A
Reduction to absurdity:
Suppose there is no such finite harmonious set, and:
A, b∈S, where a+b ∈ s, a-b ∈ s.
Namely: a+b= either A or B.
A-b = a or b
Therefore, if an element of s is 0 and b=0, then the set S' = {a, 0} must not be a harmonic set.
But:
a+0 = a∈S’;
a-0=a∈S '
That is, s' is a harmonious set.
conflict
So, suppose there is an error and there is a finite set S.
B
Prove:
k 1,k2∈Z,k 1≠k2
Then: k 1a+k2a=(k 1+k2)a, where (k1+k2) ∈ z.
∴(k 1+k2)a∈S
Similarly: (k 1-k2)a∈S
That is, s is a harmonious set.
C
We can know that 0∈S from A's reduction to absurdity.
∴S 1∩S2 ≠ empty set is established.
D
By reducing to absurdity, suppose s1∪ S2 = R.
According to a, we can know that S 1 = {a, 0} and S2 = {b, 0} are harmonic sets, where a ≠ b.
Then there must be: s1∪ S2 = R.
This is obviously wrong and contradictory.
Therefore: S 1∪S2=R is not valid.
Option d