Is $V = \{(x,y,z)\in \mathbb{R}^3:\ x+y >1 \}$ a subspace? Prove whether the following subsets of $\mathbb{R}^3$ are subspaces :
(a)
$$V = \{(x,y,z)\ \in \mathbb{R}^3:\ x+y >1 \ \},$$ 
I think that this is not a subspace as the zero vector does not fulfill the "x+y >1".
However, how do I show that this is closed under addition and closed under scalar multiplication (or if they are not, how do I show that)?
 A: By definition a subset $V\subset U$ must fulfill all of the 3 conditions below to be called a subspace of the vector space $U$:


*

*It must contain a zero vector.

*It must be closed under addition.

*It must be closed under scalar multiplication.


You successfully pointed that the subset in question doesn't fulfill the first condition, hence it isn't a subspace, and you aren't required to see whether the other conditions are fulfilled.

Assume that — for the sake of mathematical curiosity — you wanted to see whether $U$ fulfilled the other conditions. Let $v=(v_1,v_2,v_3)\in V$ and let $u=(u_1,u_2,u_3)\in V$, then $v_1+v_2>1$ and $u_1+u_2>1$. Hence $(v_1+u_1)+(v_2+u_2)>2$. Can you then show that this means that $v+u\in V$? Next, consider $\lambda v=(\lambda v_1,\lambda v_2,\lambda v_3)$. For it to be in $V$ it must satisfy $\lambda v_1+\lambda v_2=\lambda(v_1+v_2)>1$. Is this true for all $\lambda\in\bf R$? If not, what can we conclude?
A: You are right. It is not a subspace, for the reason you mentioned.
As for being closed under scalar-multiplication: suppose you multiply a vector in your set by $-1$. What do you see?
Your set $\it{is}$ however closed under addition. Can you prove it?
