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Let $F$ be a subspace over $\mathbb{R}$ or $\mathbb{C}$ (real or complex numbers), then determine whether the following is a subspace.

$\{ (x_{1}, x_{2}, x_{3}) \in F^{3} : x_{1}x_{2}x_{3} = 0\}$.

How do I check for closure under addition in this case?

Thanks in advance

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What exactly do you mean by a "subspace"? Are you treating $\mathbb{R}$ and $\mathbb{C}$ as vector spaces? If so, then the Cartesian products $\mathbb{R} \times \mathbb{R} \times \mathbb{R}$ and $\mathbb{C} \times \mathbb{C} \times \mathbb{C}$ are well defined, but there is no product structure. What exactly does $x_1x_2x_3$ mean? Please be more specific: do you mean subspace, subset, submanifold, or what? – Fly by Night Sep 18 '12 at 20:16
@FlybyNight By subspace, I mean a subset of a vector space that has an identity in V, and is closed under addition and multiplication. – achacttn Sep 18 '12 at 20:44
So you mean a vector subspace. In that case, your set it not a subspace because it is not defined by a linear map. This definition of subspace was unclear because you didn't say that $\mathbb{R}$ and $\mathbb{C}$ we being thought of as vector spaces. These sets have more structures: they are manifolds, they are fields, they are topological spaces, etc. – Fly by Night Sep 18 '12 at 20:51
up vote 2 down vote accepted

If $(x_1, x_2, x_3)$ and $(y_1, y_2, y_3)$ are such that $x_1 x_2 x_3=0$ and $y_1 y_2 y_3=0$, you would want to verify whether $(x_1, x_2, x_3)+(y_1, y_2, y_3)=(x_1+y_1, x_2+y_2, x_3+y_3)$ satisfies the same property, that is, $(x_1+y_1)(x_2+y_2)(x_3+y_3)=0$. It is easy to see this doesn't hold in general, so your set is not closed under addition.

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You can't, as $(1,0,0)\in$ this subspace, also $(0,1,1)$ but not the sum.

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Why the downvote? – Dennis Gulko Sep 18 '12 at 20:21

Let $U=\{(x_1,x_2,x_3 \in F, x_1x_2x_3=0\}$ Suppose $(1,1,0)\in U$ and $(0,0,1)\in U$, then $(1,1,0) + (0,0,1) = (1,1,1)$ not in $U$. Hence, $U$ is not a subspace of $F^3$ since it is not closed under addition property.

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