Proof
Let M be the indicated subset of R. We need to show:
0 
M.
If x, y M, then
x + y M.
For each r R and
m M, we have
rm M.
Taking n = 1, r1 = 0, and v1 any element of
V, we find 0 = r1v1
M.
M.
Suppose that
r = r1v1 + r2v2 +
··· + rnvn and
s = s1w1 + s2 w2 +
··· + smwm
are elements of M, with all vi and wj
V.
Then
r + s =
r1v1 + r2v2 +
··· + rnvn +
s1w1 + s2w2 +
··· + sm vm
also belongs to M.
V.
Then
r + s =
r1v1 + r2v2 +
··· + rnvn +
s1w1 + s2w2 +
··· + sm vm
also belongs to M.
If m =
r1v1 + r2
v2 + ···
+ rn vn is an element
of M, with vi
V, then for r R, we have
rm =
rr1v1 + rr2v2 +
··· + rrnvn,
which obviously belongs to M.