
vector space
how do you show that for every set S, S ⊂ span S?
my working:
let S be the set of real numbers in n dimension.
then S = { x1,x2,...,xn l x1,x2,...,xn are real numbers}
then span S = a1x1+ a2x2+...+anxn
for all xi ∈ S, where 1<i<n
xi = a1x1+ a2x2+.+aixi+..+anxn iff all the coefficients are 0 except ai=1
thus xi ⊂ a1x1+ a2x2+.+aixi+..+anxn
can i prove this question like this? how should i define S since the proof requires that the statement is true for all S?
thanks

So, $\displaystyle \text{span}(S)$ is the set of all linear combinations of vectors in $\displaystyle S.$ Correct? Let $\displaystyle s\in S$. Then $\displaystyle s=1\cdot s\in\text{span}(S),$ because it's a linear combination of vectors in $\displaystyle S.$
Doesn't this do it?

I don't quite understand your proof.
But to show the statement for a general set S, notice that span S consists of all finite linear combinations of elements from S. In particular, take some element x in S. Then x=1x is a finite linear combination involving elements of S, and hence x is in span S.

thanks!! i was thinking that i need to define the vectors in S and then define the span of S.

You're very welcome for my contribution, whatever that was. Have a good one!