Proving that a set of vectors isn't a subspace of $\mathbb{R}^3$ So I have the set $A = \{(x,y,z) | x\cdot y \cdot z=0\}$ and I wanna prove that this is not a valid subspace of $\mathbb{R^3}$ so I tried to prove that it's not closed under addition by taking a vector $u = \begin{bmatrix} a{_1} \cr b{_1}\cr c{_1}\end{bmatrix}$ and $v = \begin{bmatrix} a{_2} \cr b{_2}\cr c{_2}\end{bmatrix}$ then $u+v$ is equal to $\begin{bmatrix} a{_1}+a{_2} \cr b{_1}+b{_2}\cr c{_1}+c{_2}\end{bmatrix}$  and if $u+v$ is in $A$ then  $(a{_1}+a{_2})\cdot(b{_1}+b{_2})\cdot(c{_1}+c{_2})=0  $ has to be true. And this can only be true if $a{_1}=-a{_2}$, $b{_1}=-b{_2}$ or $c{_1}=-c{_2}$ and my question is where exactly do I go from here? Like this still doesn't look like a solid proof to me like what about this really proves what I'm trying to? Like what if $u$ and $v$'s only rows that hold the condition and make the vectors valid elements of $A$ are the exact solutions $a_1=-a_2$, $b_1=-b_2$ or $c_1=-c_2$. I don't even know if that made sense but basically can anyone help me understand how this is a $100\%$ proof that this set isn't closed under addition? Like just tell me what I'm missing here?
 A: If
$$
u=\begin{bmatrix}1\\0\\0\end{bmatrix}\ \ \ \ \mathrm{and}\ \ \ \ v=\begin{bmatrix}0\\1\\1\end{bmatrix},
$$
then $u$ and $v$ belong to $A$, but $u+v$ doesn't. So, $A$ is not a vector subspace of $\mathbb{R}^3$.
A: Since one of x,y,z has to be 0. just jouce  on vector with x,y≠0, z=0 and one x=0, y,z≠0 and add them
one single counterexample proofs that it is not linear.
A: Recall the following result.

Let $V$ be a vector space over a field $F$ and let $H\subseteq V$. We will say that $H$ is a vector subspace of $V$ if, and only if, the following conditions are satisfied:

*

*$\vec{0}_{V}\in H$,

*$\forall h_{1},h_{2}\in H:\quad h_{1}+h_{2}\in H$.

*$\forall \alpha \in F, \forall h\in H:\quad \alpha \cdot h\in H$.


Now, notice the importance of the universal quantifiers "$\forall$" what does it mean "for all". That is to say that if we prove that there are two vectors that do not satisfy this property, that destroys the fact that the quantifier "for all" holds and therefore the total proposition should not hold, from which it should follow that the subset is not a subspace vector of the vector space.
Now,

*

*The vector space $V=\mathbb{R}^{3}$.

*The set $A=\{(x,y,z)\in \mathbb{R}^{3}:\quad x\cdot y\cdot z=0\}$ is a subset of $V$.

Hence we can use the characterisation given in the first parte here. So we need to see if $1,2$ and $3$ are satisfied

*

*$(0,0,0)\in H$ because $0\cdot 0\cdot 0=0$, that is $\vec{0}_{V}\in H$.


*Let $h_{1}=(a_{1},b_{1},c_{1})\in H$ and $h_{2}=(a_{2},b_{2},c_{2})\in  H$ so by definition of $A$ we have
$$a_{1}\cdot b_{1}\cdot c_{1}=0,\quad a_{2}\cdot b_{2}\cdot c_{2}=0$$
Now, $h_{1}+h_{2}=(a_{1}+a_{2},b_{1}+b_{2},c_{1}+c_{2})\in H$ if $$(a_{1}+a_{2})(b_{1}+b_{2})(c_{1}+c_{2})=0$$
Expanding the RHS, we have
$$\underbrace{a_{1}b_{1}c_{1}}_{=0; \text{by hypothesis}}+\underbrace{a_{2}b_{2}c_{2}}_{=0;\text{by hypothesis}}+\color{red}{\text{other terms}}=0?$$
So we ask ourselves: is the RHS equal to $0$, always regardless of or at the choice of our vectors? I think now you can clearly see that the appearance of those $\color{red}{\text{other terms}}$ can certainly make the RHS not equal to $0$. So we provide a counterexample because it is the correct way to prove that $A$ is not a vector subspace.
Counterexample:
Let $h_{1}=(1,0,1)\in H$ and $h_{2}=(0,1,1)\in H$ but $$h_{1}+h_{2}=(1+0,0+1,1+1)$$ is not in $H$ because $$(1+0)(0+1)(1+1)=(1\cdot 0\cdot 1)+(0\cdot 1\cdot 1)+\color{red}{2}=2\not=0.$$
Therefore $A$ is not vector subspace of $\mathbb{R}^{3}$.
