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For context, I want to make it clear that I am learning introductory linear algebra, so please strive to use mathematical language that is appropriate.

When proving that a set, $V$, over a field, $F$, is a vector space, two different methods are used:

  1. Prove that the following 8 conditions hold:

(VS 1) For all $x$, $y$ in $V$, $x + y = y + x$ (commutativity of addition).

(VS 2) For all $x, y, z$ in $V, (x + y) + z = x + (y + z)$ (associativity of addition).

(VS 3) There exists an element in $V$ denoted by $0$ such that $x + 0 = x$ for each $x$ in $V$.

(VS 4) For each element $x$ in $V$ there exists an clement $y$ in $V$ such that $x + y = 0$.

(VS 5) For each clement $x$ in $V$, $1x = x$.

(VS 6) For each pair of elements $a$, $b$ in $F$ and each element $x$ in $V$, $(ab)x-a(bx)$.

(VS 7) For each clement $a$ in $F$ and each pair of elements $x$, $y$ in $V$, $a(x + y)$ = $ax + ay$.

(VS 8) For each pair of elements $o$, $b$ in $F$ and each clement $x$ in $V$, $(a + b)x$ = $ax + bx$.

  1. Prove that the set contains the $0$ vector and that the set is closed under addition and scalar multiplication:

$f(0) = 0$

$f(t) + g(t) = (f + g)(t)$

$(c)[f(t)] = (cf)(t)$

When I see people proving that a set over a field is a vector space, I usually see them prove 2 rather than 1. Initially, it seems that this is not a valid proof, since 2 is less comprehensive than 1; however, I wonder if 2 actually is a comprehensive and valid way to prove a set is a vector space? In other words, is 2 equivalent to 1 but simply shorter? As such, would it be equivalent and more efficient to simply prove 2 rather than 1?

I would greatly appreciate it if someone could please take the time to clarify this concept.

EDIT:

I just noticed that, if $(c)[f(t)] = (cf)t$ and $0 \in F$, then it must also automatically be true that $f(0) = 0$ and, therefore, $0 \in$ the set.

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    The first one is explicit in defining "addition and scalar multiplication". Addition isn't addition unless it's associative and commutative. – 2017-01-27
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    @mathematician thanks for the response. So they're both equivalent, but 2 is implicit in proving some of the conditions? – 2017-01-27
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    You have some typos, having used - instead of = for example in VS 7 and VS 8., the reason why associativity and distributivity are not always proven is because the example is working with a subset of a well known space which has already been said to satisfy those properties. For example, they would be told about $(\Bbb R^3,\Bbb R, +,\cdot)$ and are asked to prove whether or not $\{(x,y,z)\in\Bbb R^3~:~x=y+2z\}$ is a subspace. Since it is a subset, it inherits all of the rules for addition, distributivity, etc... and it need not be proven again. – 2017-01-27
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    What you write after (2.)... seems akward as well. It should read $0\in V,~~u,v\in V\implies u+v\in V,~~u\in V~\alpha\in K\implies \alpha u\in V$. I don't know why you started using functional notation... perhaps you mixed up the definition for a *function* to be linear. Regardless, these three properties can be shortened even further to only one property: $u,v\in V,~\alpha,\beta\in K\implies \alpha u+\beta v\in V$ – 2017-01-27
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    @JMoravitz For 2, I think I confused the conditions for a vector space with the conditions for a function to be a linear transformation; although, I think they're the same conditions? – 2017-01-27
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    The second line of "2" isn't even the definition of a linear function. Ignoring the weird swap from "x" to "t", this is the definition of addition of functions. If the third line is replaced by (cf)(x)=c*f(x), you get definition of scalar multiplication on the space of functions, which is an example of a vector space. – 2017-01-27
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    @mathematician I fixed the notational errors. The second line of 2 is closure under addition: https://en.wikipedia.org/wiki/Linear_map#Definition_and_first_consequences – 2017-01-27
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    And now I noticed that f(0)=0 is another rule for linear functions. I don't know if you wanted to look at linear functions or the vector space of any functions. Either way JMoravitz's comment handles everything. – 2017-01-27
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    @mathematician Indeed. Thank you all for your assistance! :) – 2017-01-27

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