$ \large \let u = \setbuild{\vec x \in \R^3}{ x_{1}+x_{2}+x_{3}=0 } $ >[!warning] i double dog dare you to PROVE that this set is closed ship it and clip it folks Prove that this [[../02 Areas/Math/Set|Set]] is [[../02 Areas/Math/Closure|Closed]] under addition. $ \begin{align} \let \vec x, \vec y &\in U\\ x_{1}+x_{2}+x_{3} &= 0\\ y_{1}+y_{2}+y_{3} &= 0\\ (x_{1}+y_{1})+(x_{2}+y_{2})+(x_{3}+y_{3}) &= (x_{1}+x_{2}+x_{3}) + (y_{1}+y_{2}+y_{3}) \\ &= 0 + 0\\ &= 0 \end{align}$ Prove that the set is [[../02 Areas/Math/Closure|Closed]] under [[../02 Areas/Math/Scalar|Scalar]] multiplication. $ \begin{align} \let \vec x &\in U, k \in \R\\ x_{1}+x_{2}+x_{3}&=0 \\ k\vec x &= k(x_{1}+x_{2}+x_{3})\\ &= k(0)\\ &= 0 \end{align} $