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In a test I was asked to give a definition to a subspace to a vector space, I wrote:

A subset $V$ is a subspace of $X$ if $0 \in V$ and $\forall u,v \in V, > \exists \thinspace V$ s.t. $ u+v = 0$

I was told that it is not correct. The correct definition is that the subset is closed under addition and closed under scalar multiplication.

Why is the second set of definition "more" correct than the definition I gave?

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Take $X=\mathbb R^2$ and $V$ be the union of the two axes. Then $V$ satisfies your definition but is not a subspace because $(1,1)=(1,0)+(0,1)$ is not in $V$.

Your definition just captures the notion of a set that is symmetric with respect to the origin.

Subspaces are symmetric with respect to the origin, but, as shown in the example above, not every set that is symmetric with respect to the origin is a subspace.

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  • $\begingroup$ Okay so a good way to check whether if a set is a subspace is to see if it is symmetric. If it is, it doesn't mean it is a subspace, you will still have to check other properties $\endgroup$ – Shamisen Expert Aug 25 '15 at 2:29
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Your definition is entirely incorrect, it's not a question of "more" or "less". For example, according to your definition, the set $V=\{-17,0,17\}$ is a vector subspace of the real vector space $\mathbb{R}$, since $0\in V$ and every element of $V$ has an additive inverse.

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