 3x-5y\in\Bbb Z\,\}$ will work as well.
A: Solution to subsection b) of my exercise.
Let $U$ be set of all polynomials $P(x)=x^2+bx+c$ with any 2 different solutions. (A set of polynomials with $\Delta_{P(x)}>0$).
Since I want $\Bbb R^{2}$ not a $\Bbb R^{3}$ I demand coeffcient of $x^2$ to always be 1 and $b,c \in \Bbb R$ so you can say $u=(b,c)$
$\forall \space \alpha \in \Bbb R: \space \alpha*P(X)\in U$  since multiplying a polynomial by a number doesn't change his solutions.
$\forall \space T(x)\in U \space \space \exists R(x)\in U: T(x)+R(x)=S(x)\notin U \space \space \space \space  (\Delta_{S(x)}<0)$
Example: Let be $\space T(x)= (x - 1) (x - 4)\space$ and $\space R(x)= (x - 5) (x - 9) \space$
Then: $S(x)=T(x)+R(x) \rightarrow S(x)=2x^2-19x+49$
$\Delta_{S(x)}=-31 \implies $ It doesn't have any (real) solutions so $S(x)\notin U$
I think that solution is kinda overkill, and you can probably do it smarter by using  somehow $y=ax+b$ but it is what I've come with.
