Linear algebra with proofs/vector space

[relskid]

Homework Statement

Suppose S denotes the set of continuous real-valued functions of a single variable f(x) having the property that f(x)=0 if |x|≥ 1. Show that S is a vector space over R by explicitly checking all the axioms.

Homework Equations

(VS1) For all x, y in V, x+y=y+x
(VS2) For all x, y, z in V, (x+y)+z=x+(y+z)
(VS3) There exists an element in V denoted by 0 such that x+0=x for each x in V
(VS4) For each element x in V there exists an element y in V such that x+y=0
(VS5) For each element x in V, 1x=x
(VS6) For each pair of elements a, b in F and each element x in V, (ab)x=a(bx)
(VS7) For each element a in F and each pair of elements x, y in V, a(x+y)=ax+ay
(VS8) For each pair of elements a, b in F and each element x in V, (a+b)x=ax+bx

The Attempt at a Solution

no real attempt yet. i am very lost on what i am supposed to do. all i really know is that I'm supposed to test S for each of those VS rules, and if they satisfy all of them, then S is a vector space over R. I'm just confused on how i am supposed to go about it.

 

[gabbagabbahey]

Well, your vector space is S and your vectors are f(x),f(y),... so if (VS1) holds for S, then f(x)+f(y)=f(y)+f(x) which is clearly true for all real valued functions... do f(x),f(y)... satisfy VS2,VS3,...?