I need help determining if the following subset is a subspace of $\mathbb{F}^{3}$:
$\{(x_{1},x_{2},x_{3}) \in \mathbb{F}^3:x_{1}x_{2}x_{3} = 0$}.
Based on what I've read in my textbook, we must test that the subset is closed under vector addition and scalar multiplication (with the zero vector included).
Let $v,w \in \mathbb{F}^3$.
If $v = (x_{1},x_{2},x_{3})$ and $w = (y_{1},y_{2},y_{3})$ where $x_{1}x_{2}x_{3} = 0$ and $y_{1}y_{2}y_{3} = 0$
then $v + w = (x_{1}+y_{1}, x_{2}+y_{2}, x_{3}+y_{3}$)
The product of all ordered triples of the sum.
$(x_{1}+y_{1})(x_{2}+y_{2})(x_{3}+y_{3}) = 0$
Multiplied out:
$x_{1}x_{2}x_{3}+x_{1}x_{3}y_{2}+x_{2}x_{3}y_{1}+y_{1}y_{2}x_{3}+x_{1}x_{2}y_{3}+x_{1}y_{2}y_{3}+x_{2}y_{1}y_{3}+y_{1}y_{2}y_{3} = 0$
simplifying,
$x_{1}x_{3}y_{2}+x_{2}x_{3}y_{1}+y_{1}y_{2}x_{3}+x_{1}x_{2}y_{3}+x_{1}y_{2}y_{3}+x_{2}y_{1}y_{3}=0$
This is about where I got stuck. My guess is that the subset is not closed under addition because the given equality is only $0$ if each individual term sums to $0$.
I already know how to check scalar multiplication for this particular problem.
Thanks for the help.