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What part of the definition of a vector space (see here) requires it to be closed under addition and multiplication by a scalar in the field? I would understand if we defined a vector space as a group of vectors rather then a set but we don't, also non of the axioms require this to be a condition?

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  • $\begingroup$ It's part of the definition, not the axioms. However, I might argue that the distributive axiom argues this... Either way, you certainly can think of a vector space as a group under addition. In fact, any group is simply a set that satisfies part, but not all, of these conditions. $\endgroup$ – Eoin Nov 8 '14 at 20:20
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I think the Wikipedia definition is tacitly noting that your set is (a) a group under addition, since the operation is defined to give $v + w \in V$ if $v$ and $w$ are in $V$, and the associativity, identity and inverse conditions give you a group. Also (b), the action of the field $F$ on $V$ is closed again by definition (since if $\alpha \in F$ and $v \in V$, we have $\alpha v \in V$.)

An equivalent—and possibly clearer—definition of vector space would foreground that the space is a group under addition and is closed under the action of the field.

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Because, addition and multiplication (by a scalar) operations are functions from $V \times V$ to $V$ and $K \times V$ to $V$, respectively.

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    $\begingroup$ I expanded your answer slightly. I think it is completely to the point, but was not easy to understand. $\endgroup$ – quid Nov 8 '14 at 20:40

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